3 Probability Topics 概率主题

本页译自 OpenStax《Introductory Statistics》第 3 章 Probability Topics。公式经本地 MathJax 渲染，自定义宏已注入。

Introduction 引言

It is often necessary to "guess" about the outcome of an event in order to make a decision. Politicians study polls to guess their likelihood of winning an election. Teachers choose a particular course of study based on what they think students can comprehend. Doctors choose the treatments needed for various diseases based on their assessment of likely results. You may have visited a casino where people play games chosen because of the belief that the likelihood of winning is good. You may have chosen your course of study based on the probable availability of jobs.

为了做出决策，常常需要"猜测"某事件的结果。政治人物研究民调以猜测自己胜选的可能性。教师根据他们认为学生能够理解的内容来选择特定的学习课程。医生根据对可能结果的评估来选择治疗各种疾病所需的疗法。你可能去过赌场，那里人们玩的赌博游戏是因其获胜可能性较大的信念而被挑选的。你可能也根据就业的可能情况来选择自己的学习方向。

You have, more than likely, used probability. In fact, you probably have an intuitive sense of probability. Probability deals with the chance of an event occurring. Whenever you weigh the odds of whether or not to do your homework or to study for an exam, you are using probability. In this chapter, you will learn how to solve probability problems using a systematic approach.

你很可能已经使用过概率。事实上，你可能已对概率有一种直觉。概率研究的是某事件发生的可能性。每当你权衡是否要做作业或复习备考的胜算时，你就在使用概率。在本章中，你将学习如何运用系统的方法来解决概率问题。

Your instructor will survey your class. Count the number of students in the class today.

你的授课教师将对你们班级进行调查。数一数今天班上有多少名学生。

Raise your hand if you have any change in your pocket or purse. Record the number of raised hands.
Raise your hand if you rode a bus within the past month. Record the number of raised hands.
Raise your hand if you answered "yes" to BOTH of the first two questions. Record the number of raised hands.

如果你口袋或钱包里有零钱，请举手。记录举手的人数。
如果你在过去一个月里乘坐过公交车，请举手。记录举手的人数。
如果你对前两个问题都回答"是"，请举手。记录举手的人数。

Use the class data as estimates of the following probabilities. *P*(change) means the probability that a randomly chosen person in your class has change in his/her pocket or purse. *P*(bus) means the probability that a randomly chosen person in your class rode a bus within the last month and so on. Discuss your answers.

用班级数据来估计以下概率。*P*(change) 表示在你的班级中随机抽取一人，其口袋或钱包里有零钱的概率。*P*(bus) 表示在你的班级中随机抽取一人，其在上个月乘坐过公交车的概率，依此类推。讨论你们的答案。

Find *P*(change).
Find *P*(bus).
Find *P*(change AND bus). Find the probability that a randomly chosen student in your class has change in his/her pocket or purse and rode a bus within the last month.
Find *P*(change\|bus). Find the probability that a randomly chosen student has change given that he or she rode a bus within the last month. Count all the students that rode a bus. From the group of students who rode a bus, count those who have change. The probability is equal to those who have change and rode a bus divided by those who rode a bus.

求 *P*(change)。
求 *P*(bus)。
求 *P*(change AND bus)。求在你的班级中随机抽取一名学生，其口袋或钱包里有零钱且在上个月乘坐过公交车的概率。
求 *P*(change\|bus)。求在已知某学生上个月乘坐过公交车的条件下，随机抽取的一名学生口袋或钱包里有零钱的概率。统计所有乘坐过公交车的学生。在乘坐过公交车的学生群体中，统计其中有零钱的人数。该概率等于"有零钱且乘坐过公交车的人数"除以"乘坐过公交车的人数"。

3.1 Terminology 3.1 术语

Probability is a measure that is associated with how certain we are of outcomes of a particular experiment or activity. An experiment is a planned operation carried out under controlled conditions. If the result is not predetermined, then the experiment is said to be a chance experiment. Flipping one fair coin twice is an example of an experiment.

概率是与我们对某一特定试验或活动结果的确信程度相关联的度量。试验是在受控条件下进行的、预先规划的操作。如果结果不是预先确定的，则称该试验为随机试验。抛一枚均匀的硬币两次就是一个试验的例子。

A result of an experiment is called an outcome. The sample space of an experiment is the set of all possible outcomes. Three ways to represent a sample space are: to list the possible outcomes, to create a tree diagram, or to create a Venn diagram. The uppercase letter *S* is used to denote the sample space. For example, if you flip one fair coin, *S* = {*H*, *T*} where *H* = heads and *T* = tails are the outcomes.

一次试验的结果称为一个结果。一次试验的样本空间是所有可能结果组成的集合。表示样本空间有三种方式：列出可能的结果、绘制树形图，或绘制韦恩图。用大写字母 *S* 表示样本空间。例如，若你抛一枚均匀的硬币，则 *S* = {*H*, *T*}，其中 *H* = 正面（heads），*T* = 反面（tails）为两个结果。

An event is any combination of outcomes. Upper case letters like *A* and *B* represent events. For example, if the experiment is to flip one fair coin, event *A* might be getting at most one head. The probability of an event *A* is written *P*(*A*).

事件是任意结果的组合。像 *A* 和 *B* 这样的大写字母表示事件。例如，若试验是抛一枚均匀的硬币，则事件 *A* 可以是"至多出现一次正面"。事件 *A* 的概率记作 *P*(*A*)。

The probability of any outcome is the long-term relative frequency of that outcome. Probabilities are between zero and one, inclusive (that is, zero and one and all numbers between these values). *P*(*A*) = 0 means the event *A* can never happen. *P*(*A*) = 1 means the event *A* always happens. *P*(*A*) = 0.5 means the event *A* is equally likely to occur or not to occur. For example, if you flip one fair coin repeatedly (from 20 to 2,000 to 20,000 times) the relative frequency of heads approaches 0.5 (the probability of heads).

任一结果的概率就是该结果的长期相对频数。概率介于 0 与 1 之间（含端点）（即 0、1 以及介于两者之间的所有数）。*P*(*A*) = 0 表示事件 *A* 不可能发生。*P*(*A*) = 1 表示事件 *A* 必然发生。*P*(*A*) = 0.5 表示事件 *A* 发生与不发生的可能性相等。例如，若你反复抛一枚均匀的硬币（从 20 次到 2,000 次再到 20,000 次），正面出现的相对频数会趋近于 0.5（即正面的概率）。

Equally likely means that each outcome of an experiment occurs with equal probability. For example, if you toss a fair, six-sided die, each face (1, 2, 3, 4, 5, or 6) is as likely to occur as any other face. If you toss a fair coin, a Head (*H*) and a Tail (*T*) are equally likely to occur. If you randomly guess the answer to a true/false question on an exam, you are equally likely to select a correct answer or an incorrect answer.

等可能是指一次试验中每个结果出现的概率相等。例如，若你投掷一枚均匀的六面骰子，则每一面（1、2、3、4、5 或 6）出现的可能性与其他面相同。若你抛一枚均匀的硬币，则正面（*H*）与反面（*T*）出现的可能性相等。若在考试中随机猜测一道判断题的答案，你选中正确答案与选中错误答案的可能性相等。

**To calculate the probability of an event *A* when all outcomes in the sample space are equally likely**, count the number of outcomes for event *A* and divide by the total number of outcomes in the sample space. For example, if you toss a fair dime and a fair nickel, the sample space is {*HH*, *TH*, *HT*, *TT*} where *T* = tails and *H* = heads. The sample space has four outcomes. *A* = getting one head. There are two outcomes that meet this condition {*HT*, *TH*}, so *P*(*A*) = $\frac{2}{4}$ = 0.5.

**当样本空间中所有结果都等可能时，计算事件 *A* 的概率**，只需数出事件 *A* 包含的结果个数，再除以样本空间中结果的总数。例如，若你同时抛出一枚均匀的 10 分硬币和一枚均匀的 5 分硬币，样本空间为 {*HH*, *TH*, *HT*, *TT*}，其中 *T* = 反面，*H* = 正面。样本空间共有 4 个结果。令 *A* = 恰好出现一枚正面。满足该条件的结果有两个 {*HT*, *TH*}，因此 *P*(*A*) = $\frac{2}{4}$ = 0.5。

Suppose you roll one fair six-sided die, with the numbers {1, 2, 3, 4, 5, 6} on its faces. Let event *E* = rolling a number that is at least five. There are two outcomes {5, 6}. *P*(*E*) = $\frac{2}{6}$. If you were to roll the die only a few times, you would not be surprised if your observed results did not match the probability. If you were to roll the die a very large number of times, you would expect that, overall, $\frac{2}{6}$ of the rolls would result in an outcome of "at least five". You would not expect exactly $\frac{2}{6}$. The long-term relative frequency of obtaining this result would approach the theoretical probability of $\frac{2}{6}$ as the number of repetitions grows larger and larger.

假设你掷一枚均匀的六面骰子，其各面上标有数字 {1, 2, 3, 4, 5, 6}。令事件 *E* = 掷出的点数至少为 5。共有两个结果 {5, 6}。*P*(*E*) = $\frac{2}{6}$。如果你只掷几次骰子，观察到的结果与概率不符也不足为奇。如果你掷很多很多次，则总体上，你可预期有 $\frac{2}{6}$ 的投掷会得到"至少为 5"的结果。但你不会恰好得到 $\frac{2}{6}$。随着重复次数越来越多，得到这一结果的长期相对频数会趋近于理论概率 $\frac{2}{6}$。

This important characteristic of probability experiments is known as the law of large numbers which states that as the number of repetitions of an experiment is increased, the relative frequency obtained in the experiment tends to become closer and closer to the theoretical probability. Even though the outcomes do not happen according to any set pattern or order, overall, the long-term observed relative frequency will approach the theoretical probability. (The word empirical is often used instead of the word observed.)

概率试验的这一重要特性被称为大数定律，它指出：随着试验重复次数的增加，试验中得到的相对频数会越来越接近于理论概率。尽管结果的出现并不遵循任何固定的模式或顺序，但总体上，长期观测到的相对频数会趋近于理论概率。（常用经验一词来代替"观测到"的说法。）

It is important to realize that in many situations, the outcomes are not equally likely. A coin or die may be unfair, or biased. Two math professors in Europe had their statistics students test the Belgian one Euro coin and discovered that in 250 trials, a head was obtained 56% of the time and a tail was obtained 44% of the time. The data seem to show that the coin is not a fair coin; more repetitions would be helpful to draw a more accurate conclusion about such bias. Some dice may be biased. Look at the dice in a game you have at home; the spots on each face are usually small holes carved out and then painted to make the spots visible. Your dice may or may not be biased; it is possible that the outcomes may be affected by the slight weight differences due to the different numbers of holes in the faces. Gambling casinos make a lot of money depending on outcomes from rolling dice, so casino dice are made differently to eliminate bias. Casino dice have flat faces; the holes are completely filled with paint having the same density as the material that the dice are made out of so that each face is equally likely to occur. Later we will learn techniques to use to work with probabilities for events that are not equally likely.

重要的是要认识到，在许多情形下结果并非等可能。硬币或骰子可能不均匀，或者说有偏倚。欧洲的两位数学教授让他们的统计学学生检验了一枚比利时的一欧元硬币，发现经过 250 次试验，正面出现了 56% 的次数，反面出现了 44% 的次数。这些数据似乎表明这枚硬币并非均匀硬币；进行更多重复试验有助于对这种偏倚得出更准确的结论。有些骰子也可能有偏倚。看看你家里游戏用的骰子；每个面上的点数通常是被挖出的小孔，再涂上颜料以使点数可见。你的骰子可能有偏倚，也可能没有；由于各面上的孔数不同，造成的轻微重量差异有可能影响结果。赌博赌场靠掷骰子的结果赚取大量金钱，因此赌场用的骰子制作方式不同，以消除偏倚。赌场骰子各面平整；孔被完全填满，所用颜料与骰子材料的密度相同，从而每个面出现的可能性相等。稍后我们将学习处理非等可能事件概率的方法。

"OR" Event:An outcome is in the event *A* OR *B* if the outcome is in *A* or is in *B* or is in both *A* and *B*. For example, let *A* = {1, 2, 3, 4, 5} and *B* = {4, 5, 6, 7, 8}. *A* OR *B* = {1, 2, 3, 4, 5, 6, 7, 8}. Notice that 4 and 5 are NOT listed twice.

"OR" 事件（并事件）：若一个结果属于 *A*，或属于 *B*，或同时属于 *A* 和 *B*，则该结果属于事件 *A* OR *B*。例如，令 *A* = {1, 2, 3, 4, 5}，*B* = {4, 5, 6, 7, 8}。则 *A* OR *B* = {1, 2, 3, 4, 5, 6, 7, 8}。注意 4 和 5 没有被列出两次。

"AND" Event:An outcome is in the event *A* AND *B* if the outcome is in both *A* and *B* at the same time. For example, let *A* and *B* be {1, 2, 3, 4, 5} and {4, 5, 6, 7, 8}, respectively. Then *A* AND *B* = {4, 5}.

"AND" 事件（交事件）：若一个结果同时既属于 *A* 又属于 *B*，则该结果属于事件 *A* AND *B*。例如，分别令 *A* = {1, 2, 3, 4, 5}，*B* = {4, 5, 6, 7, 8}。则 *A* AND *B* = {4, 5}。

The complement of event *A* is denoted *A′* (read "*A* prime"). *A′* consists of all outcomes that are NOT in *A*. Notice that *P*(*A*) + *P*(*A′*) = 1. For example, let *S* = {1, 2, 3, 4, 5, 6} and let *A* = {1, 2, 3, 4}. Then, *A′* = {5, 6}. *P*(*A*) = $\frac{4}{6}$, *P*(*A′*) = $\frac{2}{6}$, and *P*(*A*) + *P*(*A′*) = $\frac{4}{6} + \frac{2}{6}$ = 1

事件 *A* 的补事件记作 *A′*（读作"*A* prime"）。*A′* 由所有不属于 *A* 的结果组成。注意 *P*(*A*) + *P*(*A′*) = 1。例如，令 *S* = {1, 2, 3, 4, 5, 6}，令 *A* = {1, 2, 3, 4}。则 *A′* = {5, 6}。*P*(*A*) = $\frac{4}{6}$，*P*(*A′*) = $\frac{2}{6}$，且 *P*(*A*) + *P*(*A′*) = $\frac{4}{6} + \frac{2}{6}$ = 1

The conditional probability of *A* given *B* is written *P*(*A*\|*B*). *P*(*A*\|*B*) is the probability that event *A* will occur given that the event *B* has already occurred. A conditional reduces the sample space. We calculate the probability of *A* from the reduced sample space *B*. The formula to calculate *P*(*A*\|*B*) is *P*(*A*\|*B*) = $\frac{P(A\text{AND}B)}{P(B)}$ where *P*(*B*) is greater than zero.

在已知 *B* 的条件下 *A* 的条件概率记作 *P*(*A*\|*B*)。*P*(*A*\|*B*) 是在事件 *B* 已经发生的前提下，事件 *A* 发生的概率。条件概率缩小了样本空间。我们从缩小后的样本空间 *B* 出发来计算 *A* 的概率。计算 *P*(*A*\|*B*) 的公式为 *P*(*A*\|*B*) = $\frac{P(A\text{AND}B)}{P(B)}$，其中 *P*(*B*) 大于零。

For example, suppose we toss one fair, six-sided die. The sample space *S* = {1, 2, 3, 4, 5, 6}. Let *A* = face is 2 or 3 and *B* = face is even (2, 4, 6). To calculate *P*(*A*\|*B*), we count the number of outcomes 2 or 3 in the sample space *B* = {2, 4, 6}. Then we divide that by the number of outcomes *B* (rather than *S*).

例如，假设我们投掷一枚均匀的六面骰子。样本空间 *S* = {1, 2, 3, 4, 5, 6}。令 *A* = 点数为 2 或 3，*B* = 点数为偶数（2, 4, 6）。为计算 *P*(*A*\|*B*)，我们在样本空间 *B* = {2, 4, 6} 中数出结果为 2 或 3 的个数。然后用它除以 *B* 中的结果个数（而非 *S*）。

We get the same result by using the formula. Remember that *S* has six outcomes.

利用公式我们也能得到相同的结果。记住 *S* 共有六个结果。

*P*(*A*\|*B*) = $\frac{P(A\mspace{360mu}\text{AND}\mspace{360mu} B)}{P(B)} = \frac{\frac{(\text{the number of outcomes that are 2 or 3 and even in}\mspace{360mu} S)}{6}}{\frac{(\text{the number of outcomes that are even in}\mspace{360mu} S)}{6}} = \frac{\frac{1}{6}}{\frac{3}{6}} = \frac{1}{3}$

Understanding Terminology and SymbolsIt is important to read each problem carefully to think about and understand what the events are. Understanding the wording is the first very important step in solving probability problems. Reread the problem several times if necessary. Clearly identify the event of interest. Determine whether there is a condition stated in the wording that would indicate that the probability is conditional; carefully identify the condition, if any.

理解术语与符号：仔细阅读每一道题，思考并理解事件是什么，这一点很重要。理解题意是解决概率问题的首要关键步骤。如有必要，可将题目反复阅读几遍。明确识别所关注的事件。判断题意中是否陈述了某个表明该概率为条件概率的条件；若有，务必仔细识别该条件。

Problem 问题

The sample space *S* is the whole numbers starting at one and less than 20.

样本空间 *S* 是从 1 开始且小于 20 的整数。

1. *S* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Let event *A* = the even numbers and event *B* = numbers greater than 13.

令事件 *A* = 偶数，事件 *B* = 大于 13 的数。

2. *A* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_, *B* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

3. *P*(*A*) = \_\_\_\_\_\_\_\_\_\_\_\_\_, *P*(*B*) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

4. *A* AND *B* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_, *A* OR *B* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

5. *P*(*A* AND *B*) = \_\_\_\_\_\_\_\_\_, *P*(*A* OR *B*) = \_\_\_\_\_\_\_\_\_\_\_\_\_

6. *A′* = \_\_\_\_\_\_\_\_\_\_\_\_\_, *P*(*A′*) = \_\_\_\_\_\_\_\_\_\_\_\_\_

7. *P*(*A*) + *P*(*A′*) = \_\_\_\_\_\_\_\_\_\_\_\_

8. *P*(*A*\|*B*) = \_\_\_\_\_\_\_\_\_\_\_, *P*(*B*\|*A*) = \_\_\_\_\_\_\_\_\_\_\_\_\_; are the probabilities equal?

8. *P*(*A*\|*B*) = \_\_\_\_\_\_\_\_\_\_\_, *P*(*B*\|*A*) = \_\_\_\_\_\_\_\_\_\_\_\_\_; 这两个概率相等吗？

Solution 解答

1. *S* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

1. 样本空间 *S* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19}

2. *A* = {2, 4, 6, 8, 10, 12, 14, 16, 18}, *B* = {14, 15, 16, 17, 18, 19}

2. *A* = {2, 4, 6, 8, 10, 12, 14, 16, 18}，*B* = {14, 15, 16, 17, 18, 19}

3. *P*(*A*) = $\frac{9}{19}$, *P*(*B*) = $\frac{6}{19}$

3. *P*(*A*) = $\frac{9}{19}$，*P*(*B*) = $\frac{6}{19}$

4. *A* AND *B* = {14,16,18}, *A* OR *B* = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}

4. *A* AND *B* = {14,16,18}，*A* OR *B* = {2, 4, 6, 8, 10, 12, 14, 15, 16, 17, 18, 19}

5. *P*(*A* AND *B*) = $\frac{3}{19}$, *P*(*A* OR *B*) = $\frac{12}{19}$

5. *P*(*A* AND *B*) = $\frac{3}{19}$，*P*(*A* OR *B*) = $\frac{12}{19}$

6. *A′* = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19; *P*(*A′*) = $\frac{10}{19}$

6. *A′* = 1, 3, 5, 7, 9, 11, 13, 15, 17, 19；*P*(*A′*) = $\frac{10}{19}$

7. *P*(*A*) + *P*(*A′*) = 1 ($\frac{9}{19}$ + $\frac{10}{19}$ = 1)

7. *P*(*A*) + *P*(*A′*) = 1（$\frac{9}{19}$ + $\frac{10}{19}$ = 1）

8. *P*(*A*\|*B*) = $\frac{P(A\text{AND}B)}{P(B)}$ = $\frac{3}{6}$, *P*(*B*\|*A*) = $\frac{P(A\text{AND}B)}{P(A)}$ = $\frac{3}{9}$, No

8. *P*(*A*\|*B*) = $\frac{P(A\text{AND}B)}{P(B)}$ = $\frac{3}{6}$，*P*(*B*\|*A*) = $\frac{P(A\text{AND}B)}{P(A)}$ = $\frac{3}{9}$，否

The sample space *S* is all the ordered pairs of two whole numbers, the first from one to three and the second from one to four (Example: (1, 4)).

样本空间 *S* 是所有由两个整数组成的有序数对，第一个取自 1 到 3，第二个取自 1 到 4（例如：(1, 4)）。

1. *S* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

Let event *A* = the sum is even and event *B* = the first number is prime.

设事件 *A* = 两数之和为偶数，事件 *B* = 第一个数为素数。

2. *A* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_, *B* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

3. *P*(*A*) = \_\_\_\_\_\_\_\_\_\_\_\_\_, *P*(*B*) = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

4. *A* AND *B* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_, *A* OR *B* = \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_

5. *P*(*A* AND *B*) = \_\_\_\_\_\_\_\_\_, *P*(*A* OR *B*) = \_\_\_\_\_\_\_\_\_\_\_\_\_

6. *B′* = \_\_\_\_\_\_\_\_\_\_\_\_\_, *P*(*B′*) = \_\_\_\_\_\_\_\_\_\_\_\_\_

7. *P*(*A*) + *P*(*A′*) = \_\_\_\_\_\_\_\_\_\_\_\_

8. *P*(*A*\|*B*) = \_\_\_\_\_\_\_\_\_\_\_, *P*(*B*\|*A*) = \_\_\_\_\_\_\_\_\_\_\_\_\_; are the probabilities equal?

8. *P*(*A*\|*B*) = \_\_\_\_\_\_\_\_\_\_\_, *P*(*B*\|*A*) = \_\_\_\_\_\_\_\_\_\_\_\_\_；这两个概率相等吗？

Problem 问题

A fair, six-sided die is rolled. Describe the sample space *S*, identify each of the following events with a subset of *S* and compute its probability (an outcome is the number of dots that show up).

抛一枚均匀的六面骰子。描述样本空间 *S*，用 *S* 的子集表示下列各个事件，并计算其概率（结果即朝上的点数）。

1. Event *T* = the outcome is two.

1. 事件 *T* = 结果为二。

2. Event *A* = the outcome is an even number.

2. 事件 *A* = 结果为偶数。

3. Event *B* = the outcome is less than four.

3. 事件 *B* = 结果小于四。

4. The complement of *A*.

4. *A* 的补事件。

5. *A* GIVEN *B*

5. *A* GIVEN *B*（已知 *B* 发生时 *A*）

6. *B* GIVEN *A*

6. *B* GIVEN *A*（已知 *A* 发生时 *B*）

7. *A* AND *B*

7. *A* AND *B*（*A* 与 *B* 的交事件）

8. *A* OR *B*

8. *A* OR *B*（*A* 与 *B* 的并事件）

9. *A* OR *B′*

9. *A* OR *B′*（*A* 与 *B′* 的并事件）

10. Event *N* = the outcome is a prime number.

10. 事件 *N* = 结果为素数。

11. Event *I* = the outcome is seven.

11. 事件 *I* = 结果为七。

Solution 解答

1. *T* = {2}, *P*(*T*) = $\frac{1}{6}$

2. *A* = {2, 4, 6}, *P*(*A*) = $\frac{1}{2}$

3. *B* = {1, 2, 3}, *P*(*B*) = $\frac{1}{2}$

4. *A′* = {1, 3, 5}, *P*(*A′*) = $\frac{1}{2}$

5. *A*\|*B* = {2}, *P*(*A*\|*B*) = $\frac{1}{3}$

6. *B*\|*A* = {2}, *P*(*B*\|*A*) = $\frac{1}{3}$

7. *A* AND *B* = {2}, *P*(*A* AND *B*) = $\frac{1}{6}$

8. *A* OR *B* = {1, 2, 3, 4, 6}, *P*(*A* OR *B*) = $\frac{5}{6}$

9. *A* OR *B′* = {2, 4, 5, 6}, *P*(*A* OR *B′*) = $\frac{2}{3}$

10. *N* = {2, 3, 5}, *P*(*N*) = $\frac{1}{2}$

11. A six-sided die does not have seven dots. *P*(7) = 0.

11. 六面骰子没有七点。*P*(7) = 0。

Table 3.1 describes the distribution of a random sample *S* of 100 individuals, organized by gender and whether they are right- or left-handed.

表 3.1 描述了一个由 100 人组成的随机样本 *S* 的分布，按性别以及惯用右手还是左手分类。

| | Right-handed | Left-handed |

| | 惯用右手 | 惯用左手 |

|---------|--------------|-------------|

| Males | 43 | 9 |

| 男性 | 43 | 9 |

| Females | 44 | 4 |

| 女性 | 44 | 4 |

Table 3.1

表 3.1

Problem 问题

Let’s denote the events *M* = the subject is male, *F* = the subject is female, *R* = the subject is right-handed, *L* = the subject is left-handed. Compute the following probabilities:

记事件 *M* = 受试者为男性，*F* = 受试者为女性，*R* = 受试者惯用右手，*L* = 受试者惯用左手。计算下列概率：

1. *P*(*M*)

2. *P*(*F*)

3. *P*(*R*)

4. *P*(*L*)

5. *P*(*M* AND *R*)

6. *P*(*F* AND *L*)

7. *P*(*M* OR *F*)

8. *P*(*M* OR *R*)

9. *P*(*F* OR *L*)

10. *P*(*M'*)

11. *P*(*R*\|*M*)

12. *P*(*F*\|*L*)

13. *P*(*L*\|*F*)

Solution 解答

1. *P*(*M*) = 0.52

2. *P*(*F*) = 0.48

3. *P*(*R*) = 0.87

4. *P*(*L*) = 0.13

5. *P*(*M* AND *R*) = 0.43

6. *P*(*F* AND *L*) = 0.04

7. *P*(*M* OR *F*) = 1

8. *P*(*M* OR *R*) = 0.96

9. *P*(*F* OR *L*) = 0.57

10. *P*(*M'*) = 0.48

11. *P*(*R*\|*M*) = 0.8269 (rounded to four decimal places)

11. *P*(*R*\|*M*) = 0.8269（四舍五入到四位小数）

12. *P*(*F*\|*L*) = 0.3077 (rounded to four decimal places)

12. *P*(*F*\|*L*) = 0.3077（四舍五入到四位小数）

13. *P*(*L*\|*F*) = 0.0833

---

3.2 Independent and Mutually Exclusive Events 3.2 独立事件与互斥事件

Independent and mutually exclusive do not mean the same thing.

独立事件与互斥事件并非同义词。

Independent Events 独立事件

Two events are independent if the following are true:

若满足下列条件，则两个事件相互独立：

*P*(*A*\|*B*) = *P*(*A*)
*P*(*B*\|*A*) = *P*(*B*)
*P*(*A* AND *B*) = *P*(*A*)*P*(*B*)

*P*(*A*\|*B*) = *P*(*A*)
*P*(*B*\|*A*) = *P*(*B*)
*P*(*A* AND *B*) = *P*(*A*)*P*(*B*)

Two events *A* and *B* are independent if the knowledge that one occurred does not affect the chance the other occurs. For example, the outcomes of two roles of a fair die are independent events. The outcome of the first roll does not change the probability for the outcome of the second roll. To show two events are independent, you must show only one of the above conditions. If two events are NOT independent, then we say that they are dependent.

若已知其中一个事件发生并不影响另一个事件发生的可能性，则事件 *A* 与 *B* 相互独立。例如，抛掷一枚均匀骰子两次的结果就是独立事件。第一次抛掷的结果不会改变第二次抛掷结果的概率。要证明两个事件独立，只需证明上述条件中的任意一个成立即可。若两个事件不独立，则称它们相依。

Sampling may be done with replacement or without replacement.

抽样可以是有放回抽样，也可以是无放回抽样。

With replacement: If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once. When sampling is done with replacement, then events are considered to be independent, meaning the result of the first pick will not change the probabilities for the second pick.
Without replacement: When sampling is done without replacement, each member of a population may be chosen only once. In this case, the probabilities for the second pick are affected by the result of the first pick. The events are considered to be dependent or not independent.

有放回：若总体中每个个体被抽中后都放回，则该个体有可能被多次抽中。有放回抽样时，各次抽取被视为独立事件，即第一次抽取的结果不会改变第二次抽取的概率。
无放回：无放回抽样时，总体中每个个体只能被抽中一次。此时第二次抽取的概率受第一次抽取结果的影响，各次抽取被视为相依（非独立）事件。

If it is not known whether *A* and *B* are independent or dependent, assume they are dependent until you can show otherwise.

若无法判断 *A* 与 *B* 是独立还是相依，则在能证明它们独立之前，先假定它们相依。

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, *J* (jack), *Q* (queen), *K* (king) of that suit.

你有一副洗匀的 52 张均匀扑克牌。它由四种花色组成：梅花、方块、红心、黑桃。每种花色有 13 张牌，分别为该花色的 1, 2, 3, 4, 5, 6, 7, 8, 9, 10、*J*（jack，侍从）、*Q*（queen，王后）、*K*（king，国王）。

a\. Sampling with replacement:

a\. 有放回抽样：

Suppose you pick three cards with replacement. The first card you pick out of the 52 cards is the *Q* of spades. You put this card back, reshuffle the cards and pick a second card from the 52-card deck. It is the ten of clubs. You put this card back, reshuffle the cards and pick a third card from the 52-card deck. This time, the card is the *Q* of spades again. Your picks are {*Q* of spades, ten of clubs, *Q* of spades}. You have picked the *Q* of spades twice. You pick each card from the 52-card deck.

假设有放回地抽三张牌。你从 52 张牌中抽出的第一张是黑桃 *Q*。你把这张牌放回，重新洗牌，再从 52 张牌中抽第二张，是梅花十点。你把这张牌放回，重新洗牌，再从 52 张牌中抽第三张，这一次又是黑桃 *Q*。你的抽取结果是 {黑桃 *Q*，梅花十点，黑桃 *Q*}。你抽到了两次黑桃 *Q*。每张牌都从 52 张牌中抽取。

b\. Sampling without replacement:

b\. 无放回抽样：

Suppose you pick three cards without replacement. The first card you pick out of the 52 cards is the *K* of hearts. You put this card aside and pick the second card from the 51 cards remaining in the deck. It is the three of diamonds. You put this card aside and pick the third card from the remaining 50 cards in the deck. The third card is the *J* of spades. Your picks are {*K* of hearts, three of diamonds, *J* of spades}. Because you have picked the cards without replacement, you cannot pick the same card twice.

假设无放回地抽三张牌。你从 52 张牌中抽出的第一张是红心 *K*。你把这张牌放到一旁，从牌堆剩余的 51 张牌中抽第二张，是方块三点。你把这张牌放到一旁，从剩余的 50 张牌中抽第三张，是黑桃 *J*。你的抽取结果是 {红心 *K*，方块三点，黑桃 *J*}。因为是无放回抽取，所以同一张牌不可能被抽中两次。

你有一副洗匀的 52 张均匀扑克牌。它由四种花色组成：梅花、方块、红心、黑桃。每种花色有 13 张牌，分别为该花色的 1, 2, 3, 4, 5, 6, 7, 8, 9, 10、*J*（jack）、*Q*（queen）、*K*（king）。随机抽取三张牌。

1. Suppose you know that the picked cards are *Q* of spades, *K* of hearts and *Q* of spades. Can you decide if the sampling was with or without replacement?

1. 假设已知抽到的牌是黑桃 *Q*、红心 *K* 和黑桃 *Q*。能否判断这次抽样是有放回还是无放回？

2. Suppose you know that the picked cards are *Q* of spades, *K* of hearts, and *J* of spades. Can you decide if the sampling was with or without replacement?

2. 假设已知抽到的牌是黑桃 *Q*、红心 *K* 和黑桃 *J*。能否判断这次抽样是有放回还是无放回？

Problem 问题

你有一副洗匀的 52 张均匀扑克牌。它由四种花色组成：梅花、方块、红心、黑桃。每种花色有 13 张牌，分别为该花色的 1, 2, 3, 4, 5, 6, 7, 8, 9, 10、*J*（jack）、*Q*（queen）、*K*（king）。*S* = 黑桃，*H* = 红心，*D* = 方块，*C* = 梅花。

1. Suppose you pick four cards, but do not put any cards back into the deck. Your cards are *QS*, 1*D*, 1*C*, *QD*.

1. 假设你抽四张牌，且不放回任何牌。你抽到的牌是 *QS*、1*D*、1*C*、*QD*。

2. Suppose you pick four cards and put each card back before you pick the next card. Your cards are *KH*, 7*D*, 6*D*, *KH*.

2. 假设你抽四张牌，且在抽下一张之前把每张牌都放回。你抽到的牌是 *KH*、7*D*、6*D*、*KH*。

Which of a. or b. did you sample with replacement and which did you sample without replacement?

上述 a. 与 b. 中，哪一个是放回抽样，哪一个是无放回抽样？

Solution 解答

a\. Without replacement; b. With replacement

a\. 无放回；b. 有放回

You have a fair, well-shuffled deck of 52 cards. It consists of four suits. The suits are clubs, diamonds, hearts, and spades. There are 13 cards in each suit consisting of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, *J* (jack), *Q* (queen), and *K* (king) of that suit. *S* = spades, *H* = Hearts, *D* = Diamonds, *C* = Clubs. Suppose that you sample four cards without replacement. Which of the following outcomes are possible? Answer the same question for sampling with replacement.

你有一副洗匀的 52 张均匀扑克牌。它由四种花色组成：梅花、方块、红心、黑桃。每种花色有 13 张牌，分别为该花色的 1, 2, 3, 4, 5, 6, 7, 8, 9, 10、*J*（jack）、*Q*（queen）、*K*（king）。*S* = 黑桃，*H* = 红心，*D* = 方块，*C* = 梅花。假设无放回地抽四张牌。下列哪些结果可能出现？对放回抽样回答同样的问题。

1. *QS*, 1*D*, 1*C*, *QD*

2. *KH*, 7*D*, 6*D*, *KH*

3. *QS*, 7*D*, 6*D*, *KS*

Mutually Exclusive Events 互斥事件

*A* and *B* are mutually exclusive events if they cannot occur at the same time. This means that *A* and *B* do not share any outcomes and *P*(*A* AND *B*) = 0.

若 *A* 与 *B* 不能同时发生，则它们是互斥事件。这意味着 *A* 与 *B* 没有共同结果，且 *P*(*A* AND *B*) = 0。

For example, suppose the sample space *S* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Let *A* = {1, 2, 3, 4, 5}, *B* = {4, 5, 6, 7, 8}, and *C* = {7, 9}. *A* AND *B* = {4, 5}. *P*(*A* AND *B*) = $\frac{2}{10}$ and is not equal to zero. Therefore, *A* and *B* are not mutually exclusive. *A* and *C* do not have any numbers in common so *P*(*A* AND *C*) = 0. Therefore, *A* and *C* are mutually exclusive.

例如，假设样本空间 *S* = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}。令 *A* = {1, 2, 3, 4, 5}，*B* = {4, 5, 6, 7, 8}，*C* = {7, 9}。*A* AND *B* = {4, 5}。*P*(*A* AND *B*) = $\frac{2}{10}$，且不等于零，因此 *A* 与 *B* 不互斥。*A* 与 *C* 没有共同的数字，故 *P*(*A* AND *C*) = 0，因此 *A* 与 *C* 互斥。

If it is not known whether *A* and *B* are mutually exclusive, assume they are not until you can show otherwise. The following examples illustrate these definitions and terms.

若无法判断 *A* 与 *B* 是否互斥，则在能证明它们互斥之前，先假定它们不互斥。下面的例子说明这些定义与术语。

Flip two fair coins. (This is an experiment.)

抛两枚均匀硬币。（这是一次试验。）

The sample space is {*HH*, *HT*, *TH*, *TT*} where *T* = tails and *H* = heads. The outcomes are *HH*, *HT*, *TH*, and *TT*. The outcomes HT and TH are different. The *HT* means that the first coin showed heads and the second coin showed tails. The *TH* means that the first coin showed tails and the second coin showed heads.

样本空间为 {*HH*, *HT*, *TH*, *TT*}，其中 *T* = 反面，*H* = 正面。结果为 *HH*、*HT*、*TH*、*TT*。结果 HT 与 TH 不同。*HT* 表示第一枚硬币正面朝上、第二枚反面朝上；*TH* 表示第一枚反面朝上、第二枚正面朝上。

Let *A* = the event of getting at most one tail. (At most one tail means zero or one tail.) Then *A* can be written as {*HH*, *HT*, *TH*}. The outcome *HH* shows zero tails. *HT* and *TH* each show one tail.
Let *B* = the event of getting all tails. *B* can be written as {*TT*}. *B* is the complement of *A*, so *B* = *A′*. Also, *P*(*A*) + *P*(*B*) = *P*(*A*) + *P*(*A′*) = 1.
The probabilities for *A* and for *B* are *P*(*A*) = $\frac{3}{4}$ and *P*(*B*) = $\frac{1}{4}$.
Let *C* = the event of getting all heads. *C* = {*HH*}. Since *B* = {*TT*}, *P*(*B* AND *C*) = 0. *B* and *C* are mutually exclusive. (*B* and *C* have no members in common because you cannot have all tails and all heads at the same time.)
Let *D* = event of getting more than one tail. *D* = {*TT*}. *P*(*D*) = $\frac{1}{4}$
Let *E* = event of getting a head on the first roll. (This implies you can get either a head or tail on the second roll.) *E* = {*HT*, *HH*}. *P*(*E*) = $\frac{2}{4}$
Find the probability of getting at least one (one or two) tail in two flips. Let *F* = event of getting at least one tail in two flips. *F* = {*HT*, *TH*, *TT*}. *P*(*F*) = $\frac{3}{4}$

设 *A* = 至多出现一条反面的事件（至多一条反面即零条或一条反面）。则 *A* 可写为 {*HH*, *HT*, *TH*}。结果 *HH* 显示零条反面，*HT* 与 *TH* 各显示一条反面。
设 *B* = 出现全为反面的事件。*B* 可写为 {*TT*}。*B* 是 *A* 的补事件，故 *B* = *A′*。且 *P*(*A*) + *P*(*B*) = *P*(*A*) + *P*(*A′*) = 1。
*A* 与 *B* 的概率分别为 *P*(*A*) = $\frac{3}{4}$，*P*(*B*) = $\frac{1}{4}$。
设 *C* = 出现全为正面的事件。*C* = {*HH*}。由于 *B* = {*TT*}，*P*(*B* AND *C*) = 0。*B* 与 *C* 互斥（*B* 与 *C* 没有公共成员，因为不可能同时全为反面又全为正面）。
设 *D* = 出现多于一条反面的事件。*D* = {*TT*}。*P*(*D*) = $\frac{1}{4}$
设 *E* = 第一次抛掷为正面的事件（这意味着第二次可以是正面或反面）。*E* = {*HT*, *HH*}。*P*(*E*) = $\frac{2}{4}$
求两次抛掷中至少出现一条（一条或两条）反面的概率。设 *F* = 两次抛掷中至少出现一条反面的事件。*F* = {*HT*, *TH*, *TT*}。*P*(*F*) = $\frac{3}{4}$

Draw two cards from a standard 52-card deck with replacement. Find the probability of getting at least one black card.

从一副标准 52 张扑克牌中有放回地抽两张。求至少抽到一张黑色牌的概率。

Problem 问题

Flip two fair coins. Find the probabilities of the events.

抛两枚均匀硬币。求下列事件的概率。

1. Let *F* = the event of getting at most one tail (zero or one tail).

1. 设 *F* = 至多出现一条反面（零条或一条反面）的事件。

2. Let *G* = the event of getting two faces that are the same.

2. 设 *G* = 出现两个相同面的事件。

3. Let *H* = the event of getting a head on the first flip followed by a head or tail on the second flip.

3. 设 *H* = 第一次正面朝上、第二次正面或反面的事件。

4. Are *F* and *G* mutually exclusive?

4. *F* 与 *G* 是否互斥？

5. Let *J* = the event of getting all tails. Are *J* and *H* mutually exclusive?

5. 设 *J* = 出现全为反面的事件。*J* 与 *H* 是否互斥？

Solution 解答

Look at the sample space in Example 3.6.

查看示例 3.6 中的样本空间。

1. Zero (0) or one (1) tails occur when the outcomes *HH*, *TH*, *HT* show up. *P*(*F*) = $\frac{3}{4}$

1. 当出现结果 *HH*、*TH*、*HT* 时，出现零(0)枚或一枚(1)反面。*P*(*F*) = $\frac{3}{4}$

2. Two faces are the same if *HH* or *TT* show up. *P*(*G*) = $\frac{2}{4}$

2. 若 *HH* 或 *TT* 出现，则两面相同。*P*(*G*) = $\frac{2}{4}$

3. A head on the first flip followed by a head or tail on the second flip occurs when *HH* or *HT* show up. *P*(*H*) = $\frac{2}{4}$

3. 第一次抛得正面、第二次抛得正面或反面，在 *HH* 或 *HT* 出现时发生。*P*(*H*) = $\frac{2}{4}$

4. *F* and *G* share *HH* so *P*(*F* AND *G*) is not equal to zero (0). *F* and *G* are not mutually exclusive.

4. *F* 和 *G* 共有 *HH*，所以 *P*(*F* AND *G*) 不等于零(0)。*F* 和 *G* 不是互斥事件。

5. Getting all tails occurs when tails shows up on both coins (*TT*). *H*’s outcomes are *HH* and *HT*.

5. 两枚硬币都为反面（*TT*）时出现"全反面"。*H* 的结果为 *HH* 和 *HT*。

*J* and *H* have nothing in common so *P*(*J* AND *H*) = 0. *J* and *H* are mutually exclusive.

*J* 和 *H* 没有共同结果，因此 *P*(*J* AND *H*) = 0。*J* 和 *H* 是互斥事件。

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Find the probability of the following events:

一个盒子里有两个球，一个白球、一个红球。我们抽取一个球，放回盒中，再抽取第二个球（有放回抽样）。求下列事件的概率：

1. Let *F* = the event of getting the white ball twice.

1. 设 *F* = 两次都抽到白球这一事件。

2. Let *G* = the event of getting two balls of different colors.

2. 设 *G* = 抽到两个不同颜色球这一事件。

3. Let *H* = the event of getting white on the first pick.

3. 设 *H* = 第一次抽到白球这一事件。

4. Are *F* and *G* mutually exclusive?

4. *F* 和 *G* 互斥吗？

5. Are *G* and *H* mutually exclusive?

5. *G* 和 *H* 互斥吗？

Roll one fair, six-sided die. The sample space is {1, 2, 3, 4, 5, 6}. Let event *A* = a face is odd. Then *A* = {1, 3, 5}. Let event *B* = a face is even. Then *B* = {2, 4, 6}.

掷一枚均匀的六面骰子。样本空间为 {1, 2, 3, 4, 5, 6}。设事件 *A* = 点数为奇数。则 *A* = {1, 3, 5}。设事件 *B* = 点数为偶数。则 *B* = {2, 4, 6}。

Find the complement of *A*, *A′*. The complement of *A*, *A′*, is *B* because *A* and *B* together make up the sample space. *P*(*A*) + *P*(*B*) = *P*(*A*) + *P*(*A′*) = 1. Also, *P*(*A*) = $\frac{3}{6}$ and *P*(*B*) = $\frac{3}{6}$.
Let event *C* = odd faces larger than two. Then *C* = {3, 5}. Let event *D* = all even faces smaller than five. Then *D* = {2, 4}. *P*(*C* AND *D*) = 0 because you cannot have an odd and even face at the same time. Therefore, *C* and *D* are mutually exclusive events.
Let event *E* = all faces less than five. *E* = {1, 2, 3, 4}.

求 *A* 的补事件 *A′*。*A* 的补事件 *A′* 是 *B*，因为 *A* 和 *B* 合起来构成样本空间。*P*(*A*) + *P*(*B*) = *P*(*A*) + *P*(*A′*) = 1。此外，*P*(*A*) = $\frac{3}{6}$，*P*(*B*) = $\frac{3}{6}$。
设事件 *C* = 大于 2 的奇数点。则 *C* = {3, 5}。设事件 *D* = 小于 5 的全部偶数点。则 *D* = {2, 4}。*P*(*C* AND *D*) = 0，因为你不可能同时出现奇数点和偶数点。因此 *C* 和 *D* 是互斥事件。
设事件 *E* = 小于 5 的全部点数。*E* = {1, 2, 3, 4}。

Problem 问题

Are *C* and *E* mutually exclusive events? (Answer yes or no.) Why or why not?

*C* 和 *E* 是互斥事件吗？（回答"是"或"否"。）为什么是或为什么不是？

Solution 解答

No. *C* = {3, 5} and *E* = {1, 2, 3, 4}. *P*(*C* AND *E*) = $\frac{1}{6}$. To be mutually exclusive, *P*(*C* AND *E*) must be zero.

否。*C* = {3, 5}，*E* = {1, 2, 3, 4}。*P*(*C* AND *E*) = $\frac{1}{6}$。要互斥，*P*(*C* AND *E*) 必须为零。

Find *P*(*C*\|*A*). This is a conditional probability. Recall that the event *C* is {3, 5} and event *A* is {1, 3, 5}. To find *P*(*C*\|*A*), find the probability of *C* using the sample space *A*. You have reduced the sample space from the original sample space {1, 2, 3, 4, 5, 6} to {1, 3, 5}. So, *P*(*C*\|*A*) = $\frac{2}{3}$.

求 *P*(*C*\|*A*)。这是一个条件概率。注意事件 *C* 为 {3, 5}，事件 *A* 为 {1, 3, 5}。要求 *P*(*C*\|*A*)，就用样本空间 *A* 来求 *C* 的概率。你已把样本空间从原来的 {1, 2, 3, 4, 5, 6} 缩小到 {1, 3, 5}。于是 *P*(*C*\|*A*) = $\frac{2}{3}$。

Let event *A* = learning Spanish. Let event *B* = learning German. Then *A* AND *B* = learning Spanish and German. Suppose *P*(*A*) = 0.4 and *P*(*B*) = 0.2. *P*(*A* AND *B*) = 0.08. Are events *A* and *B* independent? Hint: You must show ONE of the following:

设事件 *A* = 学西班牙语。设事件 *B* = 学德语。则 *A* AND *B* = 同时学西班牙语和德语。假定 *P*(*A*) = 0.4，*P*(*B*) = 0.2，*P*(*A* AND *B*) = 0.08。事件 *A* 和 *B* 独立吗？提示：你必须证明下列之一成立：

*P*(*A*\|*B*) = *P*(*A*)
*P*(*B*\|*A*) = *P*(*B*)
*P*(*A* AND *B*) = *P*(*A*)*P*(*B*)

*P*(*A*\|*B*) = *P*(*A*)
*P*(*B*\|*A*) = *P*(*B*)
*P*(*A* AND *B*) = *P*(*A*)*P*(*B*)

Let event *G* = taking a math class. Let event *H* = taking a science class. Then, *G* AND *H* = taking a math class and a science class. Suppose *P*(*G*) = 0.6, *P*(*H*) = 0.5, and *P*(*G* AND *H*) = 0.3. Are *G* and *H* independent?

设事件 *G* = 上数学课。设事件 *H* = 上科学课。则 *G* AND *H* = 同时上数学课和科学课。假定 *P*(*G*) = 0.6，*P*(*H*) = 0.5，*P*(*G* AND *H*) = 0.3。*G* 和 *H* 独立吗？

If *G* and *H* are independent, then you must show ONE of the following:

若 *G* 和 *H* 独立，则你必须证明下列之一成立：

*P*(*G*\|*H*) = *P*(*G*)
*P*(*H*\|*G*) = *P*(*H*)
*P*(*G* AND *H*) = *P*(*G*)*P*(*H*)

*P*(*G*\|*H*) = *P*(*G*)
*P*(*H*\|*G*) = *P*(*H*)
*P*(*G* AND *H*) = *P*(*G*)*P*(*H*)

The choice you make depends on the information you have. You could choose any of the methods here because you have the necessary information.

你选择哪种方法取决于你所掌握的信息。你可以选择这里的任意一种方法，因为你已掌握必要的信息。

Problem 问题

a\. Show that *P*(*G*\|*H*) = *P*(*G*).

a. 证明 *P*(*G*\|*H*) = *P*(*G*)。

Solution 解答

*P*(*G*\|*H*) = $\frac{P\text{(}G\text{~AND~}H\text{)}}{P\text{(}H\text{)}}$ = $\frac{\text{0}\text{.3}}{\text{0}\text{.5}}$ = 0.6 = *P*(*G*)

Problem 问题

b\. Show *P*(*G* AND *H*) = *P*(*G*)*P*(*H*).

b. 证明 *P*(*G* AND *H*) = *P*(*G*)*P*(*H*)。

Solution 解答

*P*(*G*)*P*(*H*) = (0.6)(0.5) = 0.3 = *P*(*G* AND *H*)

Since *G* and *H* are independent, knowing that a person is taking a science class does not change the chance that he or she is taking a math class. If the two events had not been independent (that is, they are dependent) then knowing that a person is taking a science class would change the chance he or she is taking math. For practice, show that *P*(*H*\|*G*) = *P*(*H*) to show that *G* and *H* are independent events.

因为 *G* 和 *H* 独立，所以知道某人上科学课并不会改变他或她上数学课的概率。如果这两个事件不独立（即它们相依），那么知道某人上科学课就会改变他或她上数学课的概率。作为练习，证明 *P*(*H*\|*G*) = *P*(*H*)，以表明 *G* 和 *H* 是独立事件。

In a bag, there are six red marbles and four green marbles. The red marbles are marked with the numbers 1, 2, 3, 4, 5, and 6. The green marbles are marked with the numbers 1, 2, 3, and 4.

一个袋子里有 6 颗红弹珠和 4 颗绿弹珠。红弹珠标有数字 1、2、3、4、5、6。绿弹珠标有数字 1、2、3、4。

*R* = a red marble
*G* = a green marble
*O* = an odd-numbered marble
The sample space is *S* = {*R*1, *R*2, *R*3, *R*4, *R*5, *R*6, *G*1, *G*2, *G*3, *G*4}.

*R* = 红弹珠
*G* = 绿弹珠
*O* = 奇数编号的弹珠
样本空间为 *S* = {*R*1, *R*2, *R*3, *R*4, *R*5, *R*6, *G*1, *G*2, *G*3, *G*4}。

*S* has ten outcomes. What is *P*(*G* AND *O*)?

*S* 有 10 个结果。*P*(*G* AND *O*) 是多少？

Problem 问题

Let event *C* = taking an English class. Let event *D* = taking a speech class.

设事件 *C* = 上英语课。设事件 *D* = 上演讲课。

Suppose *P*(*C*) = 0.75, *P*(*D*) = 0.3, *P*(*C*\|*D*) = 0.75 and *P*(*C* AND *D*) = 0.225.

假定 *P*(*C*) = 0.75，*P*(*D*) = 0.3，*P*(*C*\|*D*) = 0.75，*P*(*C* AND *D*) = 0.225。

Justify your answers to the following questions numerically.

用数值对你的下列回答作出论证。

1. Are *C* and *D* independent?

1. *C* 和 *D* 独立吗？

2. Are *C* and *D* mutually exclusive?

2. *C* 和 *D* 互斥吗？

3. What is *P*(*D*\|*C*)?

3. *P*(*D*\|*C*) 是多少？

Solution 解答

1. Yes, because *P*(*C*\|*D*) = *P*(*C*).

1. 是，因为 *P*(*C*\|*D*) = *P*(*C*)。

2. No, because *P*(*C* AND *D*) is not equal to zero.

2. 否，因为 *P*(*C* AND *D*) 不等于零。

3. *P*(*D*\|*C*) = $\frac{P\text{(}C\text{~AND~}D\text{)}}{P\text{(}C\text{)}}$ = $\frac{\text{0}\text{.225}}{0.75}$ = 0.3

A student goes to the library. Let events *B* = the student checks out a book and *D* = the student checks out a DVD. Suppose that *P*(*B*) = 0.40, *P*(*D*) = 0.30 and *P*(*B* AND *D*) = 0.20.

一名学生去图书馆。设事件 *B* = 该学生借书，事件 *D* = 该学生借 DVD。假定 *P*(*B*) = 0.40，*P*(*D*) = 0.30，*P*(*B* AND *D*) = 0.20。

1. Find *P*(*B*\|*D*).

1. 求 *P*(*B*\|*D*)。

2. Find *P*(*D*\|*B*).

2. 求 *P*(*D*\|*B*)。

3. Are *B* and *D* independent?

3. *B* 和 *D* 独立吗？

4. Are *B* and *D* mutually exclusive?

4. *B* 和 *D* 互斥吗？

In a box there are three red cards and five blue cards. The red cards are marked with the numbers 1, 2, and 3, and the blue cards are marked with the numbers 1, 2, 3, 4, and 5. The cards are well-shuffled. You reach into the box (you cannot see into it) and draw one card.

一个盒子里有 3 张红卡片和 5 张蓝卡片。红卡片标有数字 1、2、3，蓝卡片标有数字 1、2、3、4、5。卡片被充分洗匀。你把手伸进盒子（看不见里面）抽出一张卡片。

Let *R* = red card is drawn, *B* = blue card is drawn, *E* = even-numbered card is drawn.

设 *R* = 抽到红卡片，*B* = 抽到蓝卡片，*E* = 抽到偶数编号的卡片。

The sample space *S* = *R*1, *R*2, *R*3, *B*1, *B*2, *B*3, *B*4, *B*5. *S* has eight outcomes.

样本空间 *S* = *R*1, *R*2, *R*3, *B*1, *B*2, *B*3, *B*4, *B*5。*S* 有 8 个结果。

*P*(*R*) = $\frac{3}{8}$. *P*(*B*) = $\frac{5}{8}$. *P*(*R* AND *B*) = 0. (You cannot draw one card that is both red and blue.)
*P*(*E*) = $\frac{3}{8}$. (There are three even-numbered cards, *R*2, *B*2, and *B*4.)
*P*(*E*\|*B*) = $\frac{2}{5}$. (There are five blue cards: *B*1, *B*2, *B*3, *B*4, and *B*5. Out of the blue cards, there are two even cards; *B*2 and *B*4.)
*P*(*B*\|*E*) = $\frac{2}{3}$. (There are three even-numbered cards: *R*2, *B*2, and *B*4. Out of the even-numbered cards, to are blue; *B*2 and *B*4.)
The events *R* and *B* are mutually exclusive because *P*(*R* AND *B*) = 0.
Let *G* = card with a number greater than 3. *G* = {*B*4, *B*5}. *P*(*G*) = $\frac{2}{8}$. Let *H* = blue card numbered between one and four, inclusive. *H* = {*B*1, *B*2, *B*3, *B*4}. *P*(*G*\|*H*) = $\frac{1}{4}$. (The only card in *H* that has a number greater than three is *B*4.) Since $\frac{2}{8}$ = $\frac{1}{4}$, *P*(*G*) = *P*(*G*\|*H*), which means that *G* and *H* are independent.

*P*(*R*) = $\frac{3}{8}$。*P*(*B*) = $\frac{5}{8}$。*P*(*R* AND *B*) = 0。（你不可能抽出一张既是红又是蓝的卡片。）
*P*(*E*) = $\frac{3}{8}$。（共有三张偶数编号卡片：*R*2、*B*2、*B*4。）
*P*(*E*\|*B*) = $\frac{2}{5}$。（共有五张蓝卡片：*B*1、*B*2、*B*3、*B*4、*B*5。在蓝卡片中，有两张偶数卡片：*B*2 和 *B*4。）
*P*(*B*\|*E*) = $\frac{2}{3}$。（共有三张偶数编号卡片：*R*2、*B*2、*B*4。在偶数编号卡片中，有两张是蓝色的：*B*2 和 *B*4。）
事件 *R* 和 *B* 互斥，因为 *P*(*R* AND *B*) = 0。
设 *G* = 编号大于 3 的卡片。*G* = {*B*4, *B*5}。*P*(*G*) = $\frac{2}{8}$。设 *H* = 编号在 1 到 4 之间（含）的蓝卡片。*H* = {*B*1, *B*2, *B*3, *B*4}。*P*(*G*\|*H*) = $\frac{1}{4}$。（*H* 中编号大于 3 的唯一卡片是 *B*4。）由于 $\frac{2}{8}$ = $\frac{1}{4}$，*P*(*G*) = *P*(*G*\|*H*)，这意味着 *G* 和 *H* 独立。

In a basketball arena,

在一个篮球场馆里，

70% of the fans are rooting for the home team.
25% of the fans are wearing blue.
20% of the fans are wearing blue and are rooting for the away team.
Of the fans rooting for the away team, 67% are wearing blue.

70% 的观众支持主队。
25% 的观众穿着蓝色。
20% 的观众穿着蓝色且支持客队。
在支持客队的观众中，67% 穿着蓝色。

Let *A* be the event that a fan is rooting for the away team.

设 *A* = 观众支持客队这一事件。

Let *B* be the event that a fan is wearing blue.

设 *B* = 观众穿着蓝色这一事件。

Are the events of rooting for the away team and wearing blue independent? Are they mutually exclusive?

支持客队与穿着蓝色这两个事件独立吗？它们互斥吗？

Problem 问题

In a particular college class, 60% of the students are female. Fifty percent of all students in the class have long hair. Forty-five percent of the students are female and have long hair. Of the female students, 75% have long hair. Let *F* be the event that a student is female. Let *L* be the event that a student has long hair. One student is picked randomly. Are the events of being female and having long hair independent?

在某大学课堂上，60% 的学生是女性。全班所有学生中 50% 留长发。45% 的学生是女性且留长发。在女学生中，75% 留长发。设 *F* = 学生为女性这一事件。设 *L* = 学生留长发这一事件。随机抽取一名学生。身为女性与留长发这两个事件独立吗？

The following probabilities are given in this example:
*P*(*F*) = 0.60; *P*(*L*) = 0.50
*P*(*F* AND *L*) = 0.45
*P*(*L*\|*F*) = 0.75

本例给出下列概率：
*P*(*F*) = 0.60；*P*(*L*) = 0.50
*P*(*F* AND *L*) = 0.45
*P*(*L*\|*F*) = 0.75

The choice you make depends on the information you have. You could use the first or last condition on the list for this example. You do not know *P*(*F*\|*L*) yet, so you cannot use the second condition.

你选择哪种方法取决于你所掌握的信息。本例你可以使用列表中的第一个或最后一个条件。你还不知道 *P*(*F*\|*L*)，所以不能使用第二个条件。

Solution 解答

>Check whether *P*(*F* AND *L*) = *P*(*F*)*P*(*L*). We are given that *P*(*F* AND *L*) = 0.45, but *P*(*F*)*P*(*L*) = (0.60)(0.50) = 0.30. The events of being female and having long hair are not independent because *P*(*F* AND *L*) does not equal *P*(*F*)*P*(*L*).

检验 *P*(*F* AND *L*) = *P*(*F*)*P*(*L*) 是否成立。已知 *P*(*F* AND *L*) = 0.45，但 *P*(*F*)*P*(*L*) = (0.60)(0.50) = 0.30。身为女性与留长发这两个事件不独立，因为 *P*(*F* AND *L*) 不等于 *P*(*F*)*P*(*L*)。

Check whether *P*(*L*\|*F*) equals *P*(*L*). We are given that *P*(*L*\|*F*) = 0.75, but *P*(*L*) = 0.50; they are not equal. The events of being female and having long hair are not independent.

检验 *P*(*L*\|*F*) 是否等于 *P*(*L*)。已知 *P*(*L*\|*F*) = 0.75，但 *P*(*L*) = 0.50；二者不相等。身为女性与留长发这两个事件不独立。

The events of being female and having long hair are not independent; knowing that a student is female changes the probability that a student has long hair.

身为女性与留长发这两个事件不独立；知道一名学生为女性会改变该学生留长发的概率。

Mark is deciding which route to take to work. His choices are *I* = the Interstate and *F* = Fifth Street.

马克正在决定上班走哪条路线。他的选择是 *I* = 州际公路，*F* = 第五大街。

*P*(*I*) = 0.44 and *P*(*F*) = 0.56
*P*(*I* AND *F*) = 0 because Mark will take only one route to work.

*P*(*I*) = 0.44，*P*(*F*) = 0.56
*P*(*I* AND *F*) = 0，因为马克上班只走一条路线。

What is the probability of *P*(*I* OR *F*)?

*P*(*I* OR *F*) 的概率是多少？

Problem 问题

1. Toss one fair coin (the coin has two sides, *H* and *T*). The outcomes are \_\_\_\_\_\_\_\_. Count the outcomes. There are \_\_\_\_ outcomes.

1. 抛一枚均匀的硬币（硬币有两面，*H* 和 *T*）。结果为 \_\_\_\_\_\_\_\_。数一数结果。共有 \_\_\_\_ 个结果。

2. Toss one fair, six-sided die (the die has 1, 2, 3, 4, 5 or 6 dots on a side). The outcomes are \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. Count the outcomes. There are \_\_\_ outcomes.

2. 掷一枚均匀的六面骰子（骰子的一面有 1、2、3、4、5 或 6 个点）。结果为 \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_。数一数结果。共有 \_\_\_ 个结果。

3. Multiply the two numbers of outcomes. The answer is \_\_\_\_\_\_\_.

3. 把两个结果数相乘。答案是 \_\_\_\_\_\_\_。

4. If you flip one fair coin and follow it with the toss of one fair, six-sided die, the answer in part c. is the number of outcomes (size of the sample space). What are the outcomes? (Hint: Two of the outcomes are *H*1 and *T*6.)

4. 如果你先抛一枚均匀的硬币，再掷一枚均匀的六面骰子，那么 c 部分的答案就是结果数（样本空间的大小）。这些结果是什么？（提示：其中两个结果是 *H*1 和 *T*6。）

5. Event *A* = heads (*H*) on the coin followed by an even number (2, 4, 6) on the die.

5. 事件 *A* = 硬币为正面（*H*）且骰子为偶数（2、4、6）。

*A* = {\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}. Find *P*(*A*).

*A* = {\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_}。求 *P*(*A*)。

6. Event *B* = heads on the coin followed by a three on the die. *B* = {\_\_\_\_\_\_\_\_}. Find *P*(*B*).

6. 事件 *B* = 硬币为正面且骰子为 3。*B* = {\_\_\_\_\_\_\_\_}。求 *P*(*B*)。

7. Are *A* and *B* mutually exclusive? (Hint: What is *P*(*A* AND *B*)? If *P*(*A* AND *B*) = 0, then *A* and *B* are mutually exclusive.)

7. *A* 和 *B* 互斥吗？（提示：*P*(*A* AND *B*) 是多少？若 *P*(*A* AND *B*) = 0，则 *A* 和 *B* 互斥。）

8. Are *A* and *B* independent? (Hint: Is *P*(*A* AND *B*) = *P*(*A*)*P*(*B*)? If *P*(*A* AND *B*) = *P*(*A*)*P*(*B*), then *A* and *B* are independent. If not, then they are dependent).

8. *A* 和 *B* 独立吗？（提示：是否有 *P*(*A* AND *B*) = *P*(*A*)*P*(*B*)？若 *P*(*A* AND *B*) = *P*(*A*)*P*(*B*)，则 *A* 和 *B* 独立；否则相依。）

Solution 解答

1. *H* and *T*; 2

1. *H* 和 *T*；2

2. 1, 2, 3, 4, 5, 6; 6

2. 1、2、3、4、5、6；6

3. 2(6) = 12

4. *T*1, *T*2, *T*3, *T*4, *T*5, *T*6, *H*1, *H*2, *H*3, *H*4, *H*5, *H*6

4. *T*1、*T*2、*T*3、*T*4、*T*5、*T*6、*H*1、*H*2、*H*3、*H*4、*H*5、*H*6

5. *A* = {*H*2, *H*4, *H*6}; *P*(*A*) = $\frac{\text{3}}{12}$

5. *A* = {*H*2, *H*4, *H*6}；*P*(*A*) = $\frac{\text{3}}{12}$

6. *B* = {*H*3}; *P*(*B*) = $\frac{\text{1}}{12}$

6. *B* = {*H*3}；*P*(*B*) = $\frac{\text{1}}{12}$

7. Yes, because *P*(*A* AND *B*) = 0

7. 是，因为 *P*(*A* AND *B*) = 0

8. *P*(*A* AND *B*) = 0.*P*(*A*)*P*(*B*) = $\left( \frac{\text{3}}{12} \right)$$\left( \frac{\text{1}}{12} \right)$. *P*(*A* AND *B*) does not equal *P*(*A*)*P*(*B*), so *A* and *B* are dependent.

8. *P*(*A* AND *B*) = 0。*P*(*A*)*P*(*B*) = $\left( \frac{\text{3}}{12} \right)$$\left( \frac{\text{1}}{12} \right)$。*P*(*A* AND *B*) 不等于 *P*(*A*)*P*(*B*)，所以 *A* 和 *B* 相依。

A box has two balls, one white and one red. We select one ball, put it back in the box, and select a second ball (sampling with replacement). Let *T* be the event of getting the white ball twice, *F* the event of picking the white ball first, *S* the event of picking the white ball in the second drawing.

一个盒子里有两个球，一个白球、一个红球。我们抽取一个球，放回盒中，再抽取第二个球（有放回抽样）。设 *T* = 两次都抽到白球这一事件，*F* = 第一次抽到白球这一事件，*S* = 第二次抽到白球这一事件。

1. Compute *P*(*T*).

1. 计算 *P*(*T*)。

2. Compute *P*(*T*\|*F*).

2. 计算 *P*(*T*\|*F*)。

3. Are *T* and *F* independent?.

3. *T* 和 *F* 独立吗？

4. Are *F* and *S* mutually exclusive?

4. *F* 和 *S* 互斥吗？

5. Are *F* and *S* independent?

5. *F* 和 *S* 独立吗？

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3.3 Two Basic Rules of Probability 3.3 概率的两大基本法则

When calculating probability, there are two rules to consider when determining if two events are independent or dependent and if they are mutually exclusive or not.

在计算概率时，有两条法则可用于判断两个事件是独立还是相依，以及是否互斥。

The Multiplication Rule 乘法法则

If *A* and *B* are two events defined on a sample space, then: *P*(*A* AND *B*) = *P*(*B*)*P*(*A*\|*B*).

若 *A* 和 *B* 是样本空间上定义的两个事件，则：*P*(*A* AND *B*) = *P*(*B*)*P*(*A*\|*B*)。

This rule may also be written as: *P*(*A*\|*B*) = $\frac{P(A\text{~AND~}B)}{P(B)}$

这条法则也可写为：*P*(*A*\|*B*) = $\frac{P(A\text{~AND~}B)}{P(B)}$

(The probability of *A* given *B* equals the probability of *A* and *B* divided by the probability of *B*.)

（在 *B* 发生的条件下 *A* 的概率，等于 *A* 与 *B* 同时发生的概率除以 *B* 发生的概率。）

If *A* and *B* are independent, then *P*(*A*\|*B*) = *P*(*A*). Then *P*(*A* AND *B*) = *P*(*A*\|*B*)*P*(*B*) becomes *P*(*A* AND *B*) = *P*(*A*)*P*(*B*).

若 *A* 和 *B* 独立，则 *P*(*A*\|*B*) = *P*(*A*)。于是 *P*(*A* AND *B*) = *P*(*A*\|*B*)*P*(*B*) 变为 *P*(*A* AND *B*) = *P*(*A*)*P*(*B*)。

The Addition Rule 加法法则

If *A* and *B* are defined on a sample space, then: *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*).

若 *A* 与 *B* 定义在同一个样本空间上，则：*P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*)。

If *A* and *B* are mutually exclusive, then *P*(*A* AND *B*) = 0. Then *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*) becomes *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*).

若 *A* 与 *B* 互斥，则 *P*(*A* AND *B*) = 0。于是 *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*) 变为 *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*)。

Klaus is trying to choose where to go on vacation. His two choices are: *A* = New Zealand and *B* = Alaska

Klaus 正在选择去哪里度假。他的两个选择是：*A* = 新西兰，*B* = 阿拉斯加。

Klaus can only afford one vacation. The probability that he chooses *A* is *P*(*A*) = 0.6 and the probability that he chooses *B* is *P*(*B*) = 0.35.
*P*(*A* AND *B*) = 0 because Klaus can only afford to take one vacation
Therefore, the probability that he chooses either New Zealand or Alaska is *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) = 0.6 + 0.35 = 0.95. Note that the probability that he does not choose to go anywhere on vacation must be 0.05.

Klaus 只负担得起一次度假。他选择 *A* 的概率为 *P*(*A*) = 0.6，选择 *B* 的概率为 *P*(*B*) = 0.35。
*P*(*A* AND *B*) = 0，因为 Klaus 只负担得起一次度假。
因此，他选择新西兰或阿拉斯加（二者之一）的概率为 *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) = 0.6 + 0.35 = 0.95。注意，他哪里都不去度假的概率必为 0.05。

Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. *A* = the event Carlos is successful on his first attempt. *P*(*A*) = 0.65. *B* = the event Carlos is successful on his second attempt. *P*(*B*) = 0.65. Carlos tends to shoot in streaks. The probability that he makes the second goal GIVEN that he made the first goal is 0.90.

Carlos 踢大学足球。他射门进球的概率为 65%。Carlos 在下一场比赛中将连续尝试两次射门。*A* = Carlos 第一次尝试成功这一事件。*P*(*A*) = 0.65。*B* = Carlos 第二次尝试成功这一事件。*P*(*B*) = 0.65。Carlos 往往连续进球。他在第一次进球条件下第二次也进球的概率为 0.90。

Problem 问题

a\. What is the probability that he makes both goals?

a. 他两次都进球的概率是多少？

b\. What is the probability that Carlos makes either the first goal or the second goal?

b. Carlos 进第一次球或第二次球（二者之一）的概率是多少？

c\. Are *A* and *B* independent?

c. *A* 与 *B* 是否独立？

d\. Are *A* and *B* mutually exclusive?

d. *A* 与 *B* 是否互斥？

Solution 解答

a\. The problem is asking you to find *P*(*A* AND *B*) = *P*(*B* AND *A*). Since *P*(*B*\|*A*) = 0.90: *P*(*B* AND *A*) = *P*(*B*\|*A*) *P*(*A*) = (0.90)(0.65) = 0.585

a. 问题要求你求 *P*(*A* AND *B*) = *P*(*B* AND *A*)。因为 *P*(*B*\|*A*) = 0.90：*P*(*B* AND *A*) = *P*(*B*\|*A*) *P*(*A*) = (0.90)(0.65) = 0.585

Carlos makes the first and second goals with probability 0.585.

Carlos 第一、二次都进球的概率为 0.585。

b\. The problem is asking you to find *P*(*A* OR *B*).

b. 问题要求你求 *P*(*A* OR *B*)。

*P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*) = 0.65 + 0.65 - 0.585 = 0.715

Carlos makes either the first goal or the second goal with probability 0.715.

Carlos 进第一次球或第二次球（二者之一）的概率为 0.715。

c\. No, they are not, because *P*(*B* AND *A*) = 0.585.

c. 否，它们不独立，因为 *P*(*B* AND *A*) = 0.585。

*P*(*B*)*P*(*A*) = (0.65)(0.65) = 0.423

0.423 ≠ 0.585 = *P*(*B* AND *A*)

So, *P*(*B* AND *A*) is not equal to *P*(*B*)*P*(*A*).

因此，*P*(*B* AND *A*) 不等于 *P*(*B*)*P*(*A*)。

d\. No, they are not because *P*(*A* and *B*) = 0.585.

d. 否，它们不互斥，因为 *P*(*A* and *B*) = 0.585。

To be mutually exclusive, *P*(*A* AND *B*) must equal zero.

要互斥，*P*(*A* AND *B*) 必须等于零。

Helen plays basketball. For free throws, she makes the shot 75% of the time. Helen must now attempt two free throws. *C* = the event that Helen makes the first shot. *P*(*C*) = 0.75. *D* = the event Helen makes the second shot. *P*(*D*) = 0.75. The probability that Helen makes the second free throw given that she made the first is 0.85. What is the probability that Helen makes both free throws?

Helen 打篮球。在罚球中，她投中的概率为 75%。Helen 现在要尝试两次罚球。*C* = Helen 第一次投中这一事件。*P*(*C*) = 0.75。*D* = Helen 第二次投中这一事件。*P*(*D*) = 0.75。她在第一次投中条件下第二次也投中的概率为 0.85。Helen 两次罚球都投中的概率是多少？

A community swim team has 150 members. Seventy-five of the members are advanced swimmers. Forty-seven of the members are intermediate swimmers. The remainder are novice swimmers. Forty of the advanced swimmers practice four times a week. Thirty of the intermediate swimmers practice four times a week. Ten of the novice swimmers practice four times a week. Suppose one member of the swim team is chosen randomly.

一个社区游泳队有 150 名成员。其中 75 名是高级游泳者。47 名是中级游泳者。其余是初级游泳者。40 名高级游泳者每周练习四次。30 名中级游泳者每周练习四次。10 名初级游泳者每周练习四次。假设随机选取该游泳队的一名成员。

Problem 问题

a\. What is the probability that the member is a novice swimmer?

a. 该成员是初级游泳者的概率是多少？

b\. What is the probability that the member practices four times a week?

b. 该成员每周练习四次的概率是多少？

c\. What is the probability that the member is an advanced swimmer and practices four times a week?

c. 该成员是高级游泳者且每周练习四次的概率是多少？

d\. What is the probability that a member is an advanced swimmer and an intermediate swimmer? Are being an advanced swimmer and an intermediate swimmer mutually exclusive? Why or why not?

d. 一名成员既是高级游泳者又是中级游泳者的概率是多少？成为高级游泳者与成为中级游泳者是否互斥？为什么是或为什么不是？

e\. Are being a novice swimmer and practicing four times a week independent events? Why or why not?

e. 成为初级游泳者与每周练习四次是否为独立事件？为什么是或为什么不是？

Solution 解答

a\. $\frac{28}{150}$

a. $\frac{28}{150}$

b\. $\frac{80}{150}$

b. $\frac{80}{150}$

c\. $\frac{40}{150}$

c. $\frac{40}{150}$

d\. *P*(advanced AND intermediate) = 0, so these are mutually exclusive events. A swimmer cannot be an advanced swimmer and an intermediate swimmer at the same time.

d. *P*(高级 AND 中级) = 0，因此这些事件是互斥事件。一名游泳者不可能同时既是高级游泳者又是中级游泳者。

e\. No, these are not independent events.

e. 否，这些不是独立事件。

*P*(novice AND practices four times per week) = 0.0667

*P*(初级 AND 每周练习四次) = 0.0667

*P*(novice)*P*(practices four times per week) = 0.0996

*P*(初级)*P*(每周练习四次) = 0.0996

0.0667 ≠ 0.0996

A school has 200 seniors of whom 140 will be going to college next year. Forty will be going directly to work. The remainder are taking a gap year. Fifty of the seniors going to college play sports. Thirty of the seniors going directly to work play sports. Five of the seniors taking a gap year play sports. What is the probability that a senior is taking a gap year?

一所学校有 200 名高三学生，其中 140 名明年将上大学。40 名将直接工作。其余正在间隔年。上大学的高三学生中有 50 名参加体育运动。直接工作的高三学生中有 30 名参加体育运动。间隔年的高三学生中有 5 名参加体育运动。一名高三学生正在间隔年的概率是多少？

Felicity attends Modesto JC in Modesto, CA. The probability that Felicity enrolls in a math class is 0.2 and the probability that she enrolls in a speech class is 0.65. The probability that she enrolls in a math class GIVEN that she enrolls in speech class is 0.25.

Felicity 就读于加州莫德斯托的 Modesto JC。Felicity 选修数学课的概率为 0.2，选修演讲课的概率为 0.65。她在选修演讲课条件下选修数学课的概率为 0.25。

Let: *M* = math class, *S* = speech class, *M*\|*S* = math given speech

令：*M* = 数学课，*S* = 演讲课，*M*\|*S* = 给定演讲课时的数学课

Problem 问题

1. What is the probability that Felicity enrolls in math and speech?

1. Felicity 选修数学课和演讲课的概率是多少？

Find *P*(*M* AND *S*) = *P*(*M*\|*S*)*P*(*S*).

求 *P*(*M* AND *S*) = *P*(*M*\|*S*)*P*(*S*)。

2. What is the probability that Felicity enrolls in math or speech classes?

2. Felicity 选修数学课或演讲课的概率是多少？

Find *P*(*M* OR *S*) = *P*(*M*) + *P*(*S*) - *P*(*M* AND *S*).

求 *P*(*M* OR *S*) = *P*(*M*) + *P*(*S*) - *P*(*M* AND *S*)。

3. Are *M* and *S* independent? Is *P*(*M*\|*S*) = *P*(*M*)?

3. *M* 与 *S* 是否独立？是否有 *P*(*M*\|*S*) = *P*(*M*)？

4. Are *M* and *S* mutually exclusive? Is *P*(*M* AND *S*) = 0?

4. *M* 与 *S* 是否互斥？是否有 *P*(*M* AND *S*) = 0？

Solution 解答

a\. 0.1625, b. 0.6875, c. No, d. No

a. 0.1625，b. 0.6875，c. 否，d. 否

A student goes to the library. Let events *B* = the student checks out a book and *D* = the student checks out a DVD. Suppose that *P*(*B*) = 0.40, *P*(*D*) = 0.30 and *P*(*D*\|*B*) = 0.5.

一名学生去图书馆。令事件 *B* = 学生借出一本书，*D* = 学生借出一张 DVD。假设 *P*(*B*) = 0.40，*P*(*D*) = 0.30，且 *P*(*D*\|*B*) = 0.5。

1. Find *P*(*B* AND *D*).

1. 求 *P*(*B* AND *D*)。

2. Find *P*(*B* OR *D*).

2. 求 *P*(*B* OR *D*)。

Studies show that about one woman in seven (approximately 14.3%) who live to be 90 will develop breast cancer. Suppose that of those women who develop breast cancer, a test is negative 2% of the time. Also suppose that in the general population of women, the test for breast cancer is negative about 85% of the time. Let *B* = woman develops breast cancer and let *N* = tests negative. Suppose one woman is selected at random.

研究表明，活到 90 岁的女性中约有七分之一（约 14.3%）会患乳腺癌。假设在那些患乳腺癌的女性中，检测为阴性的情况占 2%。另假设在女性总体中，乳腺癌检测约为 85% 的时间为阴性。令 *B* = 女性患乳腺癌，*N* = 检测为阴性。假设随机选取一名女性。

Problem 问题

a\. What is the probability that the woman develops breast cancer? What is the probability that woman tests negative?

a. 该女性患乳腺癌的概率是多少？该女性检测为阴性的概率是多少？

b\. Given that the woman has breast cancer, what is the probability that she tests negative?

b. 已知该女性患有乳腺癌，她检测为阴性的概率是多少？

c\. What is the probability that the woman has breast cancer AND tests negative?

c. 该女性患有乳腺癌且检测为阴性的概率是多少？

d\. What is the probability that the woman has breast cancer or tests negative?

d. 该女性患有乳腺癌或检测为阴性的概率是多少？

e\. Are having breast cancer and testing negative independent events?

e. 患乳腺癌与检测为阴性是否为独立事件？

f\. Are having breast cancer and testing negative mutually exclusive?

f. 患乳腺癌与检测为阴性是否互斥？

Solution 解答

a\. *P*(*B*) = 0.143; *P*(*N*) = 0.85

a. *P*(*B*) = 0.143；*P*(*N*) = 0.85

b\. *P*(*N*\|*B*) = 0.02

b. *P*(*N*\|*B*) = 0.02

c\. *P*(*B* AND *N*) = *P*(*B*)*P*(*N*\|*B*) = (0.143)(0.02) = 0.0029

c. *P*(*B* AND *N*) = *P*(*B*)*P*(*N*\|*B*) = (0.143)(0.02) = 0.0029

d\. *P*(*B* OR *N*) = *P*(*B*) + *P*(*N*) - *P*(*B* AND *N*) = 0.143 + 0.85 - 0.0029 = 0.9901

d. *P*(*B* OR *N*) = *P*(*B*) + *P*(*N*) - *P*(*B* AND *N*) = 0.143 + 0.85 - 0.0029 = 0.9901

e\. No. *P*(*N*) = 0.85; *P*(*N*\|*B*) = 0.02. So, *P*(*N*\|*B*) does not equal *P*(*N*).

e. 否。*P*(*N*) = 0.85；*P*(*N*\|*B*) = 0.02。因此，*P*(*N*\|*B*) 不等于 *P*(*N*)。

f\. No. *P*(*B* AND *N*) = 0.0029. For *B* and *N* to be mutually exclusive, *P*(*B* AND *N*) must be zero.

f. 否。*P*(*B* AND *N*) = 0.0029。要使 *B* 与 *N* 互斥，*P*(*B* AND *N*) 必须为零。

一所学校有 200 名高三学生，其中 140 名明年将上大学。40 名将直接工作。其余正在间隔年。上大学的高三学生中有 50 名参加体育运动。直接工作的高三学生中有 30 名参加体育运动。间隔年的高三学生中有 5 名参加体育运动。一名高三学生上大学且参加体育运动的概率是多少？

Problem 问题

Refer to the information in Example 3.18. *P* = tests positive.

参考示例 3.18 中的信息。*P* = 检测为阳性。

1. Given that a woman develops breast cancer, what is the probability that she tests positive. Find *P*(*P*\|*B*) = 1 - *P*(*N*\|*B*).

1. 已知一名女性患乳腺癌，她检测为阳性的概率是多少。求 *P*(*P*\|*B*) = 1 - *P*(*N*\|*B*)。

2. What is the probability that a woman develops breast cancer and tests positive. Find *P*(*B* AND *P*) = *P*(*P*\|*B*)*P*(*B*).

2. 一名女性患乳腺癌且检测为阳性的概率是多少。求 *P*(*B* AND *P*) = *P*(*P*\|*B*)*P*(*B*)。

3. What is the probability that a woman does not develop breast cancer. Find *P*(*B′*) = 1 - *P*(*B*).

3. 一名女性不患乳腺癌的概率是多少。求 *P*(*B′*) = 1 - *P*(*B*)。

4. What is the probability that a woman tests positive for breast cancer. Find *P*(*P*) = 1 - *P*(*N*).

4. 一名女性乳腺癌检测为阳性的概率是多少。求 *P*(*P*) = 1 - *P*(*N*)。

Solution 解答

a\. 0.98; b. 0.1401; c. 0.857; d. 0.15

a. 0.98；b. 0.1401；c. 0.857；d. 0.15

A student goes to the library. Let events *B* = the student checks out a book and *D* = the student checks out a DVD. Suppose that *P*(*B*) = 0.40, *P*(*D*) = 0.30 and *P*(*D*\|*B*) = 0.5.

一名学生去图书馆。令事件 *B* = 学生借出一本书，*D* = 学生借出一张 DVD。假设 *P*(*B*) = 0.40，*P*(*D*) = 0.30，且 *P*(*D*\|*B*) = 0.5。

1. Find *P*(*B′*).

1. 求 *P*(*B′*)。

2. Find *P*(*D* AND *B*).

2. 求 *P*(*D* AND *B*)。

3. Find *P*(*B*\|*D*).

3. 求 *P*(*B*\|*D*)。

4. Find *P*(*D* AND *B′*).

4. 求 *P*(*D* AND *B′*)。

5. Find *P*(*D*\|*B′*).

5. 求 *P*(*D*\|*B′*)。

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3.4 Contingency Tables 3.4 列联表

A contingency table provides a way of portraying data that can facilitate calculating probabilities. The table helps in determining conditional probabilities quite easily. The table displays sample values in relation to two different variables that may be dependent or contingent on one another. Later on, we will use contingency tables again, but in another manner.

列联表提供了一种刻画数据的方式，便于计算概率。该表有助于相当容易地确定条件概率。表中展示样本取值相对于两个可能相依或互为条件的不同变量的关系。稍后我们将再次使用列联表，但方式不同。

Problem 问题

Suppose a study of speeding violations and drivers who use cell phones produced the following fictional data:

假设一项关于超速违章与使用手机驾驶者的研究产生了以下虚构数据：

|---------------------------------------|-------------------------------------|----------------------------------------|-------|

| Uses cell phone while driving | 25 | 280 | 305 |

| 驾驶时使用手机 | 25 | 280 | 305 |

| Does not use cell phone while driving | 45 | 405 | 450 |

| 驾驶时不使用手机 | 45 | 405 | 450 |

| Total | 70 | 685 | 755 |

| 合计 | 70 | 685 | 755 |

Table 3.2

表 3.2

The total number of people in the sample is 755. The row totals are 305 and 450. The column totals are 70 and 685. Notice that 305 + 450 = 755 and 70 + 685 = 755.

样本中总人数为 755。行合计为 305 和 450。列合计为 70 和 685。注意 305 + 450 = 755，且 70 + 685 = 755。

Calculate the following probabilities using the table.

用该表计算以下概率。

a\. Find *P*(Driver is a cell phone user).

a. 求 *P*(驾驶者是手机用户)。

b. Find *P*(driver had no violation in the last year).

b. 求 *P*(驾驶者上一年无违章)。

c. Find *P*(Driver had no violation in the last year AND was a cell phone user).

c. 求 *P*(驾驶者上一年无违章 AND 是手机用户)。

d. Find *P*(Driver is a cell phone user OR driver had no violation in the last year).

d. 求 *P*(驾驶者是手机用户 OR 驾驶者上一年无违章)。

e. Find *P*(Driver is a cell phone user GIVEN driver had a violation in the last year).

e. 求 *P*(驾驶者是手机用户 GIVEN 驾驶者上一年有违章)。

f. Find *P*(Driver had no violation last year GIVEN driver was not a cell phone user)

f. 求 *P*(驾驶者上一年无违章 GIVEN 驾驶者不是手机用户)

Solution 解答

a\. $\frac{\text{number~of~cell~phone~users}}{\text{total~number~in~study}}\ = \ \frac{305}{755}$

a. $\frac{\text{number~of~cell~phone~users}}{\text{total~number~in~study}}\ = \ \frac{305}{755}$

b. $\frac{\text{number~that~had~no~violation}}{\text{total~number~in~study}}\ = \ \frac{685}{755}$

c\. $\frac{280}{755}$

c. $\frac{280}{755}$

d\. $\left( {\frac{305}{755}\ + \ \frac{685}{755}} \right)\ - \ \frac{280}{755}\ = \ \frac{710}{755}$

d. $\left( {\frac{305}{755}\ + \ \frac{685}{755}} \right)\ - \ \frac{280}{755}\ = \ \frac{710}{755}$

e\. $\frac{25}{70}$ (The sample space is reduced to the number of drivers who had a violation.)

e. $\frac{25}{70}$（样本空间缩小为有过违章的驾驶者人数。）

f\. $\frac{405}{450}$ (The sample space is reduced to the number of drivers who were not cell phone users.)

f. $\frac{405}{450}$（样本空间缩小为不是手机用户的驾驶者人数。）

Table 3.3 shows the number of athletes who stretch before exercising and how many had injuries within the past year.

表 3.3 展示了运动前拉伸的运动员人数，以及过去一年中有伤的人数。

|------------------|---------------------|------------------------|-------|

| Stretches | 55 | 295 | 350 |

| 拉伸 | 55 | 295 | 350 |

| Does not stretch | 231 | 219 | 450 |

| 不拉伸 | 231 | 219 | 450 |

| Total | 286 | 514 | 800 |

| 合计 | 286 | 514 | 800 |

Table 3.3

表 3.3

1. What is *P*(athlete stretches before exercising)?

1. *P*(运动员运动前拉伸) 是多少？

2. What is *P*(athlete stretches before exercising\|no injury in the last year)?

2. *P*(运动员运动前拉伸\|上一年未受伤) 是多少？

Table 3.4 shows a random sample of 100 hikers and the areas of hiking they prefer.

表 3.4 展示了 100 名徒步者的随机样本及其偏好的徒步区域。

| 性别 | 海岸线 | 近湖泊与溪流 | 山峰上 | 合计 |

|--------|---------------|------------------------|-------------------|-------|

| Female | 18 | 16 | | 45 |

| 女性 | 18 | 16 | | 45 |

| Male | | | 14 | 55 |

| 男性 | | | 14 | 55 |

| Total | | 41 | | |

| 合计 | | 41 | | |

Table 3.4 Hiking Area Preference

表 3.4 徒步区域偏好

Problem 问题

a\. Complete the table.

a. 补全该表。

Solution 解答

a\.

| 性别 | 海岸线 | 近湖泊与溪流 | 山峰上 | 合计 |

|--------|---------------|------------------------|-------------------|---------|

| Female | 18 | 16 | 11 | 45 |

| 女性 | 18 | 16 | 11 | 45 |

| Male | 16 | 25 | 14 | 55 |

| 男性 | 16 | 25 | 14 | 55 |

| Total | 34 | 41 | 25 | 100 |

| 合计 | 34 | 41 | 25 | 100 |

Table 3.5 Hiking Area Preference

表 3.5 徒步区域偏好

Problem 问题

b\. Are the events "being female" and "preferring the coastline" independent events?

b. "是女性"与"偏好海岸线"这两个事件是否为独立事件？

Let *F* = being female and let *C* = preferring the coastline.

令 *F* = 是女性，*C* = 偏好海岸线。

1. Find *P*(*F* AND *C*).

1. 求 *P*(*F* AND *C*)。

2. Find *P*(*F*)*P*(*C*)

2. 求 *P*(*F*)*P*(*C*)

Are these two numbers the same? If they are, then *F* and *C* are independent. If they are not, then *F* and *C* are not independent.

这两个数相同吗？若相同，则 *F* 与 *C* 独立；若不同，则 *F* 与 *C* 不独立。

Solution 解答

b\.

1. *P*(*F* AND *C*) = $\frac{18}{100}$ = 0.18

2. *P*(*F*)*P*(*C*) = $\left( \frac{45}{100} \right)\left( \frac{34}{100} \right)$ = (0.45)(0.34) = 0.153

*P*(*F* AND *C*) ≠ *P*(*F*)*P*(*C*), so the events *F* and *C* are not independent.

*P*(*F* AND *C*) ≠ *P*(*F*)*P*(*C*)，因此事件 *F* 和 *C* 不是独立事件。

Problem 问题

c\. Find the probability that a person is male given that the person prefers hiking near lakes and streams. Let *M* = being male, and let *L* = prefers hiking near lakes and streams.

c. 求一个人是男性的概率，已知该人偏好在湖边和溪流附近徒步。令 *M* = 男性，*L* = 偏好在湖边和溪流附近徒步。

1. What word tells you this is a conditional?

1. 哪个词提示你这是一个条件概率？

2. Fill in the blanks and calculate the probability: *P*(\_\_\_\|\_\_\_) = \_\_\_.

2. 填空并计算概率：*P*(\_\_\_\|\_\_\_) = \_\_\_。

3. Is the sample space for this problem all 100 hikers? If not, what is it?

3. 本问题的样本空间是否为全部 100 名徒步者？如果不是，它是什么？

Solution 解答

c\.

1. The word 'given' tells you that this is a conditional.

1. "given"（给定）一词提示你这是一个条件概率。

2. *P*(*M*\|*L*) = $\frac{25}{41}$

3. No, the sample space for this problem is the 41 hikers who prefer lakes and streams.

3. 不是，本问题的样本空间是偏好湖边和溪流附近的 41 名徒步者。

Problem 问题

d\. Find the probability that a person is female or prefers hiking on mountain peaks. Let *F* = being female, and let *P* = prefers mountain peaks.

d. 求一个人是女性或偏好在山巅徒步的概率。令 *F* = 女性，*P* = 偏好山巅。

1. Find *P*(*F*).

1. 求 *P*(*F*)。

2. Find *P*(*P*).

2. 求 *P*(*P*)。

3. Find *P*(*F* AND *P*).

3. 求 *P*(*F* AND *P*)。

4. Find *P*(*F* OR *P*).

4. 求 *P*(*F* OR *P*)。

Solution 解答

d\.

1. *P*(*F*) = $\frac{45}{100}$

2. *P*(*P*) = $\frac{25}{100}$

3. *P*(*F* AND *P*) = $\frac{11}{100}$

4. *P*(*F* OR *P*) = $\frac{45}{100}$ + $\frac{25}{100}$ - $\frac{11}{100}$ = $\frac{59}{100}$

Table 3.6 shows a random sample of 200 cyclists and the routes they prefer. Let *M* = males and *H* = hilly path.

Table 3.6 展示了一个由 200 名骑行者及其偏好路线组成的随机样本。令 *M* = 男性，*H* = 山路。

| 性别 | 湖边路 | 山路 | 林间路 | 总计 |

|--------|-----------|------------|-------------|-------|

| Female | 45 | 38 | 27 | 110 |

| 女性 | 45 | 38 | 27 | 110 |

| Male | 26 | 52 | 12 | 90 |

| 男性 | 26 | 52 | 12 | 90 |

| Total | 71 | 90 | 39 | 200 |

| 总计 | 71 | 90 | 39 | 200 |

Table 3.6

1. Out of the males, what is the probability that the cyclist prefers a hilly path?

1. 在男性中，骑行者偏好山路的概率是多少？

2. Are the events "being male" and "preferring the hilly path" independent events?

2. "男性"与"偏好山路"这两个事件是否为独立事件？

Muddy Mouse lives in a cage with three doors. If Muddy goes out the first door, the probability that he gets caught by Alissa the cat is $\frac{1}{5}$ and the probability he is not caught is $\frac{4}{5}$. If he goes out the second door, the probability he gets caught by Alissa is $\frac{1}{4}$ and the probability he is not caught is $\frac{3}{4}$. The probability that Alissa catches Muddy coming out of the third door is $\frac{1}{2}$ and the probability she does not catch Muddy is $\frac{1}{2}$. It is equally likely that Muddy will choose any of the three doors so the probability of choosing each door is $\frac{1}{3}$.

泥泞鼠（Muddy Mouse）住在一个有三扇门的笼子里。如果泥泞鼠从第一扇门出去，它被猫 Alissa 抓住的概率是 $\frac{1}{5}$，没被抓住的概率是 $\frac{4}{5}$。如果它从第二扇门出去，被 Alissa 抓住的概率是 $\frac{1}{4}$，没被抓住的概率是 $\frac{3}{4}$。Alissa 从第三扇门抓住泥泞鼠的概率是 $\frac{1}{2}$，没抓住的概率也是 $\frac{1}{2}$。泥泞鼠选择任意一扇门是等可能的，因此选择每扇门的概率都是 $\frac{1}{3}$。

| 被抓或未抓 | 门一 | 门二 | 门三 | 总计 |

|---------------|----------------|----------------|---------------|-------|

| Caught | $\frac{1}{15}$ | $\frac{1}{12}$ | $\frac{1}{6}$ | |

| 被抓 | $\frac{1}{15}$ | $\frac{1}{12}$ | $\frac{1}{6}$ | |

| Not Caught | $\frac{4}{15}$ | $\frac{3}{12}$ | $\frac{1}{6}$ | |

| 未抓 | $\frac{4}{15}$ | $\frac{3}{12}$ | $\frac{1}{6}$ | |

| Total | | | | 1 |

| 总计 | | | | 1 |

Table 3.7 Door Choice

Table 3.7 门的选择

The first entry $\frac{1}{15} = \left( \frac{1}{5} \right)\left( \frac{1}{3} \right)$ is *P*(Door One AND Caught)
The entry $\frac{4}{15} = \left( \frac{4}{5} \right)\left( \frac{1}{3} \right)$ is *P*(Door One AND Not Caught)

第一项 $\frac{1}{15} = \left( \frac{1}{5} \right)\left( \frac{1}{3} \right)$ 是 *P*(Door One AND Caught)（门一且被抓）。
该项 $\frac{4}{15} = \left( \frac{4}{5} \right)\left( \frac{1}{3} \right)$ 是 *P*(Door One AND Not Caught)（门一且未抓）。

Verify the remaining entries.

验证其余各项。

Problem 问题

a\. Complete the probability contingency table. Calculate the entries for the totals. Verify that the lower-right corner entry is 1.

a. 完成该概率列联表。计算总计的各项。验证右下角元素为 1。

b\. What is the probability that Alissa does not catch Muddy?

b. Alissa 没有抓住泥泞鼠的概率是多少？

c\. What is the probability that Muddy chooses Door One OR Door Two given that Muddy is caught by Alissa?

c. 已知泥泞鼠被 Alissa 抓住，它选择门一或门二的概率是多少？

Solution 解答

a\.

| 被抓或未抓 | 门一 | 门二 | 门三 | 总计 |

|---------------|--------------------|--------------------|-------------------|---------------------|

| Caught | $\frac{1}{15}$ | $\frac{1}{12}$ | $\frac{1}{6}$ | $\frac{19}{60}$ |

| 被抓 | $\frac{1}{15}$ | $\frac{1}{12}$ | $\frac{1}{6}$ | $\frac{19}{60}$ |

| Not Caught | $\frac{4}{15}$ | $\frac{3}{12}$ | $\frac{1}{6}$ | $\frac{41}{60}$ |

| 未抓 | $\frac{4}{15}$ | $\frac{3}{12}$ | $\frac{1}{6}$ | $\frac{41}{60}$ |

| Total | $\frac{5}{15}$ | $\frac{4}{12}$ | $\frac{2}{6}$ | 1 |

| 总计 | $\frac{5}{15}$ | $\frac{4}{12}$ | $\frac{2}{6}$ | 1 |

Table 3.8 Door Choice

Table 3.8 门的选择

b\. $\frac{41}{60}$

b. $\frac{41}{60}$

c\. $\frac{9}{19}$

c. $\frac{9}{19}$

Table 3.9 contains the number of crimes per 100,000 inhabitants from 2008 to 2011 in the U.S.

Table 3.9 列出了 2008 年至 2011 年美国每 100,000 名居民中的犯罪数量。

| 年份 | 抢劫 | 入室盗窃 | 强奸 | 机动车 | 总计 |

|-------|---------|----------|------|---------|-------|

| 2008 | 145.7 | 732.1 | 29.7 | 314.7 | |

| 2009 | 133.1 | 717.7 | 29.1 | 259.2 | |

| 2010 | 119.3 | 701 | 27.7 | 239.1 | |

| 2011 | 113.7 | 702.2 | 26.8 | 229.6 | |

| 总计 | | | | | |

Table 3.9 United States Crime Index Rates Per 100,000 Inhabitants 2008–2011

Table 3.9 美国犯罪指数率（每 100,000 名居民，2008–2011）

Problem 问题

TOTAL each column and each row. Total data = 4,520.7

对每一列和每一行求总计。总数据 = 4,520.7。

1. Find *P*(2009 AND Robbery).

1. 求 *P*(2009 AND Robbery)。

2. Find *P*(2010 AND Burglary).

2. 求 *P*(2010 AND Burglary)。

3. Find *P*(2010 OR Burglary).

3. 求 *P*(2010 OR Burglary)。

4. Find *P*(2011\|Rape).

4. 求 *P*(2011\|Rape)。

5. Find *P*(Vehicle\|2008).

5. 求 *P*(Vehicle\|2008)。

Solution 解答

a\. 0.0294, b. 0.1551, c. 0.7165, d. 0.2365, e. 0.2575

a. 0.0294，b. 0.1551，c. 0.7165，d. 0.2365，e. 0.2575

Table 3.10 relates the weights and heights of a group of individuals participating in an observational study.

Table 3.10 给出了一个参与观察性研究的群体的体重与身高关系。

| 体重/身高 | 高 | 中 | 矮 | 总计 |

|---------------|------|--------|-------|--------|

| Obese | 18 | 28 | 14 | |

| 肥胖 | 18 | 28 | 14 | |

| Normal | 20 | 51 | 28 | |

| 正常 | 20 | 51 | 28 | |

| Underweight | 12 | 25 | 9 | |

| 偏瘦 | 12 | 25 | 9 | |

| 总计 | | | | |

Table 3.10

1. Find the total for each row and column

1. 求每一行和每一列的总计

2. Find the probability that a randomly chosen individual from this group is Tall.

2. 求从该群体中随机抽取一人，其身高为高的概率。

3. Find the probability that a randomly chosen individual from this group is Obese and Tall.

3. 求从该群体中随机抽取一人，其既肥胖又高的概率。

4. Find the probability that a randomly chosen individual from this group is Tall given that the idividual is Obese.

4. 求从该群体中随机抽取一人，其身高为高，已知该人肥胖的概率。

5. Find the probability that a randomly chosen individual from this group is Obese given that the individual is Tall.

5. 求从该群体中随机抽取一人，其肥胖，已知该人身高的概率。

6. Find the probability a randomly chosen individual from this group is Tall and Underweight.

6. 求从该群体中随机抽取一人，其既高又偏瘦的概率。

7. Are the events Obese and Tall independent?

7. 肥胖与高这两个事件是否独立？

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3.5 Tree and Venn Diagrams 3.5 树形图与韦恩图

Sometimes, when the probability problems are complex, it can be helpful to graph the situation. Tree diagrams and Venn diagrams are two tools that can be used to visualize and solve conditional probabilities.

有时，当概率问题比较复杂时，把情况用图形表示会很有帮助。树形图和韦恩图是两种可用于直观呈现并求解条件概率的工具。

Tree Diagrams 树形图

A tree diagram is a special type of graph used to determine the outcomes of an experiment. It consists of "branches" that are labeled with either frequencies or probabilities. Tree diagrams can make some probability problems easier to visualize and solve. The following example illustrates how to use a tree diagram.

树形图是一种特殊类型的图形，用于确定试验的结果。它由标有频数或概率的"分支"组成。树形图可以使某些概率问题更易于直观呈现和求解。下面的示例说明了如何使用树形图。

In an urn, there are 11 balls. Three balls are red (*R*) and eight balls are blue (*B*). Draw two balls, one at a time, with replacement. "With replacement" means that you put the first ball back in the urn before you select the second ball. The tree diagram using frequencies that show all the possible outcomes follows.

一个瓮中有 11 个球。其中 3 个红球（*R*），8 个蓝球（*B*）。一次抽取一个球，共抽两次，有放回。"有放回"是指你在抽取第二个球之前，把第一个球放回瓮中。下面是用频数表示所有可能结果的树形图。

The first set of branches represents the first draw. The second set of branches represents the second draw. Each of the outcomes is distinct. In fact, we can list each red ball as *R*1, *R*2, and *R*3 and each blue ball as *B*1, *B*2, *B*3, *B*4, *B*5, *B*6, *B*7, and *B*8. Then the nine *RR* outcomes can be written as:

第一组分支表示第一次抽取。第二组分支表示第二次抽取。每个结果都是不同的。事实上，我们可以把每个红球记为 *R*1、*R*2、*R*3，每个蓝球记为 *B*1、*B*2、*B*3、*B*4、*B*5、*B*6、*B*7、*B*8。那么九个 *RR* 结果可以写成：

*R*1*R*1; *R*1*R*2; *R*1*R*3; *R*2*R*1; *R*2*R*2; *R*2*R*3; *R*3*R*1; *R*3*R*2; *R*3*R*3

The other outcomes are similar.

其余结果类似。

There are a total of 11 balls in the urn. Draw two balls, one at a time, with replacement. There are 11(11) = 121 outcomes, the size of the sample space.

瓮中共有 11 个球。一次抽一个，有放回地抽两次。共有 11(11) = 121 种结果，即样本空间的大小。

Problem 问题

a\. List the 24 *BR* outcomes: *B*1*R*1, *B*1*R*2, *B*1*R*3, ...

a. 列出 24 个 *BR* 结果：*B*1*R*1, *B*1*R*2, *B*1*R*3, ...

b\. Using the tree diagram, calculate *P*(*RR*).

b. 用树形图计算 *P*(*RR*)。

c\. Using the tree diagram, calculate *P*(*RB* OR *BR*).

c. 用树形图计算 *P*(*RB* OR *BR*)。

d\. Using the tree diagram, calculate *P*(*R* on 1st draw AND *B* on 2nd draw).

d. 用树形图计算 *P*(*R* 在第一次抽取 AND *B* 在第二次抽取)。

e\. Using the tree diagram, calculate *P*(*R* on 2nd draw GIVEN *B* on 1st draw).

e. 用树形图计算 *P*(*R* 在第二次抽取 GIVEN *B* 在第一次抽取)。

f\. Using the tree diagram, calculate *P*(*BB*).

f. 用树形图计算 *P*(*BB*)。

g\. Using the tree diagram, calculate *P*(*B* on the 2nd draw given *R* on the first draw).

g. 用树形图计算 *P*(*B* 在第二次抽取 given *R* 在第一次抽取)。

Solution 解答

a\. *B*1*R*1; *B*1*R*2; *B*1*R*3; *B*2*R*1; *B*2*R*2; *B*2*R*3; *B*3*R*1; *B*3*R*2; *B*3*R*3; *B*4*R*1; *B*4*R*2; *B*4*R*3; *B*5*R*1; *B*5*R*2; *B*5*R*3; *B*6*R*1; *B*6*R*2; *B*6*R*3; *B*7*R*1; *B*7*R*2; *B*7*R*3; *B*8*R*1; *B*8*R*2; *B*8*R*3

a. *B*1*R*1; *B*1*R*2; *B*1*R*3; *B*2*R*1; *B*2*R*2; *B*2*R*3; *B*3*R*1; *B*3*R*2; *B*3*R*3; *B*4*R*1; *B*4*R*2; *B*4*R*3; *B*5*R*1; *B*5*R*2; *B*5*R*3; *B*6*R*1; *B*6*R*2; *B*6*R*3; *B*7*R*1; *B*7*R*2; *B*7*R*3; *B*8*R*1; *B*8*R*2; *B*8*R*3

b\. *P*(*RR*) = $\left( \frac{3}{11} \right)\left( \frac{3}{11} \right)$ = $\frac{9}{121}$

b. *P*(*RR*) = $\left( \frac{3}{11} \right)\left( \frac{3}{11} \right)$ = $\frac{9}{121}$

c\. *P*(*RB* OR *BR*) = $\left( \frac{3}{11} \right)\left( \frac{8}{11} \right)$ + $\left( \frac{8}{11} \right)\left( \frac{3}{11} \right)$ = $\frac{48}{121}$

c. *P*(*RB* OR *BR*) = $\left( \frac{3}{11} \right)\left( \frac{8}{11} \right)$ + $\left( \frac{8}{11} \right)\left( \frac{3}{11} \right)$ = $\frac{48}{121}$

d\. *P*(*R* on 1st draw AND *B* on 2nd draw) = *P*(*RB*) = $\left( \frac{3}{11} \right)\left( \frac{8}{11} \right)$ = $\frac{24}{121}$

d. *P*(*R* 在第一次抽取 AND *B* 在第二次抽取) = *P*(*RB*) = $\left( \frac{3}{11} \right)\left( \frac{8}{11} \right)$ = $\frac{24}{121}$

e\. *P*(*R* on 2nd draw GIVEN *B* on 1st draw) = *P*(*R* on 2nd\|*B* on 1st) = $\frac{24}{88}$ = $\frac{3}{11}$

e. *P*(*R* 在第二次抽取 GIVEN *B* 在第一次抽取) = *P*(*R* 第二次\|*B* 第一次) = $\frac{24}{88}$ = $\frac{3}{11}$

This problem is a conditional one. The sample space has been reduced to those outcomes that already have a blue on the first draw. There are 24 + 64 = 88 possible outcomes (24 *BR* and 64 *BB*). Twenty-four of the 88 possible outcomes are *BR*. $\frac{24}{88}$ = $\frac{3}{11}$.

这是一个条件概率问题。样本空间已缩减为那些第一次抽到蓝球的结果。共有 24 + 64 = 88 种可能结果（24 个 *BR* 和 64 个 *BB*）。88 种结果中有 24 种是 *BR*。$\frac{24}{88}$ = $\frac{3}{11}$。

f\. *P*(*BB*) = $\frac{64}{121}$

f. *P*(*BB*) = $\frac{64}{121}$

g\. *P*(*B* on 2nd draw\|*R* on 1st draw) = $\frac{8}{11}$

g. *P*(*B* 在第二次抽取\|*R* 在第一次抽取) = $\frac{8}{11}$

There are 9 + 24 outcomes that have *R* on the first draw (9 *RR* and 24 *RB*). The sample space is then 9 + 24 = 33. 24 of the 33 outcomes have *B* on the second draw. The probability is then $\frac{24}{33}$.

第一次抽到 *R* 的结果共有 9 + 24 种（9 个 *RR* 和 24 个 *RB*）。因此样本空间为 9 + 24 = 33。这 33 种结果中有 24 种第二次抽到 *B*。于是概率为 $\frac{24}{33}$。

In a standard deck, there are 52 cards. 12 cards are face cards (event *F*) and 40 cards are not face cards (event *N*). Draw two cards, one at a time, with replacement. All possible outcomes are shown in the tree diagram as frequencies. Using the tree diagram, calculate *P*(*FF*).

在一副标准扑克牌中，有 52 张牌。其中 12 张是花牌（事件 *F*），40 张不是花牌（事件 *N*）。一次抽一张，有放回地抽两次。树形图用频数表示了所有可能结果。用树形图计算 *P*(*FF*)。

An urn has three red marbles and eight blue marbles in it. Draw two marbles, one at a time, this time without replacement, from the urn. "Without replacement" means that you do not put the first ball back before you select the second marble. Following is a tree diagram for this situation. The branches are labeled with probabilities instead of frequencies. The numbers at the ends of the branches are calculated by multiplying the numbers on the two corresponding branches, for example, $\left( \frac{3}{11} \right)\left( \frac{2}{10} \right) = \frac{6}{110}$.

一个瓮中有 3 个红球和 8 个蓝球。这次从瓮中一次取一个，无放回地取两次。"无放回"是指你在取第二个球之前不把第一个球放回。下面是这种情况的树形图。分支上标的不是频数而是概率。分支末端的数由对应两条分支上的数相乘得到，例如 $\left( \frac{3}{11} \right)\left( \frac{2}{10} \right) = \frac{6}{110}$。

If you draw a red on the first draw from the three red possibilities, there are two red marbles left to draw on the second draw. You do not put back or replace the first marble after you have drawn it. You draw without replacement, so that on the second draw there are ten marbles left in the urn.

如果你第一次从三个红球中抽到一个红球，那么第二次抽取时还剩两个红球可抽。你抽完第一个球后不把它放回或替换。你进行的是无放回抽取，因此第二次抽取时瓮中还剩 10 个球。

Calculate the following probabilities using the tree diagram.

用树形图计算下列概率。

Problem 问题

a\. *P*(*RR*) = \_\_\_\_\_\_\_\_

a. *P*(*RR*) = \_\_\_\_\_\_\_\_

b\. Fill in the blanks:

b. 填空：

*P*(*RB* OR *BR*) = $\left( \frac{3}{11} \right)\left( \frac{8}{10} \right)\ + \text{~(\_\_\_)(\_\_\_)~} = \ \frac{48}{110}$

c\. *P*(*R* on 2nd\|*B* on 1st) =

c. *P*(*R* 在第二次抽取\|*B* 在第一次抽取) =

d\. Fill in the blanks.

d. 填空。

*P*(*R* on 1st AND *B* on 2nd) = *P*(*RB*) = (\_\_\_)(\_\_\_) = $\frac{24}{110}$

*P*(*R* 在第一次抽取 AND *B* 在第二次抽取) = *P*(*RB*) = (\_\_\_)(\_\_\_) = $\frac{24}{110}$

e\. Find *P*(*BB*).

e. 求 *P*(*BB*)。

f\. Find *P*(*B* on 2nd\|*R* on 1st).

f. 求 *P*(*B* 在第二次抽取\|*R* 在第一次抽取)。

Solution 解答

a\. *P*(*RR*) = $\left( \frac{3}{11} \right)\left( \frac{2}{10} \right) = \frac{6}{110}$

a. *P*(*RR*) = $\left( \frac{3}{11} \right)\left( \frac{2}{10} \right) = \frac{6}{110}$

b\. *P*(*RB* OR *BR*) = $\left( \frac{3}{11} \right)\left( \frac{8}{10} \right)$ + $\left( \frac{8}{11} \right)\left( \frac{3}{10} \right)$ = $\frac{48}{110}$

b. *P*(*RB* OR *BR*) = $\left( \frac{3}{11} \right)\left( \frac{8}{10} \right)$ + $\left( \frac{8}{11} \right)\left( \frac{3}{10} \right)$ = $\frac{48}{110}$

c\. *P*(*R* on 2nd\|*B* on 1st) = $\frac{3}{10}$

c. *P*(*R* 在第二次抽取\|*B* 在第一次抽取) = $\frac{3}{10}$

d\. *P*(*R* on 1st AND *B* on 2nd) = *P*(*RB*) = $\left( \frac{3}{11} \right)\left( \frac{8}{10} \right)$ = $\frac{24}{110}$

d. *P*(*R* 在第一次抽取 AND *B* 在第二次抽取) = *P*(*RB*) = $\left( \frac{3}{11} \right)\left( \frac{8}{10} \right)$ = $\frac{24}{110}$

e\. *P*(*BB*) = $\left( \frac{8}{11} \right)\left( \frac{7}{10} \right)$

e. *P*(*BB*) = $\left( \frac{8}{11} \right)\left( \frac{7}{10} \right)$

f\. Using the tree diagram, *P*(*B* on 2nd\|*R* on 1st) = *P*(*R*\|*B*) = $\frac{8}{10}$.

f. 用树形图，*P*(*B* 在第二次抽取\|*R* 在第一次抽取) = *P*(*R*\|*B*) = $\frac{8}{10}$。

If we are using probabilities, we can label the tree in the following general way.

如果我们使用概率，可以用下面的一般方式来标注这棵树。

*P*(*R*\|*R*) here means *P*(*R* on 2nd\|*R* on 1st)
*P*(*B*\|*R*) here means *P*(*B* on 2nd\|*R* on 1st)
*P*(*R*\|*B*) here means *P*(*R* on 2nd\|*B* on 1st)
*P*(*B*\|*B*) here means *P*(*B* on 2nd\|*B* on 1st)

*P*(*R*\|*R*) 这里表示 *P*(*R* 在第二次抽取\|*R* 在第一次抽取)
*P*(*B*\|*R*) 这里表示 *P*(*B* 在第二次抽取\|*R* 在第一次抽取)
*P*(*R*\|*B*) 这里表示 *P*(*R* 在第二次抽取\|*B* 在第一次抽取)
*P*(*B*\|*B*) 这里表示 *P*(*B* 在第二次抽取\|*B* 在第一次抽取)

In a standard deck, there are 52 cards. Twelve cards are face cards (*F*) and 40 cards are not face cards (*N*). Draw two cards, one at a time, without replacement. The tree diagram is labeled with all possible probabilities.

在一副标准扑克牌中，有 52 张牌。其中 12 张是花牌（*F*），40 张不是花牌（*N*）。一次抽一张，无放回地抽两次。树形图上标出了所有可能的概率。

1. Find *P*(*FN* OR *NF*).

1. 求 *P*(*FN* OR *NF*)。

2. Find *P*(*N*\|*F*).

2. 求 *P*(*N*\|*F*)。

3. Find *P*(at most one face card).

3. 求 *P*(至多一张花牌)。

Hint: "At most one face card" means zero or one face card.

提示："至多一张花牌"指零张或一张花牌。

4. Find *P*(at least one face card).

4. 求 *P*(至少一张花牌)。

Hint: "At least one face card" means one or two face cards.

提示："至少一张花牌"指一张或两张花牌。

A litter of kittens available for adoption at the Humane Society has four tabby kittens and five black kittens. A family comes in and randomly selects two kittens (without replacement) for adoption.

动物保护协会有一窝可供领养的小猫，其中 4 只是虎斑猫，5 只是黑猫。一个家庭前来随机挑选两只小猫（无放回）领养。

Problem 问题

1. What is the probability that both kittens are tabby?

1. 两只小猫都是虎斑猫的概率是多少？

a.$\left( \frac{1}{2} \right)\left( \frac{1}{2} \right)$ b.$\left( \frac{4}{9} \right)\left( \frac{4}{9} \right)$ c.$\left( \frac{4}{9} \right)\left( \frac{3}{8} \right)$ d.$\left( \frac{4}{9} \right)\left( \frac{5}{9} \right)$

2. What is the probability that one kitten of each coloring is selected?

2. 选到一只各为不同颜色的小猫的概率是多少？

a.$\left( \frac{4}{9} \right)\left( \frac{5}{9} \right)$ b.$\left( \frac{4}{9} \right)\left( \frac{5}{8} \right)$ c.$\left( \frac{4}{9} \right)\left( \frac{5}{9} \right) + \left( \frac{5}{9} \right)\left( \frac{4}{9} \right)$ d.$\left( \frac{4}{9} \right)\left( \frac{5}{8} \right) + \left( \frac{5}{9} \right)\left( \frac{4}{8} \right)$

3. What is the probability that a tabby is chosen as the second kitten when a black kitten was chosen as the first?

3. 若第一只选中的是黑色小猫，那么第二只选中的是虎斑猫的概率是多少？

4. What is the probability of choosing two kittens of the same color?

4. 选到两只同色小猫的概率是多少？

Solution 解答

a\. c, b. d, c. $\frac{4}{8}$, d. $\frac{32}{72}$

a. c，b. d，c. $\frac{4}{8}$，d. $\frac{32}{72}$

Suppose there are four red balls and three yellow balls in a box. Two balls are drawn from the box without replacement. What is the probability that one ball of each coloring is selected?

假设盒中有 4 个红球和 3 个黄球。从盒中无放回地取出两个球。问取到一个红球和一个黄球的概率是多少？

Venn Diagram 韦恩图

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space S together with circles or ovals. The circles or ovals represent events.

韦恩图是一种表示试验结果的图形。它通常由一个代表样本空间 S 的矩形，连同若干圆形或椭圆形组成。这些圆形或椭圆形代表事件。

Suppose an experiment has the outcomes 1, 2, 3, ... , 12 where each outcome has an equal chance of occurring. Let event *A* = {1, 2, 3, 4, 5, 6} and event *B* = {6, 7, 8, 9}. Then *A* AND *B* = {6} and *A* OR *B* = {1, 2, 3, 4, 5, 6, 7, 8, 9}. The Venn diagram is as follows:

假设一次试验的结果为 1, 2, 3, … , 12，且每个结果出现的可能性相等。令事件 *A* = {1, 2, 3, 4, 5, 6}，事件 *B* = {6, 7, 8, 9}。则 *A* AND *B* = {6}，*A* OR *B* = {1, 2, 3, 4, 5, 6, 7, 8, 9}。韦恩图如下：

Suppose an experiment has outcomes black, white, red, orange, yellow, green, blue, and purple, where each outcome has an equal chance of occurring. Let event *C* = {green, blue, purple} and event *P* = {red, yellow, blue}. Then *C* AND *P* = {blue} and *C* OR *P* = {green, blue, purple, red, yellow}. Draw a Venn diagram representing this situation.

假设一次试验的结果为黑、白、红、橙、黄、绿、蓝、紫，且每个结果出现的可能性相等。令事件 *C* = {绿, 蓝, 紫}，事件 *P* = {红, 黄, 蓝}。则 *C* AND *P* = {蓝}，*C* OR *P* = {绿, 蓝, 紫, 红, 黄}。画出表示这种情形的韦恩图。

Flip two fair coins. Let *A* = tails on the first coin. Let *B* = tails on the second coin. Then *A* = {*TT*, *TH*} and *B* = {*TT*, *HT*}. Therefore, *A* AND *B* = {*TT*}. *A* OR *B* = {*TH*, *TT*, *HT*}.

抛两枚均匀的硬币。令 *A* = 第一枚硬币反面朝上。令 *B* = 第二枚硬币反面朝上。则 *A* = {*TT*, *TH*}，*B* = {*TT*, *HT*}。因此 *A* AND *B* = {*TT*}，*A* OR *B* = {*TH*, *TT*, *HT*}。

The sample space when you flip two fair coins is *X* = {*HH*, *HT*, *TH*, *TT*}. The outcome *HH* is in NEITHER *A* NOR *B*. The Venn diagram is as follows:

抛两枚均匀硬币时的样本空间为 *X* = {*HH*, *HT*, *TH*, *TT*}。结果 *HH* 既不在 *A* 中也不在 *B* 中。韦恩图如下：

Roll a fair, six-sided die. Let *A* = a prime number of dots is rolled. Let *B* = an odd number of dots is rolled. Then *A* = {2, 3, 5} and *B* = {1, 3, 5}. Therefore, *A* AND *B* = {3, 5}. *A* OR *B* = {1, 2, 3, 5}. The sample space for rolling a fair die is *S* = {1, 2, 3, 4, 5, 6}. Draw a Venn diagram representing this situation.

掷一枚均匀的六面骰子。令 *A* = 掷出质数个点。令 *B* = 掷出奇数个点。则 *A* = {2, 3, 5}，*B* = {1, 3, 5}。因此 *A* AND *B* = {3, 5}，*A* OR *B* = {1, 2, 3, 5}。掷一枚均匀骰子的样本空间为 *S* = {1, 2, 3, 4, 5, 6}。画出表示这种情形的韦恩图。

Forty percent of the students at a local college belong to a club and 50% work part time. Five percent of the students work part time and belong to a club. Draw a Venn diagram showing the relationships. Let *C* = student belongs to a club and *PT* = student works part time.

本地一所高校中，40% 的学生属于某个俱乐部，50% 的学生兼职工作。5% 的学生既兼职工作又属于某个俱乐部。画出表示这些关系的韦恩图。令 *C* = 学生属于某个俱乐部，*PT* = 学生兼职工作。

If a student is selected at random, find

若随机抽取一名学生，求

the probability that the student belongs to a club. *P*(*C*) = 0.40
the probability that the student works part time. *P*(*PT*) = 0.50
the probability that the student belongs to a club AND works part time. *P*(*C* AND *PT*) = 0.05
the probability that the student belongs to a club given that the student works part time. $P\text{(}C\text{|}PT\text{)}~ = ~\frac{P\text{(}C\text{~AND~}PT\text{)}}{P\text{(}PT\text{)}}~ = ~\frac{0.05}{0.50}~ = ~0.1$
the probability that the student belongs to a club OR works part time. *P*(*C* OR *PT*) = *P*(*C*) + *P*(*PT*) - *P*(*C* AND *PT*) = 0.40 + 0.50 - 0.05 = 0.85

该学生属于某个俱乐部的概率。*P*(*C*) = 0.40
该学生兼职工作的概率。*P*(*PT*) = 0.50
该学生既属于俱乐部又兼职工作的概率。*P*(*C* AND *PT*) = 0.05
已知该学生兼职工作，其属于某个俱乐部（条件）的概率。 $P\text{(}C\text{|}PT\text{)}~ = ~\frac{P\text{(}C\text{~AND~}PT\text{)}}{P\text{(}PT\text{)}}~ = ~\frac{0.05}{0.50}~ = ~0.1$
该学生属于俱乐部或兼职工作的概率。*P*(*C* OR *PT*) = *P*(*C*) + *P*(*PT*) - *P*(*C* AND *PT*) = 0.40 + 0.50 - 0.05 = 0.85

Fifty percent of the workers at a factory work a second job, 25% have a spouse who also works, 5% work a second job and have a spouse who also works. Draw a Venn diagram showing the relationships. Let *W* = works a second job and *S* = spouse also works.

某工厂 50% 的工人从事第二份工作，25% 的工人其配偶也有工作，5% 的工人既从事第二份工作其配偶也有工作。画出表示这些关系的韦恩图。令 *W* = 从事第二份工作，*S* = 配偶也有工作。

Problem 问题

A person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any blood type. Four percent of African Americans have type O blood and a negative RH factor, 5−10% of African Americans have the Rh- factor, and 51% have type O blood.

血型为 O 型且 Rh 因子阴性（Rh-）的人可以向任何血型的人献血。4% 的非裔美国人具有 O 型血且 Rh 因子阴性；5%—10% 的非裔美国人具有 Rh- 因子；51% 具有 O 型血。

The “O” circle represents the African Americans with type O blood. The “Rh-“ oval represents the African Americans with the Rh- factor.

“O” 圆代表具有 O 型血的非裔美国人。“Rh-” 椭圆代表具有 Rh- 因子的非裔美国人。

We will take the average of 5% and 10% and use 7.5% as the percent of African Americans who have the Rh- factor. Let *O* = African American with Type O blood and *R* = African American with Rh- factor.

我们取 5% 与 10% 的平均数，用 7.5% 作为具有 Rh- 因子的非裔美国人所占的百分比。令 *O* = 具有 O 型血的非裔美国人，*R* = 具有 Rh- 因子的非裔美国人。

1. *P*(*O*) = \_\_\_\_\_\_\_\_\_\_\_

2. *P*(*R*) = \_\_\_\_\_\_\_\_\_\_\_

3. *P*(*O* AND *R*) = \_\_\_\_\_\_\_\_\_\_\_

4. *P*(*O* OR *R*) = \_\_\_\_\_\_\_\_\_\_\_\_

5. In the Venn Diagram, describe the overlapping area using a complete sentence.

5. 用一句完整的话描述韦恩图中重叠的部分。

6. In the Venn Diagram, describe the area in the rectangle but outside both the circle and the oval using a complete sentence.

6. 用一句完整的话描述矩形内、但同时在圆和椭圆之外的区域。

Solution 解答

a\. 0.51; b. 0.075; c. 0.04; d. 0.545; e. The area represents the African Americans that have type O blood and the Rh- factor. f. The area represents the African Americans that have neither type O blood nor the Rh- factor.

a. 0.51；b. 0.075；c. 0.04；d. 0.545；e. 该区域代表具有 O 型血且兼具 Rh- 因子的非裔美国人。f. 该区域代表既不具有 O 型血也不具有 Rh- 因子的非裔美国人。

In a bookstore, the probability that the customer buys a novel is 0.6, and the probability that the customer buys a non-fiction book is 0.4. Suppose that the probability that the customer buys both is 0.2.

在一家书店里，顾客购买小说的概率是 0.6，购买非小说类图书的概率是 0.4。假设顾客两样都买的概率是 0.2。

1. Draw a Venn diagram representing the situation.

1. 画出表示这种情形的韦恩图。

2. Find the probability that the customer buys either a novel or a non-fiction book.

2. 求顾客购买小说或非小说类图书（二者之一）的概率。

3. In the Venn diagram, describe the overlapping area using a complete sentence.

3. 用一句完整的话描述韦恩图中重叠的部分。

4. Suppose that some customers buy only compact disks. Draw an oval in your Venn diagram representing this event.

4. 假设有些顾客只购买压缩光盘。在你的韦恩图中画一个椭圆来表示这一事件。

3.6 Probability Topics 3.6 概率主题

Probability Topics 概率主题

Class time:

上课时间：

Names:

姓名：

Student Learning Outcomes

学生学习目标

The student will use theoretical and empirical methods to estimate probabilities.
The student will appraise the differences between the two estimates.
The student will demonstrate an understanding of long-term relative frequencies.

学生将运用理论方法和经验方法估计概率。
学生将评估两种估计之间的差异。
学生将展示对长期相对频率的理解。

Do the Experiment Count out 40 mixed-color M&Ms® which is approximately one small bag’s worth. Record the number of each color in Table 3.11. Use the information from this table to complete Table 3.12. Next, put the M&Ms in a cup. The experiment is to pick two M&Ms, one at a time. Do not look at them as you pick them. The first time through, replace the first M&M before picking the second one. Record the results in the “With Replacement” column of Table 3.13. Do this 24 times. The second time through, after picking the first M&M, do not replace it before picking the second one. Then, pick the second one. Record the results in the “Without Replacement” column section of Table 3.14. After you record the pick, put both M&Ms back. Do this a total of 24 times, also. Use the data from Table 3.14 to calculate the empirical probability questions. Leave your answers in unreduced fractional form. Do not multiply out any fractions.

做实验数出 40 颗混色 M&Ms® 巧克力豆，大约相当于一小袋的量。把每种颜色的数量记录在表 3.11 中。利用该表的信息完成表 3.12。接着把 M&Ms 放入杯中。试验为一次取两颗 M&Ms，每次取一颗。取的时候不要看它们。第一轮，在取第二颗之前把第一颗放回。把结果记录在表 3.13 的“有放回”一栏中。重复 24 次。第二轮，取完第一颗后，在取第二颗之前不要放回。然后取第二颗。把结果记录在表 3.14 的“无放回”一栏中。记录完所取的结果后，把两颗 M&Ms 都放回去。这一轮也总共重复 24 次。利用表 3.14 的数据计算经验概率问题。答案保留未约分的分数形式。不要把任何分数乘开。

| Color | Quantity |

| 颜色 | 数量 |

|--------------|----------|

| Yellow (*Y*) | |

| 黄色 (*Y*) | |

| Green (*G*) | |

| 绿色 (*G*) | |

| Blue (*BL*) | |

| 蓝色 (*BL*) | |

| Brown (*B*) | |

| 棕色 (*B*) | |

| Orange (*O*) | |

| 橙色 (*O*) | |

| Red (*R*) | |

| 红色 (*R*) | |

Table 3.11 Population

表 3.11 总体

| | With Replacement | Without Replacement |

| | 有放回 | 无放回 |

|-----------------------------------------------------------------------|------------------|---------------------|

| *P*(2 reds) | | |

| *P*(2 红) | | |

| *P*(*R*₁*B*₂ OR *B*₁*R*₂) | | |

| *P*(*R*₁ AND *G*₂) | | |

| *P*(*G*₂\|*R*₁) | | |

| *P*(no yellows) | | |

| *P*(没有黄色) | | |

| *P*(doubles) | | |

| *P*(同色对) | | |

| *P*(no doubles) | | |

| *P*(非同色对) | | |

Table 3.12 Theoretical Probabilities

表 3.12 理论概率

*G*₂ = green on second pick; *R*₁ = red on first pick; *B*₁ = brown on first pick; *B*₂ = brown on second pick; doubles = both picks are the same colour.

*G*₂ = 第二次取到绿色；*R*₁ = 第一次取到红色；*B*₁ = 第一次取到棕色；*B*₂ = 第二次取到棕色；doubles = 两次取到的颜色相同。

| With Replacement | Without Replacement |

| 有放回 | 无放回 |

|---------------------------------|---------------------------------|

| ( \_\_ , \_\_ ) ( \_\_ , \_\_ ) | ( \_\_ , \_\_ ) ( \_\_ , \_\_ ) |

Table 3.13 Empirical Results

表 3.13 经验结果

| | With Replacement | Without Replacement |

| | 有放回 | 无放回 |

|-----------------------------------------------------------------------|------------------|---------------------|

| *P*(2 reds) | | |

| *P*(2 红) | | |

| *P*(*R*₁*B*₂ OR *B*₁*R*₂) | | |

| *P*(*R*₁ AND *G*₂) | | |

| *P*(*G*₂\|*R*₁) | | |

| *P*(no yellows) | | |

| *P*(没有黄色) | | |

| *P*(doubles) | | |

| *P*(同色对) | | |

| *P*(no doubles) | | |

| *P*(非同色对) | | |

Table 3.14 Empirical Probabilities

表 3.14 经验概率

Discussion Questions

讨论问题

1. Why are the “With Replacement” and “Without Replacement” probabilities different?

1. 为什么“有放回”与“无放回”的概率不同？

2. Convert *P*(no yellows) to decimal format for both Theoretical “With Replacement” and for Empirical “With Replacement”. Round to four decimal places.

2. 把 *P*(没有黄色) 分别换算成理论“有放回”与经验“有放回”的小数形式，保留四位小数。

1. Theoretical “With Replacement”: *P*(no yellows) = \_\_\_\_\_\_\_

1. 理论“有放回”：*P*(没有黄色) = \_\_\_\_\_\_\_

2. Empirical “With Replacement”: *P*(no yellows) = \_\_\_\_\_\_\_

2. 经验“有放回”：*P*(没有黄色) = \_\_\_\_\_\_\_

3. Are the decimal values “close”? Did you expect them to be closer together or farther apart? Why?

3. 这些小数值“接近”吗？你原先预期它们是更接近还是更疏远？为什么？

3. If you increased the number of times you picked two M&Ms to 240 times, why would empirical probability values change?

3. 如果你把取两颗 M&Ms 的次数增加到 240 次，为什么经验概率值会变化？

4. Would this change (see part 3) cause the empirical probabilities and theoretical probabilities to be closer together or farther apart? How do you know?

4. 这一变化（见第 3 问）会使经验概率与理论概率变得更接近还是更疏远？你是如何判断的？

5. Explain the differences in what *P*(*G*₁ AND *R*₂) and *P*(*R*₁\|*G*₂) represent. Hint: Think about the sample space for each probability.

5. 解释 *P*(*G*₁ AND *R*₂) 与 *P*(*R*₁\|*G*₂) 所表示含义的不同。提示：想一想每种概率对应的样本空间。

Key Terms 关键术语

Conditional Probability

条件概率

the likelihood that an event will occur given that another event has already occurred

在另一事件已经发生的条件下，某事件发生的可能性。

contingency table

列联表

the method of displaying a frequency distribution as a table with rows and columns to show how two variables may be dependent (contingent) upon each other; the table provides an easy way to calculate conditional probabilities.

将频数分布以带行与列的表格形式展示的方法，用以显示两个变量如何相互依存（随条件变化）；该表提供了一种便捷的方式来计算条件概率。

Dependent Events

相依事件（非独立事件）

If two events are NOT independent, then we say that they are dependent.

若两个事件不独立，则称它们为相依事件。

Equally Likely

等可能

Each outcome of an experiment has the same probability.

试验中每个结果具有相同的概率。

Event

事件

a subset of the set of all outcomes of an experiment; the set of all outcomes of an experiment is called a sample space and is usually denoted by *S*. An event is an arbitrary subset in *S*. It can contain one outcome, two outcomes, no outcomes (empty subset), the entire sample space, and the like. Standard notations for events are capital letters such as *A*, *B*, *C*, and so on.

试验全部结果集合的一个子集；试验的全部结果集合称为样本空间，通常记作 *S*。事件是 *S* 中的任意子集，它可以包含一个结果、两个结果、不含任何结果（空集）、整个样本空间等等。事件的常用记号是 *A*、*B*、*C* 等大写字母。

Experiment

试验

a planned activity carried out under controlled conditions

在受控条件下进行的有计划的活动。

Independent Events

独立事件

The occurrence of one event has no effect on the probability of the occurrence of another event. Events *A* and *B* are independent if one of the following is true:

一个事件的发生不影响另一事件发生的概率。若下列条件之一成立，则事件 *A* 与 *B* 相互独立：

1. *P*(*A*\|*B*) = *P*(*A*)

2. *P*(*B*\|*A*) = *P*(*B*)

3. *P*(*A* AND *B*) = *P*(*A*)*P*(*B*)

Mutually Exclusive

互斥事件

Two events are mutually exclusive if the probability that they both happen at the same time is zero. If events *A* and *B* are mutually exclusive, then *P*(*A* AND *B*) = 0.

若两个事件同时发生的概率为零，则它们互斥。若事件 *A* 与 *B* 互斥，则 *P*(*A* AND *B*) = 0。

Outcome

结果

a particular result of an experiment

试验的一个特定结果。

Probability

概率

a number between zero and one, inclusive, that gives the likelihood that a specific event will occur; the foundation of statistics is given by the following 3 axioms (by A.N. Kolmogorov, 1930’s): Let *S* denote the sample space and *A* and *B* are two events in *S*. Then:

一个介于零与一之间（含端点）的数，表示某特定事件发生的可能性；统计学的基石是如下三条公理（A.N. Kolmogorov，1930 年代）：设 *S* 表示样本空间，*A* 与 *B* 为 *S* 中的两个事件。则：

0 ≤ *P*(*A*) ≤ 1
If *A* and *B* are any two mutually exclusive events, then *P*(*A* OR *B*) = *P*(*A*) + *P*(B).
*P*(*S*) = 1

0 ≤ *P*(*A*) ≤ 1
若 *A* 与 *B* 为任意两个互斥事件，则 *P*(*A* OR *B*) = *P*(*A*) + *P*(B)。
*P*(*S*) = 1

Sample Space

样本空间

the set of all possible outcomes of an experiment

试验所有可能结果的集合。

Sampling with Replacement

有放回抽样

If each member of a population is replaced after it is picked, then that member has the possibility of being chosen more than once.

若总体中每个个体被抽取后都放回，则该个体有可能被重复选中。

Sampling without Replacement

无放回抽样

When sampling is done without replacement, each member of a population may be chosen only once.

无放回抽样时，总体中每个个体至多只能被选中一次。

The AND Event

交事件（AND 事件）

An outcome is in the event *A* AND *B* if the outcome is in both *A* AND *B* at the same time.

若一个结果同时属于 *A* 与 *B*，则该结果属于事件 *A* AND *B*。

The Complement Event

补事件（余事件，A′）

The complement of event *A* consists of all outcomes that are NOT in *A*.

事件 *A* 的补事件由所有不属于 *A* 的结果组成。

The Conditional Probability of *A* GIVEN *B*

*A* 给定 *B* 的条件概率

*P*(*A*\|*B*) is the probability that event *A* will occur given that the event *B* has already occurred.

*P*(*A*\|*B*) 是在事件 *B* 已经发生的条件下事件 *A* 发生的概率。

The Conditional Probability of One Event Given Another Event

一个事件给定另一个事件的条件概率

*P*(*A*\|*B*) is the probability that event *A* will occur given that the event *B* has already occurred.

*P*(*A*\|*B*) 是在事件 *B* 已经发生的条件下事件 *A* 发生的概率。

The Or Event

并事件（OR 事件）

An outcome is in the event *A* OR *B* if the outcome is in *A* or is in *B* or is in both *A* and *B*.

若一个结果属于 *A*，或属于 *B*，或同时属于二者，则该结果属于事件 *A* OR *B*。

The OR of Two Events

两个事件的并

An outcome is in the event *A* OR *B* if the outcome is in *A*, is in *B*, or is in both *A* and *B*.

若一个结果属于 *A*、属于 *B*，或同时属于二者，则该结果属于事件 *A* OR *B*。

Tree Diagram

树形图

the useful visual representation of a sample space and events in the form of a “tree” with branches marked by possible outcomes together with associated probabilities (frequencies, relative frequencies)

一种有用的可视化表示，以"树"的形式展示样本空间与事件，其分支标注可能的结果以及相应的概率（频数、相对频数）。

Venn Diagram

韦恩图

the visual representation of a sample space and events in the form of circles or ovals showing their intersections

以圆形或椭圆形表示样本空间与事件、并展示其交集的可视化表示。

Chapter Review 本章回顾

3.1 Terminology 3.1 术语

In this module we learned the basic terminology of probability. The set of all possible outcomes of an experiment is called the sample space. Events are subsets of the sample space, and they are assigned a probability that is a number between zero and one, inclusive.

本模块我们学习了概率的基本术语。试验所有可能结果的集合称为样本空间。事件是样本空间的子集，并被赋予一个介于零与一之间（含端点）的概率。

3.2 Independent and Mutually Exclusive Events 3.2 独立事件与互斥事件

Two events *A* and *B* are independent if the knowledge that one occurred does not affect the chance the other occurs. If two events are not independent, then we say that they are dependent.

若已知其中一个事件发生并不影响另一事件发生的机会，则事件 *A* 与 *B* 相互独立。若两个事件不独立，则称它们为相依事件。

In sampling with replacement, each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered not to be independent. When events do not share outcomes, they are mutually exclusive of each other.

有放回抽样时，总体中每个个体被抽取后都会放回，因而该个体有可能被重复选中，这些事件被认为是独立的。无放回抽样时，总体中每个个体至多被选中一次，这些事件被认为是非独立的。当事件不共享任何结果时，它们彼此互斥。

3.3 Two Basic Rules of Probability 3.3 概率的两条基本法则

The multiplication rule and the addition rule are used for computing the probability of *A* and *B*, as well as the probability of *A* or *B* for two given events *A*, *B* defined on the sample space. In sampling with replacement each member of a population is replaced after it is picked, so that member has the possibility of being chosen more than once, and the events are considered to be independent. In sampling without replacement, each member of a population may be chosen only once, and the events are considered to be not independent. The events *A* and *B* are mutually exclusive events when they do not have any outcomes in common.

乘法法则与加法法则用于计算 *A* 与 *B* 的概率，以及定义在样本空间上的两个给定事件 *A*、*B* 中 *A* 或 *B* 的概率。有放回抽样时，总体中每个个体被抽取后都会放回，因而该个体有可能被重复选中，这些事件被认为是独立的。无放回抽样时，总体中每个个体至多被选中一次，这些事件被认为是非独立的。当事件 *A* 与 *B* 没有任何公共结果时，它们是互斥事件。

3.4 Contingency Tables 3.4 列联表

There are several tools you can use to help organize and sort data when calculating probabilities. Contingency tables help display data and are particularly useful when calculating probabilites that have multiple dependent variables.

在计算概率时，有若干工具可帮助你整理与归类数据。列联表有助于展示数据，在计算涉及多个相依变量的概率时尤其有用。

3.5 Tree and Venn Diagrams 3.5 树形图与韦恩图

A tree diagram use branches to show the different outcomes of experiments and makes complex probability questions easy to visualize.

树形图用分支展示试验的不同结果，使复杂的概率问题易于可视化。

A Venn diagram is a picture that represents the outcomes of an experiment. It generally consists of a box that represents the sample space *S* together with circles or ovals. The circles or ovals represent events. A Venn diagram is especially helpful for visualizing the OR event, the AND event, and the complement of an event and for understanding conditional probabilities.

韦恩图是表示试验结果的图形。它通常由一个表示样本空间 *S* 的方框，连同若干圆形或椭圆形构成。这些圆形或椭圆形表示事件。韦恩图对可视化 OR 事件、AND 事件和事件的补事件，以及理解条件概率特别有帮助。

Formula Review 公式回顾

3.1 Terminology 3.1 术语

*A* and *B* are events

*A* 与 *B* 为事件。

*P*(*S*) = 1 where *S* is the sample space

*P*(*S*) = 1，其中 *S* 为样本空间。

0 ≤ *P*(*A*) ≤ 1

*P*(*A*\|*B*) = $\frac{P\text{(}A\text{AND}B\text{)}}{P\text{(}B\text{)}}$

3.2 Independent and Mutually Exclusive Events 3.2 独立事件与互斥事件

If *A* and *B* are independent, *P*(*A* AND *B*) = *P*(*A*)*P*(*B*), *P*(*A*\|*B*) = *P*(*A*) and *P*(*B*\|*A*) = *P*(*B*).

若 *A* 与 *B* 独立，则 *P*(*A* AND *B*) = *P*(*A*)*P*(*B*)，*P*(*A*\|*B*) = *P*(*A*)，且 *P*(*B*\|*A*) = *P*(*B*)。

If *A* and *B* are mutually exclusive, *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) and *P*(*A* AND *B*) = 0.

若 *A* 与 *B* 互斥，则 *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*)，且 *P*(*A* AND *B*) = 0。

3.3 Two Basic Rules of Probability 3.3 概率的两大基本法则

The multiplication rule: *P*(*A* AND *B*) = *P*(*A*\|*B*)*P*(*B*)

乘法法则：*P*(*A* AND *B*) = *P*(*A*\|*B*)*P*(*B*)

The addition rule: *P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*)

加法法则：*P*(*A* OR *B*) = *P*(*A*) + *P*(*B*) - *P*(*A* AND *B*)

Practice 练习

3.1 Terminology 3.1 术语

In a particular college class, there are male and female students. Some students have long hair and some students have short hair. Write the symbols for the probabilities of the events for parts a through j. (Note that you cannot find numerical answers here. You were not given enough information to find any probability values yet; concentrate on understanding the symbols.)

在某个大学课堂上，既有男生也有女生。有些学生留长发，有些留短发。请写出 a 到 j 各部分所对应事件的概率记号。（注意：此处无法得出具体数值，因为尚未给出足够信息求任何概率值；重点在于理解记号。）

Let *F* be the event that a student is female.
Let *M* be the event that a student is male.
Let *S* be the event that a student has short hair.
Let *L* be the event that a student has long hair.

设 *F* 表示学生为女性的事件。
设 *M* 表示学生为男性的事件。
设 *S* 表示学生留短发的事件。
设 *L* 表示学生留长发的事件。

1. The probability that a student does not have long hair.

1. 学生不留长发的概率。

2. The probability that a student is male or has short hair.

2. 学生是男性或留短发的概率。

3. The probability that a student is a female and has long hair.

3. 学生是女性且留长发的概率。

4. The probability that a student is male, given that the student has long hair.

4. 已知学生留长发，该学生是男性的概率。

5. The probability that a student has long hair, given that the student is male.

5. 已知学生是男性，该学生留长发的概率。

6. Of all the female students, the probability that a student has short hair.

6. 在所有女生中，学生留短发的概率。

7. Of all students with long hair, the probability that a student is female.

7. 在所有留长发的学生中，学生是女性的概率。

8. The probability that a student is female or has long hair.

8. 学生是女性或留长发的概率。

9. The probability that a randomly selected student is a male student with short hair.

9. 随机抽取的学生是留短发的男性的概率。

10. The probability that a student is female.

10. 学生是女性的概率。

*Use the following information to answer the next four exercises.* A box is filled with several party favors. It contains 12 hats, 15 noisemakers, ten finger traps, and five bags of confetti. One party favor is chosen from the box at random.

*用以下信息回答接下来的四道习题。* 一个盒子里装了若干派对小礼物，其中有 12 顶帽子、15 个响笛、10 个夹指玩具和 5 袋彩纸。从盒中随机取出一件派对小礼物。

Let *H* = the event of getting a hat.

设 *H* = 抽到帽子的事件。

Let *N* = the event of getting a noisemaker.

设 *N* = 抽到响笛的事件。

Let *F* = the event of getting a finger trap.

设 *F* = 抽到夹指玩具的事件。

Let *C* = the event of getting a bag of confetti.

设 *C* = 抽到一袋彩纸的事件。

2\.

Find *P*(*H*).

求 *P*(*H*)。

Find *P*(*N*).

求 *P*(*N*)。

4\.

Find *P*(*F*).

求 *P*(*F*)。

Find *P*(*C*).

求 *P*(*C*)。

*Use the following information to answer the next six exercises.* A jar of 150 jelly beans contains 22 red jelly beans, 38 yellow, 20 green, 28 purple, 26 blue, and the rest are orange. One jelly bean is chosen from the box at random.

*用以下信息回答接下来的六道习题。* 一罐共 150 颗软糖，其中有 22 颗红色、38 颗黄色、20 颗绿色、28 颗紫色、26 颗蓝色，其余为橙色。从罐中随机取出一颗软糖。

Let *B* = the event of getting a blue jelly bean

设 *B* = 取到蓝色软糖的事件

Let *G* = the event of getting a green jelly bean.

设 *G* = 取到绿色软糖的事件。

Let *O* = the event of getting an orange jelly bean.

设 *O* = 取到橙色软糖的事件。

Let *P* = the event of getting a purple jelly bean.

设 *P* = 取到紫色软糖的事件。

Let *R* = the event of getting a red jelly bean.

设 *R* = 取到红色软糖的事件。

Let *Y* = the event of getting a yellow jelly bean.

设 *Y* = 取到黄色软糖的事件。

6\.

Find *P*(*B*).

求 *P*(*B*)。

Find *P*(*G*).

求 *P*(*G*)。

8\.

Find *P*(*P*).

求 *P*(*P*)。

Find *P*(*R*).

求 *P*(*R*)。

10\.

10.

Find *P*(*Y*).

求 *P*(*Y*)。

11.

Find *P*(*O*).

求 *P*(*O*)。

*Use the following information to answer the next six exercises.* There are 23 countries in North America, 12 countries in South America, 47 countries in Europe, 44 countries in Asia, 54 countries in Africa, and 14 in Oceania (Pacific Ocean region).

*用以下信息回答接下来的六道习题。* 北美洲有 23 个国家，南美洲有 12 个，欧洲有 47 个，亚洲有 44 个，非洲有 54 个，大洋洲（太平洋地区）有 14 个。

Let *A* = the event that a country is in Asia.

设 *A* = 某国位于亚洲的事件。

Let *E* = the event that a country is in Europe.

设 *E* = 某国位于欧洲的事件。

Let *F* = the event that a country is in Africa.

设 *F* = 某国位于非洲的事件。

Let *N* = the event that a country is in North America.

设 *N* = 某国位于北美洲的事件。

Let *O* = the event that a country is in Oceania.

设 *O* = 某国位于大洋洲的事件。

Let *S* = the event that a country is in South America.

设 *S* = 某国位于南美洲的事件。

12\.

12.

Find *P*(*A*).

求 *P*(*A*)。

13.

Find *P*(*E*).

求 *P*(*E*)。

14\.

14.

Find *P*(*F*).

求 *P*(*F*)。

15.

Find *P*(*N*).

求 *P*(*N*)。

16\.

16.

Find *P*(*O*).

求 *P*(*O*)。

17.

Find *P*(*S*).

求 *P*(*S*)。

18\.

18.

What is the probability of drawing a red card in a standard deck of 52 cards?

从一副标准 52 张扑克牌中抽出一张红牌的概率是多少？

19.

What is the probability of drawing a club in a standard deck of 52 cards?

从一副标准 52 张扑克牌中抽出一张梅花的概率是多少？

20\.

20.

What is the probability of rolling an even number of dots with a fair, six-sided die numbered one through six?

掷一枚均匀的正六面体骰子（点数从 1 到 6），出现偶数点的概率是多少？

21.

What is the probability of rolling a prime number of dots with a fair, six-sided die numbered one through six?

掷一枚均匀的正六面体骰子（点数从 1 到 6），出现素数点的概率是多少？

*Use the following information to answer the next two exercises.* You see a game at a local fair. You have to throw a dart at a color wheel. Each section on the color wheel is equal in area.

*用以下信息回答接下来的两道习题。* 你在当地集市上看到一种游戏，需向一个彩色转盘投掷飞镖。转盘上每个区域的面积相等。

Let *B* = the event of landing on blue.

设 *B* = 飞镖落在蓝色的事件。

Let *R* = the event of landing on red.

设 *R* = 飞镖落在红色的事件。

Let *G* = the event of landing on green.

设 *G* = 飞镖落在绿色的事件。

Let *Y* = the event of landing on yellow.

设 *Y* = 飞镖落在黄色的事件。

22\.

22.

If you land on *Y*, you get the biggest prize. Find *P*(*Y*).

若飞镖落在 *Y*（黄色）上，你可得最大奖。求 *P*(*Y*)。

23.

If you land on red, you don’t get a prize. What is *P*(*R*)?

若飞镖落在红色上，你不会得奖。*P*(*R*) 是多少？

*Use the following information to answer the next ten exercises.* On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters.

*用以下信息回答接下来的十道习题。* 在一支棒球队中，有内野手和外野手。有些球员是优秀击球手，有些不是。

Let *I* = the event that a player in an infielder.

设 *I* = 球员是内野手的事件。

Let *O* = the event that a player is an outfielder.

设 *O* = 球员是外野手的事件。

Let *H* = the event that a player is a great hitter.

设 *H* = 球员是优秀击球手的事件。

Let *N* = the event that a player is not a great hitter.

设 *N* = 球员不是优秀击球手的事件。

24\.

24.

Write the symbols for the probability that a player is not an outfielder.

写出"球员不是外野手"这一事件的概率记号。

25.

Write the symbols for the probability that a player is an outfielder or is a great hitter.

写出"球员是外野手或是优秀击球手"的概率记号。

26\.

26.

Write the symbols for the probability that a player is an infielder and is not a great hitter.

写出"球员是内野手且不是优秀击球手"的概率记号。

27.

Write the symbols for the probability that a player is a great hitter, given that the player is an infielder.

写出"已知球员是内野手，其为优秀击球手"的条件概率记号。

28\.

28.

Write the symbols for the probability that a player is an infielder, given that the player is a great hitter.

写出"已知球员是优秀击球手，其为内野手"的条件概率记号。

29.

Write the symbols for the probability that of all the outfielders, a player is not a great hitter.

写出"在所有外野手中，球员不是优秀击球手"的概率记号。

30\.

30.

Write the symbols for the probability that of all the great hitters, a player is an outfielder.

写出"在所有优秀击球手中，球员是外野手"的概率记号。

31.

Write the symbols for the probability that a player is an infielder or is not a great hitter.

写出"球员是内野手或不是优秀击球手"的概率记号。

32\.

32.

Write the symbols for the probability that a player is an outfielder and is a great hitter.

写出"球员是外野手且是优秀击球手"的概率记号。

33.

Write the symbols for the probability that a player is an infielder.

写出"球员是内野手"的概率记号。

34\.

34.

What is the word for the set of all possible outcomes?

"所有可能结果组成的集合"叫什么？

35.

What is conditional probability?

什么是条件概率？

36\.

36.

A shelf holds 12 books. Eight are fiction and the rest are nonfiction. Each is a different book with a unique title. The fiction books are numbered one to eight. The nonfiction books are numbered one to four. Randomly select one book

一架子上有 12 本书，其中 8 本是小说，其余是非小说类。每本书都不同且书名唯一。小说编号为 1 到 8，非小说类编号为 1 到 4。随机抽取一本书

Let *F* = event that book is fiction

设 *F* = 抽到小说类书籍的事件

Let *N* = event that book is nonfiction

设 *N* = 抽到非小说类书籍的事件

What is the sample space?

样本空间是什么？

37.

What is the sum of the probabilities of an event and its complement?

一个事件与其补事件的概率之和是多少？

*Use the following information to answer the next two exercises.* You are rolling a fair, six-sided number cube. Let *E* = the event that it lands on an even number. Let *M* = the event that it lands on a multiple of three.

*用以下信息回答接下来的两道习题。* 你正在掷一枚均匀的正六面体数字骰。设 *E* = 掷出偶数的事件，*M* = 掷出 3 的倍数的事件。

38\.

38.

What does *P*(*E*\|*M*) mean in words?

*P*(*E*\|*M*) 用语言表示是什么意思？

39.

What does *P*(*E* OR *M*) mean in words?

*P*(*E* OR *M*) 用语言表示是什么意思？

3.2 Independent and Mutually Exclusive Events 3.2 独立事件与互斥事件

40\.

40.

*E* and *F* are mutually exclusive events. *P*(*E*) = 0.4; *P*(*F*) = 0.5. Find *P*(*E*∣*F*).

*E* 与 *F* 是互斥事件，*P*(*E*) = 0.4，*P*(*F*) = 0.5。求 *P*(*E*∣*F*)。

41.

*J* and *K* are independent events. *P*(*J*\|*K*) = 0.3. Find *P*(*J*).

*J* 与 *K* 是独立事件，*P*(*J*\|*K*) = 0.3。求 *P*(*J*)。

42\.

42.

*U* and *V* are mutually exclusive events. *P*(*U*) = 0.26; *P*(*V*) = 0.37. Find:

*U* 与 *V* 是互斥事件，*P*(*U*) = 0.26，*P*(*V*) = 0.37。求：

1. *P*(*U* AND *V*) =

2. *P*(*U*\|*V*) =

3. *P*(*U* OR *V*) =

43.

*Q* and *R* are independent events. *P*(*Q*) = 0.4 and *P*(*Q* AND *R*) = 0.1. Find *P*(*R*).

*Q* 与 *R* 是独立事件，*P*(*Q*) = 0.4 且 *P*(*Q* AND *R*) = 0.1。求 *P*(*R*)。

3.3 Two Basic Rules of Probability 3.3 概率的两大基本法则

*Use the following information to answer the next ten exercises.* Forty-eight percent of all Californians registered voters prefer life in prison without parole over the death penalty for a person convicted of first degree murder. Among Latino California registered voters, 55% prefer life in prison without parole over the death penalty for a person convicted of first degree murder. 37.6% of all Californians are Latino.

*用以下信息回答接下来的十道习题。* 在所有加利福尼亚登记选民中，48% 的人对一级谋杀定罪者更倾向于无期徒刑而非死刑。在加利福尼亚拉丁裔登记选民中，55% 的人持此偏好。所有加利福尼亚人中 37.6% 为拉丁裔。

In this problem, let:

本题中，设：

*C* = Californians (registered voters) preferring life in prison without parole over the death penalty for a person convicted of first degree murder.
*L* = Latino Californians

*C* = 加利福尼亚（登记选民）中，对一级谋杀定罪者更倾向于无期徒刑而非死刑的人。
*L* = 加利福尼亚拉丁裔

Suppose that one Californian is randomly selected.

假设随机抽取一名加利福尼亚人。

44\.

44.

Find *P*(*C*).

求 *P*(*C*)。

45.

Find *P*(*L*).

求 *P*(*L*)。

46\.

46.

Find *P*(*C*\|*L*).

求 *P*(*C*\|*L*)。

47.

In words, what is *C*\|*L*?

用语言表述，*C*\|*L* 表示什么？

48\.

48.

Find *P*(*L* AND *C*).

求 *P*(*L* AND *C*)。

49.

In words, what is *L* AND *C*?

用语言表述，*L* AND *C* 表示什么？

50\.

50.

Are *L* and *C* independent events? Show why or why not.

*L* 与 *C* 是独立事件吗？说明原因。

51.

Find *P*(*L* OR *C*).

求 *P*(*L* OR *C*)。

52\.

52.

In words, what is *L* OR *C*?

用语言表述，*L* OR *C* 表示什么？

53.

Are *L* and *C* mutually exclusive events? Show why or why not.

*L* 与 *C* 是互斥事件吗？说明原因。

3.4 Contingency Tables 3.4 列联表

*Use the following information to answer the next four exercises.* Table 3.15 shows a random sample of musicians and how they learned to play their instruments.

*用以下信息回答接下来的四道习题。* 表 3.15 给出一组音乐家随机样本，以及他们学习演奏乐器的方式。

| 性别 | 自学 | 在校学习 | 私人授课 | 合计 |

|--------|-------------|-------------------|---------------------|-------|

| Female | 12 | 38 | 22 | 72 |

| 女性 | 12 | 38 | 22 | 72 |

| Male | 19 | 24 | 15 | 58 |

| 男性 | 19 | 24 | 15 | 58 |

| Total | 31 | 62 | 37 | 130 |

| 合计 | 31 | 62 | 37 | 130 |

Table 3.15 54.

表 3.15 54.

Find *P*(musician is a female).

求 *P*（音乐家为女性）。

55.

Find *P*(musician is a male AND had private instruction).

求 *P*（音乐家为男性且接受过私人授课）。

56\.

56.

Find *P*(musician is a female OR is self taught).

求 *P*（音乐家为女性或自学成才）。

57.

Are the events “being a female musician” and “learning music in school” mutually exclusive events?

"身为女性音乐家"与"在学校学习音乐"这两个事件是否互斥？

3.5 Tree and Venn Diagrams 3.5 树形图与韦恩图

58.

The probability that a man develops some form of cancer in his lifetime is 0.4567. The probability that a man has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Let: *C* = a man develops cancer in his lifetime; *P* = man has at least one false positive. Construct a tree diagram of the situation.

一名男性在其一生中罹患某种癌症的概率为 0.4567。一名男性至少有一次假阳性检测结果（即该男性实际未患癌症，但检测却回报为癌症）的概率为 0.51。令：*C* = 一名男性在其一生中罹患癌症；*P* = 该男性至少有一次假阳性。请绘出该情形的树形图。

Bringing It Together: Practice 综合应用：练习

Use the following information to answer the next seven exercises. An article in the New England Journal of Medicine, reported about a study of smokers in California and Hawaii. In one part of the report, the self-reported ethnicity and smoking levels per day were given. Of the people smoking at most ten cigarettes per day, there were 9,886 African American people, 2,745 Native Hawaiian people, 12,831 Latino people, 8,378 Japanese people, and 7,650 White people. Of the people smoking 11 to 20 cigarettes per day, there were 6,514 African American people, 3,062 Native Hawaiian people, 4,932 Latino people, 10,680 Japanese people, and 9,877 White people. Of the people smoking 21 to 30 cigarettes per day, there were 1,671 African American people, 1,419 Native Hawaiian people, 1,406 Latino people, 4,715 Japanese people, and 6,062 White people. Of the people smoking at least 31 cigarettes per day, there were 759 African American people, 788 Native Hawaiian people, 800 Latino people, 2,305 Japanese people, and 3,970 White people.

使用以下信息回答后续七道习题。《新英格兰医学杂志》上的一篇文章报道了在加利福尼亚和夏威夷对吸烟者的一项研究。报告的一部分给出了自我报告的族裔以及每日吸烟量。在每天吸烟至多 10 支的人群中，有 9,886 名非裔美国人、2,745 名夏威夷原住民、12,831 名拉美裔、8,378 名日本人以及 7,650 名白人。在每天吸烟 11 至 20 支的人群中，有 6,514 名非裔美国人、3,062 名夏威夷原住民、4,932 名拉美裔、10,680 名日本人以及 9,877 名白人。在每天吸烟 21 至 30 支的人群中，有 1,671 名非裔美国人、1,419 名夏威夷原住民、1,406 名拉美裔、4,715 名日本人以及 6,062 名白人。在每天吸烟至少 31 支的人群中，有 759 名非裔美国人、788 名夏威夷原住民、800 名拉美裔、2,305 名日本人以及 3,970 名白人。

59\.

59.

Complete the table using the data provided.

使用所给数据补全表格。

| 吸烟水平 | 非裔美国人 | 夏威夷原住民 | 拉美裔 | 日本人 | 白人 | 合计 |

|---------------|------------------|-----------------|--------|-----------------|-------|--------|

| 1–10 | | | | | | |

| 11–20 | | | | | | |

| 21–30 | | | | | | |

| 31+ | | | | | | |

| 合计 | | | | | | |

Table 3.16 Smoking Levels by Ethnicity

表 3.16 按族裔划分的吸烟水平

60.

Suppose that one person from the study is randomly selected. Find the probability that person smoked 11 to 20 cigarettes per day.

假设从研究中随机抽取一人。求该人每天吸 11 至 20 支烟的概率。

61\.

61.

Find the probability that the person was Latino.

求该人为拉美裔的概率。

62.

In words, explain what it means to pick one person from the study who is “Japanese American AND smokes 21 to 30 cigarettes per day.” Also, find the probability.

用语言解释：从研究中抽取一人，该人为“日裔美国人且每天吸 21 至 30 支烟”是什么意思。并求该概率。

63\.

63.

In words, explain what it means to pick one person from the study who is “Japanese American OR smokes 21 to 30 cigarettes per day.” Also, find the probability.

用语言解释：从研究中抽取一人，该人为“日裔美国人或每天吸 21 至 30 支烟”是什么意思。并求该概率。

64.

In words, explain what it means to pick one person from the study who is “Japanese American GIVEN that person smokes 21 to 30 cigarettes per day.” Also, find the probability.

用语言解释：从研究中抽取一人，该人为“日裔美国人给定该人每天吸 21 至 30 支烟”是什么意思。并求该概率。

65\.

65.

Prove that smoking level/day and ethnicity are dependent events.

证明吸烟水平/每日吸烟量与族裔是相依事件。

Homework 作业

3.1 Terminology 3.1 术语

66\.

66.

The graph in Figure 3.11 displays the sample sizes and percentages of people in different age and gender groups who were polled concerning their approval of Mayor Ford’s actions in office. The total number in the sample of all the age groups is 1,045.

图 3.11 中的图形显示了针对不同年龄与性别群体、就其就市长福特（Mayor Ford）执政表现表示认可者所做的民意调查的样本量与百分比。所有年龄组样本的总人数为 1,045。

1. Define three events in the graph.

1. 在图中定义三个事件。

2. Describe in words what the entry 40 means.

2. 用语言描述条目 40 的含义。

3. Describe in words the complement of the entry in question 2.

3. 用语言描述第 2 题条目的补事件。

4. Describe in words what the entry 30 means.

4. 用语言描述条目 30 的含义。

5. Out of the males and females, what percent are males?

5. 在男性与女性中，男性占百分之多少？

6. Out of the females, what percent disapprove of Mayor Ford?

6. 在女性中，不认可福特市长的占百分之多少？

7. Out of all the age groups, what percent approve of Mayor Ford?

7. 在所有年龄组中，认可福特市长的占百分之多少？

8. Find *P*(Approve\|Male).

8. 求 *P*(Approve\|Male)（在男性中认可的概率）。

9. Out of the age groups, what percent are more than 44 years old?

9. 在各年龄组中，年龄超过 44 岁者占百分之多少？

10. Find *P*(Approve\|Age < 35).

10. 求 *P*(Approve\|Age < 35)（在年龄小于 35 岁者中认可的概率）。

67.

Explain what is wrong with the following statements. Use complete sentences.

解释下列说法的错误之处。请用完整的句子作答。

1. If there is a 60% chance of rain on Saturday and a 70% chance of rain on Sunday, then there is a 130% chance of rain over the weekend.

1. 如果周六下雨的概率为 60%，周日下雨的概率为 70%，那么整个周末下雨的概率就是 130%。

2. The probability that a baseball player hits a home run is greater than the probability that he gets a successful hit.

2. 棒球运动员击出本垒打的概率大于他成功击球的概率。

3.2 Independent and Mutually Exclusive Events 3.2 独立事件与互斥事件

*Use the following information to answer the next 12 exercises.* The graph shown is based on more than 170,000 interviews done by Gallup that took place from January through December 2012. The sample consists of employed Americans 18 years of age or older. The Emotional Health Index Scores are the sample space. We randomly sample one Emotional Health Index Score.

*使用以下信息回答后续 12 道习题。* 所示图形基于盖洛普（Gallup）在 2012 年 1 月至 12 月期间进行的逾 170,000 次访谈。样本由 18 岁及以上的在业美国人构成。情绪健康指数得分（Emotional Health Index Scores）即为样本空间。我们随机抽取一个情绪健康指数得分。

68\.

68.

Find the probability that an Emotional Health Index Score is 82.7.

求某个情绪健康指数得分为 82.7 的概率。

69.

Find the probability that an Emotional Health Index Score is 81.0.

求某个情绪健康指数得分为 81.0 的概率。

70\.

70.

Find the probability that an Emotional Health Index Score is more than 81?

求某个情绪健康指数得分大于 81 的概率？

71.

Find the probability that an Emotional Health Index Score is between 80.5 and 82?

求某个情绪健康指数得分在 80.5 与 82 之间的概率？

72\.

72.

If we know an Emotional Health Index Score is 81.5 or more, what is the probability that it is 82.7?

若已知一个情绪健康指数得分为 81.5 或以上，则它是 82.7 的概率是多少？

73.

What is the probability that an Emotional Health Index Score is 80.7 or 82.7?

某个情绪健康指数得分为 80.7 或 82.7 的概率是多少？

74\.

74.

What is the probability that an Emotional Health Index Score is less than 80.2 given that it is already less than 81.

在已知某个情绪健康指数得分已小于 81 的条件下，它小于 80.2 的概率是多少？

75.

What occupation has the highest emotional index score?

哪种职业的情绪指数得分最高？

76\.

76.

What occupation has the lowest emotional index score?

哪种职业的情绪指数得分最低？

77.

What is the range of the data?

数据的极差是多少？

78\.

78.

Compute the average EHIS.

计算 EHIS 的平均值。

79.

If all occupations are equally likely for a certain individual, what is the probability that he or she will have an occupation with lower than average EHIS?

若对某一特定个人而言所有职业等可能，则他或她所从事职业的 EHIS 低于平均值的概率是多少？

3.3 Two Basic Rules of Probability 3.3 概率的两个基本法则

80\.

80.

On February 28, 2013, a Field Poll Survey reported that 61% of California registered voters approved of allowing two people of the same gender to marry and have regular marriage laws apply to them. Among 18 to 39 year olds (California registered voters), the approval rating was 78%. Six in ten California registered voters said that the upcoming Supreme Court’s ruling about the constitutionality of California’s Proposition 8 was either very or somewhat important to them. Out of those CA registered voters who support same-sex marriage, 75% say the ruling is important to them.

2013 年 2 月 28 日，Field Poll 的一项调查报告指出，61% 的加利福尼亚州登记选民赞成允许同性两人结婚并适用普通婚姻法。在 18 至 39 岁的加利福尼亚州登记选民中，支持率为 78%。十分之六的加利福尼亚州登记选民表示，最高法院即将对加州 8 号提案合宪性作出的裁决对他们而言非常重要或有些重要。在支持同性婚姻的加州登记选民中，75% 认为该裁决对他们很重要。

In this problem, let:

在本题中，令：

*C* = California registered voters who support same-sex marriage.
*B* = California registered voters who say the Supreme Court’s ruling about the constitutionality of California’s Proposition 8 is very or somewhat important to them
*A* = California registered voters who are 18 to 39 years old.

*C* = 赞成同性婚姻的加利福尼亚州登记选民。
*B* = 认为最高法院对加州 8 号提案合宪性的裁决对自己非常重要或有些重要的加利福尼亚州登记选民
*A* = 18 至 39 岁的加利福尼亚州登记选民。

1. Find *P*(*C*).

1. 求 *P*(*C*)。

2. Find *P*(*B*).

2. 求 *P*(*B*)。

3. Find *P*(*C*\|*A*).

3. 求 *P*(*C*\|*A*)。

4. Find *P*(*B*\|*C*).

4. 求 *P*(*B*\|*C*)。

5. In words, what is *C*\|*A*?

5. 用文字表述，*C*\|*A* 是什么？

6. In words, what is *B*\|*C*?

6. 用文字表述，*B*\|*C* 是什么？

7. Find *P*(*C* AND *B*).

7. 求 *P*(*C* AND *B*)。

8. In words, what is *C* AND *B*?

8. 用文字表述，*C* AND *B* 是什么？

9. Find *P*(*C* OR *B*).

9. 求 *P*(*C* OR *B*)。

10. Are *C* and *B* mutually exclusive events? Show why or why not.

10. *C* 和 *B* 是互斥事件吗？说明原因。

81.

After Rob Ford, the mayor of Toronto, announced his plans to cut budget costs in late 2011, the Forum Research polled 1,046 people to measure the mayor’s popularity. Everyone polled expressed either approval or disapproval. These are the results their poll produced:

2011 年末多伦多市长 Rob Ford 宣布削减预算成本的计划后，Forum Research 调查了 1,046 人，以衡量市长的受欢迎程度。每位受访者都表示赞成或反对。以下是该调查得出的结果：

In early 2011, 60 percent of the population approved of Mayor Ford’s actions in office.
In mid-2011, 57 percent of the population approved of his actions.
In late 2011, the percentage of popular approval was measured at 42 percent.

2011 年初，60% 的民众赞成 Ford 市长在任内的作为。
2011 年年中，57% 的民众赞成他的作为。
2011 年末，民众支持率测得为 42%。

1. What is the sample size for this study?

1. 这项研究的样本量是多少？

2. What proportion in the poll disapproved of Mayor Ford, according to the results from late 2011?

2. 根据 2011 年末的结果，调查中反对 Ford 市长的比例是多少？

3. How many people polled responded that they approved of Mayor Ford in late 2011?

3. 2011 年末有多少名受访者表示赞成 Ford 市长？

4. What is the probability that a person supported Mayor Ford, based on the data collected in mid-2011?

4. 根据 2011 年年中收集的数据，一个人支持 Ford 市长的概率是多少？

5. What is the probability that a person supported Mayor Ford, based on the data collected in early 2011?

5. 根据 2011 年初收集的数据，一个人支持 Ford 市长的概率是多少？

*Use the following information to answer the next three exercises.* The casino game, roulette, allows the gambler to bet on the probability of a ball, which spins in the roulette wheel, landing on a particular color, number, or range of numbers. The table used to place bets contains of 38 numbers, and each number is assigned to a color and a range.

*用以下信息回答接下来的三道习题。* 赌场游戏轮盘赌允许赌徒就轮盘上旋转的小球落在特定颜色、数字或数字范围内的概率下注。下注所用的表格包含 38 个数字，每个数字都被指定一种颜色和所属范围。

82\.

82.

1. List the sample space of the 38 possible outcomes in roulette.

1. 列出轮盘赌 38 种可能结果的样本空间。

2. You bet on red. Find *P*(red).

2. 你押红色。求 *P*(red)。

3. You bet on -1st 12- (1st Dozen). Find *P*(-1st 12-).

3. 你押"第一组 12 个"（1st Dozen）。求 *P*(-1st 12-)。

4. You bet on an even number. Find *P*(even number).

4. 你押偶数。求 *P*(even number)。

5. Is getting an odd number the complement of getting an even number? Why?

5. 得到奇数是否为得到偶数的补事件？为什么？

6. Find two mutually exclusive events.

6. 找出两个互斥事件。

7. Are the events Even and 1st Dozen independent?

7. "偶数"事件与"第一组 12 个"事件是否独立？

83.

Compute the probability of winning the following types of bets:

计算以下各类下注的获胜概率：

1. Betting on two lines that touch each other on the table as in 1-2-3-4-5-6

1. 押表格上相邻的两条线，如 1-2-3-4-5-6

2. Betting on three numbers in a line, as in 1-2-3

2. 押一线上三个数字，如 1-2-3

3. Betting on one number

3. 押一个数字

4. Betting on four numbers that touch each other to form a square, as in 10-11-13-14

4. 押相邻四个数字构成正方形，如 10-11-13-14

5. Betting on two numbers that touch each other on the table, as in 10-11 or 10-13

5. 押表格上相邻的两个数字，如 10-11 或 10-13

6. Betting on 0-00-1-2-3

6. 押 0-00-1-2-3

7. Betting on 0-1-2; or 0-00-2; or 00-2-3

7. 押 0-1-2；或 0-00-2；或 00-2-3

84\.

84.

Compute the probability of winning the following types of bets:

计算以下各类下注的获胜概率：

1. Betting on a color

1. 押一种颜色

2. Betting on one of the dozen groups

2. 押十二个一组中的某一组

3. Betting on the range of numbers from 1 to 18

3. 押 1 到 18 的数字范围

4. Betting on the range of numbers 19–36

4. 押 19–36 的数字范围

5. Betting on one of the columns

5. 押某一列

6. Betting on an even or odd number (excluding zero)

6. 押偶数或奇数（不含零）

85.

Suppose that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.

假设你有八张卡片。五张为绿色，三张为黄色。五张绿色卡片编号为 1、2、3、4、5。三张黄色卡片编号为 1、2、3。卡片已充分洗匀。你随机抽取一张卡片。

*G* = card drawn is green
*E* = card drawn is even-numbered

*G* = 抽到的卡片为绿色
*E* = 抽到的卡片为偶数编号

1. List the sample space.

1. 列出样本空间。

2. *P*(*G*) = \_\_\_\_\_

3. *P*(*G*\|*E*) = \_\_\_\_\_

4. *P*(*G* AND *E*) = \_\_\_\_\_

5. *P*(*G* OR *E*) = \_\_\_\_\_

6. Are *G* and *E* mutually exclusive? Justify your answer numerically.

6. *G* 和 *E* 互斥吗？用数值说明你的答案。

86\.

86.

Roll two fair dice separately. Each die has six faces.

分别掷两枚均匀的骰子。每枚骰子有六个面。

1. List the sample space.

1. 列出样本空间。

2. Let *A* be the event that either a three or four is rolled first, followed by an even number. Find *P*(*A*).

2. 设 *A* 为"先掷出 3 或 4，随后掷出偶数"的事件。求 *P*(*A*)。

3. Let *B* be the event that the sum of the two rolls is at most seven. Find *P*(*B*).

3. 设 *B* 为"两次掷出的点数之和至多为 7"的事件。求 *P*(*B*)。

4. In words, explain what "*P*(*A*\|*B*)" represents. Find *P*(*A*\|*B*).

4. 用文字解释"*P*(*A*\|*B*)"表示什么。求 *P*(*A*\|*B*)。

5. Are *A* and *B* mutually exclusive events? Explain your answer in one to three complete sentences, including numerical justification.

5. *A* 与 *B* 互斥吗？用一到三句完整的话说明你的答案，并给出数值依据。

6. Are *A* and *B* independent events? Explain your answer in one to three complete sentences, including numerical justification.

6. *A* 与 *B* 独立吗？用一到三句完整的话说明你的答案，并给出数值依据。

87.

A special deck of cards has ten cards. Four are green, three are blue, and three are red. When a card is picked, its color of it is recorded. An experiment consists of first picking a card and then tossing a coin.

一副特制纸牌有十张。四张绿色、三张蓝色、三张红色。抽牌时记录其颜色。一项试验为先抽一张牌，再抛一枚硬币。

1. List the sample space.

1. 列出样本空间。

2. Let *A* be the event that a blue card is picked first, followed by landing a head on the coin toss. Find *P*(*A*).

2. 设 *A* 为"先抽到蓝色牌，随后抛硬币得到正面"的事件。求 *P*(*A*)。

3. Let *B* be the event that a red or green is picked, followed by landing a head on the coin toss. Are the events *A* and *B* mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.

3. 设 *B* 为"抽到红色或绿色，随后抛硬币得到正面"的事件。*A* 与 *B* 是否互斥？用一到三句完整的话说明你的答案，并给出数值依据。

4. Let *C* be the event that a red or blue is picked, followed by landing a head on the coin toss. Are the events *A* and *C* mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.

4. 设 *C* 为"抽到红色或蓝色，随后抛硬币得到正面"的事件。*A* 与 *C* 是否互斥？用一到三句完整的话说明你的答案，并给出数值依据。

88\.

88.

An experiment consists of first rolling a die and then tossing a coin.

一项试验为先掷一枚骰子，再抛一枚硬币。

1. List the sample space.

1. 列出样本空间。

2. Let *A* be the event that either a three or a four is rolled first, followed by landing a head on the coin toss. Find *P*(*A*).

2. 设 *A* 为"先掷出 3 或 4，随后抛硬币得到正面"的事件。求 *P*(*A*)。

3. Let *B* be the event that the first and second tosses land on heads. Are the events *A* and *B* mutually exclusive? Explain your answer in one to three complete sentences, including numerical justification.

3. 设 *B* 为"第一、二次抛掷均得到正面"的事件。*A* 与 *B* 是否互斥？用一到三句完整的话说明你的答案，并给出数值依据。

89.

An experiment consists of tossing a nickel, a dime, and a quarter. Of interest is the side the coin lands on.

一项试验为抛掷一枚五分币、一角币和二角五分币。关注点是硬币落地时的那一面。

1. List the sample space.

1. 列出样本空间。

2. Let *A* be the event that there are at least two tails. Find *P*(*A*).

2. 设 *A* 为"至少出现两个反面"的事件。求 *P*(*A*)。

3. Let *B* be the event that the first and second tosses land on heads. Are the events *A* and *B* mutually exclusive? Explain your answer in one to three complete sentences, including justification.

3. 设 *B* 为"第一、二次抛掷均得到正面"的事件。*A* 与 *B* 是否互斥？用一到三句完整的话说明你的答案，并给出依据。

90\.

90.

Consider the following scenario:

考虑以下情境：

Let *P*(*C*) = 0.4.

设 *P*(*C*) = 0.4。

Let *P*(*D*) = 0.5.

设 *P*(*D*) = 0.5。

Let *P*(*C*\|*D*) = 0.6.

设 *P*(*C*\|*D*) = 0.6。

1. Find *P*(*C* AND *D*).

1. 求 *P*(*C* AND *D*)。

2. Are *C* and *D* mutually exclusive? Why or why not?

2. *C* 与 *D* 互斥吗？说明原因。

3. Are *C* and *D* independent events? Why or why not?

3. *C* 与 *D* 独立吗？说明原因。

4. Find *P*(*C* OR *D*).

4. 求 *P*(*C* OR *D*)。

5. Find *P*(*D*\|*C*).

5. 求 *P*(*D*\|*C*)。

91.

*Y* and *Z* are independent events.

*Y* 与 *Z* 是独立事件。

1. Rewrite the basic Addition Rule *P*(*Y* OR *Z*) = *P*(*Y*) + *P*(*Z*) - *P*(*Y* AND *Z*) using the information that *Y* and *Z* are independent events.

1. 利用 *Y* 与 *Z* 独立这一信息，改写基本加法法则 *P*(*Y* OR *Z*) = *P*(*Y*) + *P*(*Z*) - *P*(*Y* AND *Z*)。

2. Use the rewritten rule to find *P*(*Z*) if *P*(*Y* OR *Z*) = 0.71 and *P*(*Y*) = 0.42.

2. 若 *P*(*Y* OR *Z*) = 0.71 且 *P*(*Y*) = 0.42，用改写后的法则求 *P*(*Z*)。

92\.

92.

*G* and *H* are mutually exclusive events. *P*(*G*) = 0.5 *P*(*H*) = 0.3

*G* 与 *H* 是互斥事件。*P*(*G*) = 0.5 *P*(*H*) = 0.3

1. Explain why the following statement MUST be false: *P*(*H*\|*G*) = 0.4.

1. 解释为什么以下陈述必定为假：*P*(*H*\|*G*) = 0.4。

2. Find *P*(*H* OR *G*).

2. 求 *P*(*H* OR *G*)。

3. Are *G* and *H* independent or dependent events? Explain in a complete sentence.

3. *G* 与 *H* 是独立还是相依事件？用一句完整的话说明。

93.

Approximately 281,000,000 people over age five live in the United States. Of these people, 55,000,000 speak a language other than English at home. Of those who speak another language at home, 62.3% speak Spanish.

美国约有 2.81 亿五岁以上人口。其中 5,500 万人在家中使用英语以外的语言。在使用其他语言的人群中，62.3% 说西班牙语。

Let: *E* = speaks English at home; *E′* = speaks another language at home; *S* = speaks Spanish;

设：*E* = 在家说英语；*E′* = 在家说另一种语言；*S* = 说西班牙语；

Finish each probability statement by matching the correct answer.

通过匹配正确答案完成各概率陈述。

| Probability Statements | Answers |

| 概率陈述 | 答案 |

|-------------------------|--------------|

| a\. *P*(*E′*) = | i\. 0.8043 |

| b\. *P*(*E*) = | ii\. 0.623 |

| c\. *P*(*S* and *E′*) = | iii\. 0.1957 |

| d\. *P*(*S*\|*E′*) = | iv\. 0.1219 |

Table 3.17 94.

表 3.17 94.

1994, the U.S. government held a lottery to issue 55,000 Green Cards (permits for non-citizens to work legally in the U.S.). Renate Deutsch, from Germany, was one of approximately 6.5 million people who entered this lottery. Let *G* = won green card.

1994 年，美国政府举办抽签发放 55,000 张绿卡（非公民在美国合法工作的许可）。来自德国的 Renate Deutsch 是约 650 万名参与抽签者之一。设 *G* = 赢得绿卡。

1. What was Renate’s chance of winning a Green Card? Write your answer as a probability statement.

1. Renate 赢得绿卡的机会是多少？将答案写成概率陈述。

2. In the summer of 1994, Renate received a letter stating she was one of 110,000 finalists chosen. Once the finalists were chosen, assuming that each finalist had an equal chance to win, what was Renate’s chance of winning a Green Card? Write your answer as a conditional probability statement. Let *F* = was a finalist.

2. 1994 年夏，Renate 收到一封信，告知她是被选中的 110,000 名决赛选手之一。决赛选手确定后，假设每位决赛选手获胜机会相等，Renate 赢得绿卡的机会是多少？将答案写成条件概率陈述。设 *F* = 进入决赛。

3. Are *G* and *F* independent or dependent events? Justify your answer numerically and also explain why.

3. *G* 与 *F* 是独立还是相依事件？用数值说明你的答案，并解释原因。

4. Are *G* and *F* mutually exclusive events? Justify your answer numerically and explain why.

4. *G* 与 *F* 是互斥事件吗？用数值说明你的答案并解释原因。

95.

Three professors at George Washington University did an experiment to determine if economists are more selfish than other people. They dropped 64 stamped, addressed envelopes with \$10 cash in different classrooms on the George Washington campus. 44% were returned overall. From the economics classes 56% of the envelopes were returned. From the business, psychology, and history classes 31% were returned.

乔治·华盛顿大学的三位教授做了一项实验，以确定经济学家是否比其他人更自私。他们在乔治·华盛顿大学校园的不同教室中放置了 64 个贴有邮票、写有地址、内装 10 美元现金的信封。总体有 44% 被归还。经济学课堂上 56% 的信封被归还。商科、心理学和历史课堂上 31% 被归还。

Let: *R* = money returned; *E* = economics classes; *O* = other classes

设：*R* = 钱被归还；*E* = 经济学课堂；*O* = 其他课堂

1. Write a probability statement for the overall percent of money returned.

1. 写出总体归还金额百分比的概率陈述。

2. Write a probability statement for the percent of money returned out of the economics classes.

2. 写出经济学课堂中归还金额百分比的概率陈述。

3. Write a probability statement for the percent of money returned out of the other classes.

3. 写出其他课堂中归还金额百分比的概率陈述。

4. Is money being returned independent of the class? Justify your answer numerically and explain it.

4. 钱是否被归还与课堂无关（独立）吗？用数值说明你的答案并作解释。

5. Based upon this study, do you think that economists are more selfish than other people? Explain why or why not. Include numbers to justify your answer.

5. 基于这项研究，你认为经济学家比其他人更自私吗？说明原因，并引用数据支持你的答案。

96.

The following table of data obtained from www.baseball-almanac.com shows hit information for four players. Suppose that one hit from the table is randomly selected.

下表数据来自 www.baseball-almanac.com，显示了四名球员的安打信息。假设从表中随机抽取一次安打。

| 姓名 | 一垒打 | 二垒打 | 三垒打 | 本垒打 | 总安打数 |

|-----------------|--------|--------|--------|----------|------------|

| Babe Ruth | 1,517 | 506 | 136 | 714 | 2,873 |

| Jackie Robinson | 1,054 | 273 | 54 | 137 | 1,518 |

| Ty Cobb | 3,603 | 174 | 295 | 114 | 4,189 |

| Hank Aaron | 2,294 | 624 | 98 | 755 | 3,771 |

| Total | 8,471 | 1,577 | 583 | 1,720 | 12,351 |

Table 3.18

表 3.18

Are "the hit being made by Hank Aaron" and "the hit being a double" independent events?

"由 Hank Aaron 击出的安打"与"安打为二垒打"这两个事件是否独立？

1. Yes, because *P*(hit by Hank Aaron\|hit is a double) = *P*(hit by Hank Aaron)

1. 是，因为 *P*(hit by Hank Aaron\|hit is a double) = *P*(hit by Hank Aaron)

2. No, because *P*(hit by Hank Aaron\|hit is a double) ≠ *P*(hit is a double)

2. 否，因为 *P*(hit by Hank Aaron\|hit is a double) ≠ *P*(hit is a double)

3. No, because *P*(hit is by Hank Aaron\|hit is a double) ≠ *P*(hit by Hank Aaron)

3. 否，因为 *P*(hit is by Hank Aaron\|hit is a double) ≠ *P*(hit by Hank Aaron)

4. Yes, because *P*(hit is by Hank Aaron\|hit is a double) = *P*(hit is a double)

4. 是，因为 *P*(hit is by Hank Aaron\|hit is a double) = *P*(hit is a double)

97.

United Blood Services is a blood bank that serves more than 500 hospitals in 18 states. According to their website, a person with type O blood and a negative Rh factor (Rh-) can donate blood to any person with any bloodtype. Their data show that 43% of people have type O blood and 15% of people have Rh- factor; 52% of people have type O or Rh- factor.

United Blood Services 是一家血库，为 18 个州超过 500 家医院供血。根据其网站，拥有 O 型血且 Rh 因子阴性（Rh-）的人可以向任何血型的人献血。其数据显示，43% 的人为 O 型血，15% 的人为 Rh- 因子；52% 的人为 O 型或 Rh- 因子。

1. Find the probability that a person has both type O blood and the Rh- factor.

1. 求一个人同时具有 O 型血和 Rh- 因子的概率。

2. Find the probability that a person does NOT have both type O blood and the Rh- factor.

2. 求一个人不同时具有 O 型血和 Rh- 因子的概率。

98\.

98.

At a college, 72% of courses have final exams and 46% of courses require research papers. Suppose that 32% of courses have a research paper and a final exam. Let *F* be the event that a course has a final exam. Let *R* be the event that a course requires a research paper.

在一所大学中，72% 的课程有期末考试，46% 的课程要求写研究论文。假设 32% 的课程既有研究论文又有期末考试。设 *F* 为课程有期末考试的事件。设 *R* 为课程要求写研究论文的事件。

1. Find the probability that a course has a final exam or a research project.

1. 求一门课程有期末考试或研究项目的概率。

2. Find the probability that a course has NEITHER of these two requirements.

2. 求一门课程两项要求都不具备的概率。

99.

In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. In the box, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.

在一盒什锦饼干中，36% 含巧克力，12% 含坚果。盒中 8% 同时含巧克力和坚果。Sean 对巧克力和坚果都过敏。

1. Find the probability that a cookie contains chocolate or nuts (he can't eat it).

1. 求一块饼干含巧克力或坚果的概率（他不能吃）。

2. Find the probability that a cookie does not contain chocolate or nuts (he can eat it).

2. 求一块饼干不含巧克力也不含坚果的概率（他能吃）。

100\.

100.

A college finds that 10% of students have taken a distance learning class and that 40% of students are part time students. Of the part time students, 20% have taken a distance learning class. Let *D* = event that a student takes a distance learning class and *E* = event that a student is a part time student

一所大学发现，10% 的学生修过远程学习课程，40% 的学生为非全日制学生。在非全日制学生中，20% 修过远程学习课程。设 *D* = 学生修远程学习课程，*E* = 学生为非全日制学生。

1. Find *P*(*D* AND *E*).

1. 求 *P*(*D* AND *E*)。

2. Find *P*(*E*\|*D*).

2. 求 *P*(*E*\|*D*)。

3. Find *P*(*D* OR *E*).

3. 求 *P*(*D* OR *E*)。

4. Using an appropriate test, show whether *D* and *E* are independent.

4. 用适当的检验说明 *D* 与 *E* 是否独立。

5. Using an appropriate test, show whether *D* and *E* are mutually exclusive.

5. 用适当的检验说明 *D* 与 *E* 是否互斥。

3.4 Contingency Tables 3.4 列联表

*Use the information in the Table 3.19 to answer the next eight exercises.* The table shows the political party affiliation of each of 67 members of the US Senate in June 2012, and when they are up for reelection.

使用表 3.19 中的信息回答接下来的八道题。该表显示了 2012 年 6 月美国参议院 67 名成员各自的政党归属，以及他们何时面临改选。

| 改选年份： | 民主党 | 共和党 | 其他 | 合计 |

|--------------------|------------------|------------------|-------|-------|

| November 2014 | 20 | 13 | 0 | |

| 2014年11月 | 20 | 13 | 0 | |

| November 2016 | 10 | 24 | 0 | |

| 2016年11月 | 10 | 24 | 0 | |

| 合计 | | | | |

Table 3.19 101.

表 3.19 101。

What is the probability that a randomly selected senator has an “Other” affiliation?

随机选出的一名参议员具有"其他"政党归属的概率是多少？

102\.

What is the probability that a randomly selected senator is up for reelection in November 2016?

随机选出的一名参议员在 2016 年 11 月面临改选的概率是多少？

103.

What is the probability that a randomly selected senator is a Democrat and up for reelection in November 2016?

随机选出的一名参议员既是民主党人，又在 2016 年 11 月面临改选的概率是多少？

104\.

What is the probability that a randomly selected senator is a Republican or is up for reelection in November 2014?

随机选出的一名参议员是共和党人，或者在 2014 年 11 月面临改选的概率是多少？

105.

Suppose that a member of the US Senate is randomly selected. Given that the randomly selected senator is up for reelection in November 2016, what is the probability that this senator is a Democrat?

假设随机选出了一名美国参议院成员。已知该参议员在 2016 年 11 月面临改选，那么该参议员是民主党人的概率是多少？

106\.

Suppose that a member of the US Senate is randomly selected. What is the probability that the senator is up for reelection in November 2014, knowing that this senator is a Republican?

假设随机选出了一名美国参议院成员。已知该参议员是共和党人，那么该参议员在 2014 年 11 月面临改选的概率是多少？

107.

The events “Republican” and “Up for reelection in 2016” are \_\_\_\_\_\_\_\_

事件"共和党人"与"2016 年面临改选"是 \_\_\_\_\_\_\_\_

1. mutually exclusive.

1. 互斥。

2. independent.

2. 独立。

3. both mutually exclusive and independent.

3. 既互斥又独立。

4. neither mutually exclusive nor independent.

4. 既不互斥也不独立。

108\.

The events “Other” and “Up for reelection in November 2016” are \_\_\_\_\_\_\_\_

事件"其他"与"2016 年 11 月面临改选"是 \_\_\_\_\_\_\_\_

1. mutually exclusive.

1. 互斥。

2. independent.

2. 独立。

3. both mutually exclusive and independent.

3. 既互斥又独立。

4. neither mutually exclusive nor independent.

4. 既不互斥也不独立。

109.

Table 3.20 gives the number of participants in the recent National Health Interview Survey who had been treated for cancer in the previous 12 months. The results are sorted by age, race (Black or White), and sex. We are interested in possible relationships between age, race, and sex.

表 3.20 给出了最近一次全国健康访谈调查中，在过去 12 个月内接受过癌症治疗的参与者人数。结果按年龄、种族（黑人或白人）和性别分类。我们关注年龄、种族和性别之间可能存在的关系。

| Race and sex | 15–24 | 25–40 | 41–65 | Over 65 | TOTALS |

| 种族与性别 | 15–24 | 25–40 | 41–65 | 65 岁以上 | 合计 |

|---------------|-------|-------|-------|---------|--------|

| White, male | 1,165 | 2,036 | 3,703 | | 8,395 |

| 白人男性 | 1,165 | 2,036 | 3,703 | | 8,395 |

| White, female | 1,076 | 2,242 | 4,060 | | 9,129 |

| 白人女性 | 1,076 | 2,242 | 4,060 | | 9,129 |

| Black, male | 142 | 194 | 384 | | 824 |

| 黑人男性 | 142 | 194 | 384 | | 824 |

| Black, female | 131 | 290 | 486 | | 1,061 |

| 黑人女性 | 131 | 290 | 486 | | 1,061 |

| TOTALS | 2,792 | 5,279 | 9,354 | | 21,081 |

| 合计 | 2,792 | 5,279 | 9,354 | | 21,081 |

Table 3.20

表 3.20

Do not include "all others" for parts f and g.

在 f 和 g 两问中不要包含"所有其他"这一行。

1. Fill in the column for cancer treatment for individuals over age 65.

1. 填写 65 岁以上个体的癌症治疗人数这一列。

2. Fill in the row for all other races.

2. 填写所有其他种族这一行。

3. Find the probability that a randomly selected individual was a White male.

3. 求随机选出的个体为白人男性的概率。

4. Find the probability that a randomly selected individual was a Black female.

4. 求随机选出的个体为黑人女性的概率。

5. Find the probability that a randomly selected individual was Black

5. 求随机选出的个体为黑人的概率。

6. Find the probability that a randomly selected individual was male.

6. 求随机选出的个体为男性的概率。

7. Out of the individuals over age 65, find the probability that a randomly selected individual was a Black or White male.

7. 在 65 岁以上的个体中，求随机选出的个体为黑人或白人男性的概率。

*Use the following information to answer the next two exercises.* The table of data obtained from www.baseball-almanac.com shows hit information for four well known baseball players. Suppose that one hit from the table is randomly selected.

使用以下信息回答接下来的两道题。该数据表来自 www.baseball-almanac.com，给出了四位著名棒球运动员的击球信息。假设从表中随机选出一次击球。

|-----------------|--------|--------|--------|----------|------------|

| Babe Ruth | 1,517 | 506 | 136 | 714 | 2,873 |

| 贝比·鲁斯 | 1,517 | 506 | 136 | 714 | 2,873 |

| Jackie Robinson | 1,054 | 273 | 54 | 137 | 1,518 |

| 杰基·罗宾逊 | 1,054 | 273 | 54 | 137 | 1,518 |

| Ty Cobb | 3,603 | 174 | 295 | 114 | 4,189 |

| 泰·柯布 | 3,603 | 174 | 295 | 114 | 4,189 |

| Hank Aaron | 2,294 | 624 | 98 | 755 | 3,771 |

| 汉克·阿伦 | 2,294 | 624 | 98 | 755 | 3,771 |

| TOTAL | 8,471 | 1,577 | 583 | 1,720 | 12,351 |

| 合计 | 8,471 | 1,577 | 583 | 1,720 | 12,351 |

Table 3.21 110.

表 3.21 110。

Find *P*(hit was made by Babe Ruth).

求 *P*(击球由贝比·鲁斯完成)。

1. $\frac{1518}{2873}$

2. $\frac{2873}{12351}$

3. $\frac{583}{12351}$

4. $\frac{4189}{12351}$

111.

Find *P*(hit was made by Ty Cobb\|The hit was a Home Run).

求 *P*(击球由泰·柯布完成\|该击球为本垒打)。

1. $\frac{4189}{12351}$

2. $\frac{114}{1720}$

3. $\frac{1720}{4189}$

4. $\frac{114}{12351}$

112\.

Table 3.22 identifies a group of children by one of four hair colors, and by type of hair.

表 3.22 按四种发色之一和发质类型对一组儿童进行分类。

| 发质 | 棕色 | 金色 | 黑色 | 红色 | 合计 |

|-----------|-------|-------|-------|-----|--------|

| Wavy | 20 | | 15 | 3 | 43 |

| 卷发 | 20 | | 15 | 3 | 43 |

| Straight | 80 | 15 | | 12 | |

| 直发 | 80 | 15 | | 12 | |

| Totals | | 20 | | | 215 |

| 合计 | | 20 | | | 215 |

Table 3.22

表 3.22

1. Complete the table.

1. 补全该表。

2. What is the probability that a randomly selected child will have wavy hair?

2. 随机选出的儿童拥有卷发的概率是多少？

3. What is the probability that a randomly selected child will have either brown or blond hair?

3. 随机选出的儿童拥有棕色或金色头发的概率是多少？

4. What is the probability that a randomly selected child will have wavy brown hair?

4. 随机选出的儿童拥有卷曲棕发的概率是多少？

5. What is the probability that a randomly selected child will have red hair, given that he or she has straight hair?

5. 已知一名儿童是直发，随机选出的该儿童拥有红色头发的概率是多少？

6. If *B* is the event of a child having brown hair, find the probability of the complement of *B*.

6. 若 *B* 表示儿童拥有棕色头发的事件，求 *B* 的补事件的概率。

7. In words, what does the complement of *B* represent?

7. 用文字说明，*B* 的补事件代表什么？

113.

In a previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data were compiled into the following table.

在往年，《圣何塞水星报》刊登了旧金山49人队与达拉斯牛仔队成员的体重。这些真实数据被整理成下表。

| Shirt# | ≤ 210 | 211–250 | 251–290 | > 290 |

| 球衣号码 | ≤ 210 | 211–250 | 251–290 | > 290 |

|--------|-------|---------|---------|--------|

| 1–33 | 21 | 5 | 0 | 0 |

| 34–66 | 6 | 18 | 7 | 4 |

| 66–99 | 6 | 12 | 22 | 5 |

Table 3.23

表 3.23

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

对于以下各问，假设你在 49 人队或牛仔队中随机选出一名球员。

1. Find the probability that his shirt number is from 1 to 33.

1. 求他的球衣号码在 1 到 33 之间的概率。

2. Find the probability that he weighs at most 210 pounds.

2. 求他的体重不超过 210 磅的概率。

3. Find the probability that his shirt number is from 1 to 33 AND he weighs at most 210 pounds.

3. 求他的球衣号码在 1 到 33 之间，且体重不超过 210 磅的概率。

4. Find the probability that his shirt number is from 1 to 33 OR he weighs at most 210 pounds.

4. 求他的球衣号码在 1 到 33 之间，或体重不超过 210 磅的概率。

5. Find the probability that his shirt number is from 1 to 33 GIVEN that he weighs at most 210 pounds.

5. 已知他的体重不超过 210 磅，求他的球衣号码在 1 到 33 之间的概率。

3.5 Tree and Venn Diagrams 3.5 树形图与韦恩图

*Use the following information to answer the next two exercises.* This tree diagram shows the tossing of an unfair coin followed by drawing one bead from a cup containing three red (*R*), four yellow (*Y*) and five blue (*B*) beads. For the coin, *P*(*H*) = $\frac{2}{3}$ and *P*(*T*) = $\frac{1}{3}$ where *H* is heads and *T* is tails.

使用以下信息回答接下来的两道题。该树形图展示了一枚不均匀硬币的抛掷，随后从装有三颗红色（*R*）、四颗黄色（*Y*）和五颗蓝色（*B*）珠子的杯中抽取一颗珠子。对于硬币，*P*(*H*) = $\frac{2}{3}$，*P*(*T*) = $\frac{1}{3}$，其中 *H* 表示正面，*T* 表示反面。

114\.

Find *P*(tossing a Head on the coin AND a Red bead)

求 *P*(硬币抛出正面且抽到红色珠子)。

1. $\frac{2}{3}$

2. $\frac{5}{15}$

3. $\frac{6}{36}$

4. $\frac{5}{36}$

115.

Find *P*(Blue bead).

求 *P*(蓝色珠子)。

1. $\frac{15}{36}$

2. $\frac{10}{36}$

3. $\frac{10}{12}$

4. $\frac{6}{36}$

116\.

A box of cookies contains three chocolate and seven butter cookies. Miguel randomly selects a cookie and eats it. Then he randomly selects another cookie and eats it. (How many cookies did he take?)

一盒饼干中有三块巧克力和七块黄油饼干。Miguel 随机选出一块饼干吃掉，然后又随机选出另一块饼干吃掉。（他一共拿了几块饼干？）

1. Draw the tree that represents the possibilities for the cookie selections. Write the probabilities along each branch of the tree.

1. 画出表示饼干选取各种可能性的树形图。在树的每个分支上标出概率。

2. Are the probabilities for the flavor of the SECOND cookie that Miguel selects independent of his first selection? Explain.

2. Miguel 选出的第二块饼干的口味概率，是否独立于他第一次的选择？请说明。

3. For each complete path through the tree, write the event it represents and find the probabilities.

3. 对于穿过树的每条完整路径，写出它所代表的事件并求出概率。

4. Let S be the event that both cookies selected were the same flavor. Find *P*(*S*).

4. 设 S 为两次选出的饼干口味相同的事件。求 *P*(*S*)。

5. Let *T* be the event that the cookies selected were different flavors. Find *P*(*T*) by two different methods: by using the complement rule and by using the branches of the tree. Your answers should be the same with both methods.

5. 设 *T* 为选出的饼干口味不同的事件。用两种不同的方法求 *P*(*T*)：使用补事件法则，以及使用树的分支。两种方法所得答案应相同。

6. Let *U* be the event that the second cookie selected is a butter cookie. Find *P*(*U*).

6. 设 *U* 为第二次选出的饼干是黄油饼干的事件。求 *P*(*U*)。

Bringing It Together: Homework 综合应用：作业

117.

A previous year, the weights of the members of the San Francisco 49ers and the Dallas Cowboys were published in the San Jose Mercury News. The factual data are compiled into Table 3.24.

前一年，《圣何塞水星报》刊登了旧金山49人队与达拉斯牛仔队队员的体重。这些真实数据汇总于表 3.24。

| Shirt# | ≤ 210 | 211–250 | 251–290 | 290≤ |

| 球衣号码 | ≤ 210 | 211–250 | 251–290 | 290≤ |

|--------|-------|---------|---------|------|

| 1–33 | 21 | 5 | 0 | 0 |

| 34–66 | 6 | 18 | 7 | 4 |

| 66–99 | 6 | 12 | 22 | 5 |

Table 3.24

表 3.24

For the following, suppose that you randomly select one player from the 49ers or Cowboys.

对于以下各题，假设你从49人队或牛仔队中随机选取一名球员。

If having a shirt number from one to 33 and weighing at most 210 pounds were independent events, then what should be true about *P*(Shirt# 1–33\|≤ 210 pounds)?

若衬衫号码为1至33与体重至多210磅是两个独立事件，那么关于 *P*(Shirt# 1–33\|≤ 210 pounds) 应当成立什么结论？

118.

The probability that a male develops some form of cancer in his lifetime is 0.4567. The probability that a male has at least one false positive test result (meaning the test comes back for cancer when the man does not have it) is 0.51. Some of the following questions do not have enough information for you to answer them. Write “not enough information” for those answers. Let *C* = a man develops cancer in his lifetime and *P* = man has at least one false positive.

一名男性一生中罹患某种癌症的概率为0.4567。一名男性至少有一次假阳性检测结果（即该男性实际未患癌症而检测却显示患癌）的概率为0.51。以下部分问题信息不足、无法作答；此类问题请填写“not enough information”（信息不足）。令 *C* = 一名男性一生中罹患癌症，*P* = 一名男性至少有一次假阳性。

1. *P*(*C*) = \_\_\_\_\_\_

2. *P*(*P*\|*C*) = \_\_\_\_\_\_

3. *P*(*P*\|*C'*) = \_\_\_\_\_\_

4. If a test comes up positive, based upon numerical values, can you assume that man has cancer? Justify numerically and explain why or why not.

4. 若检测结果为阳性，仅凭数值能否推断该男性患有癌症？请通过数值加以论证，并说明能或不能的理由。

119.

Given events *G* and *H*: *P*(*G*) = 0.43; *P*(*H*) = 0.26; *P*(*H* AND *G*) = 0.14

已知事件 *G* 与 *H*：*P*(*G*) = 0.43；*P*(*H*) = 0.26；*P*(*H* AND *G*) = 0.14

1. Find *P*(*H* OR *G*).

1. 求 *P*(*H* OR *G*)。

2. Find the probability of the complement of event (*H* AND *G*).

2. 求事件 (*H* AND *G*) 的补事件的概率。

3. Find the probability of the complement of event (*H* OR *G*).

3. 求事件 (*H* OR *G*) 的补事件的概率。

120.

Given events *J* and *K*: *P*(*J*) = 0.18; *P*(*K*) = 0.37; *P*(*J* OR *K*) = 0.45

已知事件 *J* 与 *K*：*P*(*J*) = 0.18；*P*(*K*) = 0.37；*P*(*J* OR *K*) = 0.45

1. Find *P*(*J* AND *K*).

1. 求 *P*(*J* AND *K*)。

2. Find the probability of the complement of event (*J* AND *K*).

2. 求事件 (*J* AND *K*) 的补事件的概率。

3. Find the probability of the complement of event (*J* OR *K*).

3. 求事件 (*J* OR *K*) 的补事件的概率。

*Use the following information to answer the next two exercises.* Suppose that you have eight cards. Five are green and three are yellow. The cards are well shuffled.

*Use the following information to answer the next two exercises.* 假设你有八张卡片，其中五张为绿色、三张为黄色，且已充分洗匀。

121.

Suppose that you randomly draw two cards, one at a time, with replacement.

假设你逐张随机抽取两张卡片，有放回。

Let *G*₁ = first card is green

令 *G*₁ = 第一张卡片为绿色

Let *G*₂ = second card is green

令 *G*₂ = 第二张卡片为绿色

1. Draw a tree diagram of the situation.

1. 画出该情形的树形图。

2. Find *P*(*G*₁ AND *G*₂).

2. 求 *P*(*G*₁ AND *G*₂)。

3. Find *P*(at least one green).

3. 求 *P*(至少一张为绿色)。

4. Find *P*(*G*₂\|*G*₁).

4. 求 *P*(*G*₂\|*G*₁)。

5. Are *G*₂ and *G*₁ independent events? Explain why or why not.

5. *G*₂ 与 *G*₁ 是独立事件吗？说明理由。

122.

Suppose that you randomly draw two cards, one at a time, without replacement.

假设你逐张随机抽取两张卡片，无放回。

*G₁* = first card is green

*G₁* = 第一张卡片为绿色

*G₂* = second card is green

*G₂* = 第二张卡片为绿色

1. Draw a tree diagram of the situation.

1. 画出该情形的树形图。

2. Find *P*(*G₁* AND *G₂*).

2. 求 *P*(*G₁* AND *G₂*)。

3. Find *P*(at least one green).

3. 求 *P*(至少一张为绿色)。

4. Find *P*(*G₂*\|*G₁*).

4. 求 *P*(*G₂*\|*G₁*)。

5. Are *G₂* and *G₁* independent events? Explain why or why not.

5. *G₂* 与 *G₁* 是独立事件吗？说明理由。

*Use the following information to answer the next two exercises.* The percent of licensed U.S. drivers (from a recent year) that are female is 48.60. Of the females, 5.03% are age 19 and under; 81.36% are age 20–64; 13.61% are age 65 or over. Of the licensed U.S. male drivers, 5.04% are age 19 and under; 81.43% are age 20–64; 13.53% are age 65 or over.

*Use the following information to answer the next two exercises.* 在最近一年，美国持证驾驶员中女性占48.60%。女性中，5.03%年龄在19岁及以下；81.36%年龄在20–64岁；13.61%年龄在65岁及以上。美国男性持证驾驶员中，5.04%年龄在19岁及以下；81.43%年龄在20–64岁；13.53%年龄在65岁及以上。

123.

Complete the following.

完成下列各题。

1. Construct a table or a tree diagram of the situation.

1. 构造该情形的列联表或树形图。

2. Find *P*(driver is female).

2. 求 *P*(驾驶员为女性)。

3. Find *P*(driver is age 65 or over\|driver is female).

3. 求 *P*(驾驶员年龄在65岁及以上\|驾驶员为女性)。

4. Find *P*(driver is age 65 or over AND female).

4. 求 *P*(驾驶员年龄在65岁及以上 AND 为女性)。

5. In words, explain the difference between the probabilities in part c and part d.

5. 用文字说明 c 问与 d 问中两个概率的区别。

6. Find *P*(driver is age 65 or over).

6. 求 *P*(驾驶员年龄在65岁及以上)。

7. Are being age 65 or over and being female mutually exclusive events? How do you know?

7. 年龄在65岁及以上与为女性是互斥事件吗？你如何判断？

124.

Suppose that 10,000 U.S. licensed drivers are randomly selected.

假设随机选取10,000名美国持证驾驶员。

1. How many would you expect to be male?

1. 预期其中有多少名为男性？

2. Using the table or tree diagram, construct a contingency table of gender versus age group.

2. 利用该表或树形图，构造性别对年龄组的列联表。

3. Using the contingency table, find the probability that out of the age 20–64 group, a randomly selected driver is female.

3. 利用该列联表，求在20–64岁年龄组中，随机抽取的一名驾驶员为女性的概率。

125.

Approximately 86.5% of Americans commute to work by car, truck, or van. Out of that group, 84.6% drive alone and 15.4% drive in a carpool. Approximately 3.9% walk to work and approximately 5.3% take public transportation.

约86.5%的美国人乘小汽车、卡车或厢式车通勤。在该群体中，84.6%独自驾车，15.4%合伙乘车。约3.9%步行上班，约5.3%乘坐公共交通。

1. Construct a table or a tree diagram of the situation. Include a branch for all other modes of transportation to work.

1. 构造该情形的列联表或树形图。须包含一个代表所有其他通勤方式的分支。

2. Assuming that the walkers walk alone, what percent of all commuters travel alone to work?

2. 假设步行者均独自步行，所有通勤者中有百分之多少独自前往工作？

3. Suppose that 1,000 workers are randomly selected. How many would you expect to travel alone to work?

3. 假设随机选取1,000名工作者，预期其中有多少人独自通勤上班？

4. Suppose that 1,000 workers are randomly selected. How many would you expect to drive in a carpool?

4. 假设随机选取1,000名工作者，预期其中有多少人合伙乘车？

126.

When the Euro coin was introduced in 2002, two math professors had their statistics students test whether the Belgian one Euro coin was a fair coin. They spun the coin rather than tossing it and found that out of 250 spins, 140 showed a head (event *H*) while 110 showed a tail (event *T*). On that basis, they claimed that it is not a fair coin.

2002年欧元硬币问世时，两位数学教授让他们的统计学学生检验比利时一元欧元硬币是否为均匀硬币。他们采用旋转而非抛掷的方式，结果发现250次旋转中有140次正面朝上（事件 *H*），110次反面朝上（事件 *T*）。据此，他们声称该硬币不是均匀的。

1. Based on the given data, find *P*(*H*) and *P*(*T*).

1. 根据给定的数据，求 *P*(*H*) 与 *P*(*T*)。

2. Use a tree to find the probabilities of each possible outcome for the experiment of tossing the coin twice.

2. 用树形图求抛掷该硬币两次这一试验的各种可能结果的概率。

3. Use the tree to find the probability of obtaining exactly one head in two tosses of the coin.

3. 用树形图求两次抛掷中恰好出现一次正面的概率。

4. Use the tree to find the probability of obtaining at least one head.

4. 用树形图求至少出现一次正面的概率。

*Use the following information to answer the next two exercises.* The following are real data from Santa Clara County, CA. As of a certain time, there had been a total of 3,059 documented cases of AIDS in the county. They were grouped into the following categories:

*Use the following information to answer the next two exercises.* 以下为加利福尼亚州圣克拉拉县的真实数据。截至某一时点，该县记录在案的爱滋病病例共3,059例，归类如下：

|--------|---------------------|----------------|----------------------|-------|--------|

| Female | 0 | 70 | 136 | 49 | |

| 女性 | 0 | 70 | 136 | 49 | |

| Male | 2,146 | 463 | 60 | 135 | |

| 男性 | 2,146 | 463 | 60 | 135 | |

| 合计 | | | | | |

Table 3.25 \* includes homosexual/bisexual IV drug users

表 3.25 \* 含同性恋/双性恋静脉注射吸毒者

Suppose a person with AIDS in Santa Clara County is randomly selected.

假设从圣克拉拉县的爱滋病患者中随机选取一人。

127.

1. Find *P*(Person is female).

1. 求 *P*(此人为女性)。

2. Find *P*(Person has a risk factor heterosexual contact).

2. 求 *P*(此人具有异性接触这一危险因素)。

3. Find *P*(Person is female OR has a risk factor of IV drug user).

3. 求 *P*(此人为女性 OR 具有静脉注射吸毒者这一危险因素)。

4. Find *P*(Person is female AND has a risk factor of homosexual/bisexual).

4. 求 *P*(此人为女性 AND 具有同性恋/双性恋这一危险因素)。

5. Find *P*(Person is male AND has a risk factor of IV drug user).

5. 求 *P*(此人为男性 AND 具有静脉注射吸毒者这一危险因素)。

6. Find *P*(Person is female GIVEN person got the disease from heterosexual contact).

6. 求 *P*(此人为女性 GIVEN 此人通过异性接触染病)。

7. Construct a Venn diagram. Make one group females and the other group heterosexual contact.

7. 构造一幅韦恩图，一组为女性，另一组为异性接触。

128.

Answer these questions using probability rules. Do NOT use the contingency table. Three thousand fifty-nine cases of AIDS had been reported in Santa Clara County, CA, through a certain date. Those cases will be our population. Of those cases, 6.4% obtained the disease through heterosexual contact and 7.4% are female. Out of the females with the disease, 53.3% got the disease from heterosexual contact.

使用概率法则（不要使用列联表）回答下列各题。截至某一日期，加利福尼亚州圣克拉拉县共报告3,059例爱滋病病例，这些病例即为我们的总体。其中，6.4%通过异性接触染病，7.4%为女性。在患病的女性中，53.3%通过异性接触染病。

1. Find *P*(Person is female).

1. 求 *P*(此人为女性)。

2. Find *P*(Person obtained the disease through heterosexual contact).

2. 求 *P*(此人通过异性接触染病)。

3. Find *P*(Person is female GIVEN person got the disease from heterosexual contact)

3. 求 *P*(此人为女性 GIVEN 此人通过异性接触染病)

4. Construct a Venn diagram representing this situation. Make one group females and the other group heterosexual contact. Fill in all values as probabilities.

4. 构造一幅表示该情形的韦恩图，一组为女性，另一组为异性接触。所有数值均以概率填入。