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Solution Manual for Probability and Statistics, 4th Edition

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    S OLUTIONS M ANUAL (O NLINE ONLY ) M ARK S CHERVISH Carnegie Mellon University P ROBABILITY AND S TATISTICS F OURTH E DITION Morris DeGroot Carnegie Mellon University Mark Schervish Carnegie Mellon University

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    Contents Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi 1 Introduction to Probability 1 1.2 Interpretations of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.4 Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1.5 The Definition of Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 1.6 Finite Sample Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 1.7 Counting Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7 1.8 Combinatorial Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8 1.9 Multinomial Coefficients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 1.10 The Probability of a Union of Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 1.12 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2 Conditional Probability 25 2.1 The Definition of Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 2.2 Independent Events . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3 Bayes’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 2.4 The Gambler’s Ruin Problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40 2.5 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41 3 Random Variables and Distributions 49 3.1 Random Variables and Discrete Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 3.2 Continuous Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 3.3 The Cumulative Distribution Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53 3.4 Bivariate Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 3.5 Marginal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 3.6 Conditional Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 3.7 Multivariate Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76 3.8 Functions of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 3.9 Functions of Two or More Random Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 3.10 Markov Chains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 3.11 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 4 Expectation 107 4.1 The Expectation of a Random Variable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 4.2 Properties of Expectations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110 4.3 Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 4.4 Moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 4.5 The Mean and the Median . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118

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    iv CONTENTS 4.6 Covariance and Correlation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 4.7 Conditional Expectation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 4.8 Utility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 4.9 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134 5 Special Distributions 141 5.2 The Bernoulli and Binomial Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 141 5.3 The Hypergeometric Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 5.4 The Poisson Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 5.5 The Negative Binomial Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155 5.6 The Normal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 5.7 The Gamma Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 5.8 The Beta Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171 5.9 The Multinomial Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 5.10 The Bivariate Normal Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 5.11 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 6 Large Random Samples 187 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 6.2 The Law of Large Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 6.3 The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 6.4 The Correction for Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 6.5 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 199 7 Estimation 203 7.1 Statistical Inference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 7.2 Prior and Posterior Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 7.3 Conjugate Prior Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 7.4 Bayes Estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 7.5 Maximum Likelihood Estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 7.6 Properties of Maximum Likelihood Estimators . . . . . . . . . . . . . . . . . . . . . . . . . . 220 7.7 Sufficient Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 7.8 Jointly Sufficient Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 7.9 Improving an Estimator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 7.10 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 8 Sampling Distributions of Estimators 239 8.1 The Sampling Distribution of a Statistic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 8.2 The Chi-Square Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 8.3 Joint Distribution of the Sample Mean and Sample Variance . . . . . . . . . . . . . . . . . . 245 8.4 The t Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 247 8.5 Confidence Intervals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 250 8.6 Bayesian Analysis of Samples from a Normal Distribution . . . . . . . . . . . . . . . . . . . . 254 8.7 Unbiased Estimators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 258 8.8 Fisher Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 8.9 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267

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    CONTENTS v 9 Testing Hypotheses 273 9.1 Problems of Testing Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273 9.2 Testing Simple Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 278 9.3 Uniformly Most Powerful Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284 9.4 Two-Sided Alternatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 289 9.5 The t Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 9.6 Comparing the Means of Two Normal Distributions . . . . . . . . . . . . . . . . . . . . . . . 296 9.7 The F Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299 9.8 Bayes Test Procedures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 9.9 Foundational Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307 9.10 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309 10 Categorical Data and Nonparametric Methods 315 10.1 Tests of Goodness-of-Fit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 10.2 Goodness-of-Fit for Composite Hypotheses . . . . . . . . . . . . . . . . . . . . . . . . . . . . 317 10.3 Contingency Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 10.4 Tests of Homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 10.5 Simpson’s Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 10.6 Kolmogorov-Smirnov Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 327 10.7 Robust Estimation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 333 10.8 Sign and Rank Tests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 337 10.9 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 11 Linear Statistical Models 349 11.1 The Method of Least Squares . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 349 11.2 Regression . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 11.3 Statistical Inference in Simple Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . 356 11.4 Bayesian Inference in Simple Linear Regression . . . . . . . . . . . . . . . . . . . . . . . . . . 364 11.5 The General Linear Model and Multiple Regression . . . . . . . . . . . . . . . . . . . . . . . . 366 11.6 Analysis of Variance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373 11.7 The Two-Way Layout . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 11.8 The Two-Way Layout with Replications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 11.9 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 389 12 Simulation 399 12.1 What is Simulation? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400 12.2 Why Is Simulation Useful? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400 12.3 Simulating Specific Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404 12.4 Importance Sampling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 410 12.5 Markov Chain Monte Carlo . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414 12.6 The Bootstrap . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 12.7 Supplementary Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425 R Code For Two Text Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 432

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    vi CONTENTS Preface This manual contains solutions to all of the exercises in Probability and Statistics , 4th edition, by Morris DeGroot and myself. I have preserved most of the solutions to the exercises that existed in the 3rd edition. Certainly errors have been introduced, and I will post any errors brought to my attention on my web page http://www.stat.cmu.edu/ mark/ along with errors in the text itself. Feel free to send me comments. For instructors who are familiar with earlier editions, I hope that you will find the 4th edition at least as useful. Some new material has been added, and little has been removed. Assuming that you will be spending the same amount of time using the text as before, something will have to be skipped. I have tried to arrange the material so that instructors can choose what to cover and what not to cover based on the type of course they want. This manual contains commentary on specific sections right before the solutions for those sections. This commentary is intended to explain special features of those sections and help instructors decide which parts they want to require of their students. Special attention is given to more challenging material and how the remainder of the text does or does not depend upon it. To teach a mathematical statistics course for students with a strong calculus background, one could safely cover all of the material for which one could find time. The Bayesian sections include 4.8, 7.2, 7.3, 7.4, 8.6, 9.8, and 11.4. One can choose to skip some or all of this material if one desires, but that would be ignoring one of the unique features of the text. The more challenging material in Sections 7.7–7.9, and 9.2–9.4 is really only suitable for a mathematical statistics course. One should try to make time for some of the material in Sections 12.1–12.3 even if it meant cutting back on some of the nonparametrics and two-way ANOVA. To teach a more modern statistics course, one could skip Sections 7.7–7.9, 9.2–9.4, 10.8, and 11.7–11.8. This would leave time to discuss robust estimation (Section 10.7) and simulation (Chapter 12). Section 3.10 on Markov chains is not actually necessary even if one wishes to introduce Markov chain Monte Carlo (Section 12.5), although it is helpful for understanding what this topic is about. Using Statistical Software The text was written without reference to any particular statistical or mathematical software. However, there are several places throughout the text where references are made to what general statistical software might be able to do. This is done for at least two reasons. One is that different instructors who wish to use statistical software while teaching will generally choose different programs. I didn’t want the text to be tied to a particular program to the exclusion of others. A second reason is that there are still many instructors of mathematical probability and statistics courses who prefer not to use any software at all. Given how pervasive computing is becoming in the use of statistics, the second reason above is becoming less compelling. Given the free and multiplatform availability and the versatility of the environment R , even the first reason is becoming less compelling. Throughout this manual, I have inserted pointers to which R functions will perform many of the calculations that would formerly have been done by hand when using this text. The software can be downloaded for Unix, Windows, or Mac OS from http://www.r-project.org/ That site also has manuals for installation and use. Help is also available directly from within the R envi- ronment. Many tutorials for getting started with R are available online. At the official R site there is the detailed manual: http://cran.r-project.org/doc/manuals/R-intro.html that starts simple and has a good table of contents and lots of examples. However, reading it from start to finish is not an efficient way to get started. The sample sessions should be most helpful. One major issue with using an environment like R is that it is essentially programming. That is, students who have never programmed seriously before are going to have a steep learning curve. Without going into the philosophy of whether students should learn statistics without programming, the field is moving in the direction of requiring programming skills. People who want only to understand what a statistical analysis

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    CONTENTS vii is about can still learn that without being able to program. But anyone who actually wants to do statistics as part of their job will be seriously handicapped without programming ability. At the end of this manual is a series of heavily commented R programms that illustrate many of the features of R in the context of a specific example from the text. Mark J. Schervish

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    Chapter 1 Introduction to Probability 1.2 Interpretations of Probability Commentary It is interesting to have the students determine some of their own subjective probabilities. For example, let X denote the temperature at noon tomorrow outside the building in which the class is being held. Have each student determine a number x 1 such that the student considers the following two possible outcomes to be equally likely: X ≤ x 1 and X > x 1 . Also, have each student determine numbers x 2 and x 3 (with x 2 < x 3 ) such that the student considers the following three possible outcomes to be equally likely: X ≤ x 2 , x 2 < X < x 3 , and X ≥ x 3 . Determinations of more than three outcomes that are considered to be equally likely can also be made. The different values of x 1 determined by different members of the class should be discussed, and also the possibility of getting the class to agree on a common value of x 1 . Similar determinations of equally likely outcomes can be made by the students in the class for quantities such as the following ones which were found in the 1973 World Almanac and Book of Facts: the number of freight cars that were in use by American railways in 1960 (1,690,396), the number of banks in the United States which closed temporarily or permanently in 1931 on account of financial difficulties (2,294), and the total number of telephones which were in service in South America in 1971 (6,137,000). 1.4 Set Theory Solutions to Exercises 1. Assume that x ∈ B c . We need to show that x ∈ A c . We shall show this indirectly. Assume, to the contrary, that x ∈ A . Then x ∈ B because A ⊂ B . This contradicts x ∈ B c . Hence x ∈ A is false and x ∈ A c . 2. First, show that A ∩ ( B ∪ C ) ⊂ ( A ∩ B ) ∪ ( A ∩ C ). Let x ∈ A ∩ ( B ∪ C ). Then x ∈ A and x ∈ B ∪ C . That is, x ∈ A and either x ∈ B or x ∈ C (or both). So either ( x ∈ A and x ∈ B ) or ( x ∈ A and x ∈ C ) or both. That is, either x ∈ A ∩ B or x ∈ A ∩ C . This is what it means to say that x ∈ ( A ∩ B ) ∪ ( A ∩ C ). Thus A ∩ ( B ∪ C ) ⊂ ( A ∩ B ) ∪ ( A ∩ C ). Basically, running these steps backwards shows that ( A ∩ B ) ∪ ( A ∩ C ) ⊂ A ∩ ( B ∪ C ). 3. To prove the first result, let x ∈ ( A ∪ B ) c . This means that x is not in A ∪ B . In other words, x is neither in A nor in B . Hence x ∈ A c and x ∈ B c . So x ∈ A c ∩ B c . This proves that ( A ∪ B ) c ⊂ A c ∩ B c . Next, suppose that x ∈ A c ∩ B c . Then x ∈ A c and x ∈ B c . So x is neither in A nor in B , so it can’t be in A ∪ B . Hence x ∈ ( A ∪ B ) c . This shows that A c ∩ B c ⊂ ( A ∪ B ) c . The second result follows from the first by applying the first result to A c and B c and then taking complements of both sides.

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    2 Chapter 1. Introduction to Probability 4. To see that A ∩ B and A ∩ B c are disjoint, let x ∈ A ∩ B . Then x ∈ B , hence x  ∈ B c and so x  ∈ A ∩ B c . So no element of A ∩ B is in A ∩ B c , hence the two events are disjoint. To prove that A = ( A ∩ B ) ∪ ( A ∩ B c ), we shall show that each side is a subset of the other side. First, let x ∈ A . Either x ∈ B or x ∈ B c . If x ∈ B , then x ∈ A ∩ B . If x ∈ B c , then x ∈ A ∩ B c . Either way, x ∈ ( A ∩ B ) ∪ ( A ∩ B c ). So every element of A is an element of ( A ∩ B ) ∪ ( A ∩ B c ) and we conclude that A ⊂ ( A ∩ B ) ∪ ( A ∩ B c ). Finally, let x ∈ ( A ∩ B ) ∪ ( A ∩ B c ). Then either x ∈ A ∩ B , in which case x ∈ A , or x ∈ A ∩ B c , in which case x ∈ A . Either way x ∈ A , so every element of ( A ∩ B ) ∪ ( A ∩ B c ) is also an element of A and ( A ∩ B ) ∪ ( A ∩ B c ) ⊂ A . 5. To prove the first result, let x ∈ ( ∪ i A i ) c . This means that x is not in ∪ i A i . In other words, for every i ∈ I , x is not in A i . Hence for every i ∈ I , x ∈ A c i . So x ∈ ∩ i A c i . This proves that ( ∪ i A i ) c ⊂ ∩ i A c i . Next, suppose that x ∈ ∩ i A c i . Then x ∈ A c i for every i ∈ I . So for every i ∈ I , x is not in A i . So x can’t be in ∪ i A i . Hence x ∈ ( ∪ i A i ) c . This shows that ∩ i A c i ⊂ ( ∪ i A i ) c . The second result follows from the first by applying the first result to A c i for i ∈ I and then taking complements of both sides. 6. (a) Blue card numbered 2 or 4. (b) Blue card numbered 5, 6, 7, 8, 9, or 10. (c) Any blue card or a red card numbered 1, 2, 3, 4, 6, 8, or 10. (d) Blue card numbered 2, 4, 6, 8, or 10, or red card numbered 2 or 4. (e) Red card numbered 5, 7, or 9. 7. (a) These are the points not in A , hence they must be either below 1 or above 5. That is A c = { x : x < 1 or x > 5 } . (b) These are the points in either A or B or both. So they must be between 1 and 5 or between 3 and 7. That is, A ∪ B = { x : 1 ≤ x ≤ 7 } . (c) These are the points in B but not in C . That is BC c = { x : 3 < x ≤ 7 } . (Note that B ⊂ C c .) (d) These are the points in none of the three sets, namely A c B c C c = { x : 0 < x < 1 or x > 7 } . (e) These are the points in the answer to part (b) and in C . There are no such values and ( A ∪ B ) C = ∅ . 8. Blood type A reacts only with anti-A, so type A blood corresponds to A ∩ B c . Type B blood reacts only with anti-B, so type B blood corresponds to A c B . Type AB blood reacts with both, so A ∩ B characterizes type AB blood. Finally, type O reacts with neither antigen, so type O blood corresponds to the event A c B c . 9. (a) For each n , B n = B n +1 ∪ A n , hence B n ⊃ B n +1 for all n . For each n , C n +1 ∩ A n = C n , so C n ⊂ C n +1 . (b) Suppose that x ∈ ∩ ∞ n =1 B n . Then x ∈ B n for all n . That is, x ∈ ∪ ∞ i = n A i for all n . For n = 1, there exists i ≥ n such that x ∈ A i . Assume to the contrary that there are at most finitely many i such that x ∈ A i . Let m be the largest such i . For n = m + 1, we know that there is i ≥ n such that x ∈ A i . This contradicts m being the largest i such that x ∈ A i . Hence, x is in infinitely many A i . For the other direction, assume that x is in infinitely many A i . Then, for every n , there is a value of j > n such that x ∈ A j , hence x ∈ ∪ ∞ i = n A i = B n for every n and x ∈ ∩ ∞ n =1 B n . (c) Suppose that x ∈ ∪ ∞ n =1 C n . That is, there exists n such that x ∈ C n = ∩ ∞ i = n A i , so x ∈ A i for all i ≥ n . So, there at most finitely many i (a subset of 1 , . . . , n − 1) such that x  ∈ A i . Finally, suppose that x ∈ A i for all but finitely many i . Let k be the last i such that x  ∈ A i . Then x ∈ A i for all i ≥ k + 1, hence x ∈ ∩ ∞ i = k +1 A i = C k +1 . Hence x ∈ ∪ ∞ n =1 C n .

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    Section 1.5. The Definition of Probability 3 10. (a) All three dice show even numbers if and only if all three of A , B , and C occur. So, the event is A ∩ B ∩ C . (b) None of the three dice show even numbers if and only if all three of A c , B c , and C c occur. So, the event is A c ∩ B c ∩ C c . (c) At least one die shows an odd number if and only if at least one of A c , B c , and C c occur. So, the event is A c ∪ B c ∪ C c . (d) At most two dice show odd numbers if and only if at least one die shows an even number, so the event is A ∪ B ∪ C . This can also be expressed as the union of the three events of the form A ∩ B ∩ C c where exactly one die shows odd together with the three events of the form A ∩ B c ∩ C c where exactly two dice show odd together with the even A ∩ B ∩ C where no dice show odd. (e) We can enumerate all the sums that are no greater than 5: 1 + 1 + 1, 2 + 1 + 1, 1 + 2 + 1, 1 + 1 + 2, 2 + 2 + 1, 2 + 1 + 2, and 1 + 2 + 2. The first of these corresponds to the event A 1 ∩ B 1 ∩ C 1 , the second to A 2 ∩ B 1 ∩ C 1 , etc. The union of the seven such events is what is requested, namely ( A 1 ∩ B 1 ∩ C 1 ) ∪ ( A 2 ∩ B 1 ∩ C 1 ) ∪ ( A 1 ∩ B 2 ∩ C 1 ) ∪ ( A 1 ∩ B 1 ∩ C 2 ) ∪ ( A 2 ∩ B 2 ∩ C 1 ) ∪ ( A 2 ∩ B 1 ∩ C 2 ) ∪ ( A 1 ∩ B 2 ∩ C 2 ) . 11. (a) All of the events mentioned can be determined by knowing the voltages of the two subcells. Hence the following set can serve as a sample space S = { ( x, y ) : 0 ≤ x ≤ 5 and 0 ≤ y ≤ 5 } , where the first coordinate is the voltage of the first subcell and the second coordinate is the voltage of the second subcell. Any more complicated set from which these two voltages can be determined could serve as the sample space, so long as each outcome could at least hypothetically be learned. (b) The power cell is functional if and only if the sum of the voltages is at least 6. Hence, A = { ( x, y ) ∈ S : x + y ≥ 6 } . It is clear that B = { ( x, y ) ∈ S : x = y } and C = { ( x, y ) ∈ S : x > y } . The powercell is not functional if and only if the sum of the voltages is less than 6. It needs less than one volt to be functional if and only if the sum of the voltages is greater than 5. The intersection of these two is the event D = { ( x, y ) ∈ S : 5 < x + y < 6 } . The restriction “ ∈ S ” that appears in each of these descriptions guarantees that the set is a subset of S . One could leave off this restriction and add the two restrictions 0 ≤ x ≤ 5 and 0 ≤ y ≤ 5 to each set. (c) The description can be worded as “the power cell is not functional, and needs at least one more volt to be functional, and both subcells have the same voltage.” This is the intersection of A c , D c , and B . That is, A c ∩ D c ∩ B . The part of D c in which x + y ≥ 6 is not part of this set because of the intersection with A c . (d) We need the intersection of A c (not functional) with C c (second subcell at least as big as first) and with B c (subcells are not the same). In particular, C c ∩ B c is the event that the second subcell is strictly higher than the first. So, the event is A c ∩ B c ∩ C c . 1.5 The Definition of Probability Solutions to Exercises 1. Define the following events: A = { the selected ball is red } , B = { the selected ball is white } , C = { the selected ball is either blue, yellow, or green } .

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    4 Chapter 1. Introduction to Probability We are asked to find Pr( C ). The three events A , B , and C are disjoint and A ∪ B ∪ C = S . So 1 = Pr( A ) + Pr( B ) + Pr( C ). We are told that Pr( A ) = 1 / 5 and Pr( B ) = 2 / 5. It follows that Pr( C ) = 2 / 5. 2. Let B be the event that a boy is selected, and let G be the event that a girl is selected. We are told that B ∪ G = S , so G = B c . Since Pr( B ) = 0 . 3, it follows that Pr( G ) = 0 . 7. 3. (a) If A and B are disjoint then B ⊂ A c and BA c = B , so Pr( BA c ) = Pr( B ) = 1 / 2. (b) If A ⊂ B , then B = A ∪ ( BA c ) with A and BA c disjoint. So Pr( B ) = Pr( A ) + Pr( BA c ). That is, 1 / 2 = 1 / 3 + Pr( BA c ), so Pr( BA c ) = 1 / 6. (c) According to Theorem 1.4.11, B = ( BA ) ∪ ( BA c ). Also, BA and BA c are disjoint so, Pr( B ) = Pr( BA ) + Pr( BA c ). That is, 1 / 2 = 1 / 8 + Pr( BA c ), so Pr( BA c ) = 3 / 8. 4. Let E 1 be the event that student A fails and let E 2 be the event that student B fails. We want Pr( E 1 ∪ E 2 ). We are told that Pr( E 1 ) = 0 . 5, Pr( E 2 ) = 0 . 2, and Pr( E 1 E 2 ) = 0 . 1. According to Theorem 1.5.7, Pr( E 1 ∪ E 2 ) = 0 . 5 + 0 . 2 − 0 . 1 = 0 . 6. 5. Using the same notation as in Exercise 4, we now want Pr( E c 1 ∩ E c 2 ). According to Theorems 1.4.9 and 1.5.3, this equals 1 − Pr( E 1 ∪ E 2 ) = 0 . 4. 6. Using the same notation as in Exercise 4, we now want Pr([ E 1 ∩ E c 2 ] ∪ [ E c 1 ∩ E 2 ]). These two events are disjoint, so Pr([ E 1 ∩ E c 2 ] ∪ [ E c 1 ∩ E 2 ]) = Pr( E 1 ∩ E c 2 ) + Pr( E c 1 ∩ E 2 ) . Use the reasoning from part (c) of Exercise 3 above to conclude that Pr( E 1 ∩ E c 2 ) = Pr( E 1 ) − Pr( E 1 ∩ E 2 ) = 0 . 4 , Pr( E c 1 ∩ E 2 ) = Pr( E 2 ) − Pr( E 1 ∩ E 2 ) = 0 . 1 . It follows that the probability we want is 0.5. 7. Rearranging terms in Eq. (1.5.1) of the text, we get Pr( A ∩ B ) = Pr( A ) + Pr( B ) − Pr( A ∪ B ) = 0 . 4 + 0 . 7 − Pr( A ∪ B ) = 1 . 1 − Pr( A ∪ B ) . So Pr( A ∩ B ) is largest when Pr( A ∪ B ) is smallest and vice-versa. The smallest possible value for Pr( A ∪ B ) occurs when one of the events is a subset of the other. In the present exercise this could only happen if A ⊂ B , in which case Pr( A ∪ B ) = Pr( B ) = 0 . 7, and Pr( A ∩ B ) = 0 . 4. The largest possible value of Pr( A ∪ B ) occurs when either A and B are disjoint or when A ∪ B = S . The former is not possible since the probabilities are too large, but the latter is possible. In this case Pr( A ∪ B ) = 1 and Pr( A ∩ B ) = 0 . 1. 8. Let A be the event that a randomly selected family subscribes to the morning paper, and let B be the event that a randomly selected family subscribes to the afternoon paper. We are told that Pr( A ) = 0 . 5, Pr( B ) = 0 . 65, and Pr( A ∪ B ) = 0 . 85. We are asked to find Pr( A ∩ B ). Using Theorem 1.5.7 in the text we obtain Pr( A ∩ B ) = Pr( A ) + Pr( B ) − Pr( A ∪ B ) = 0 . 5 + 0 . 65 − 0 . 85 = 0 . 3 .

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    Section 1.5. The Definition of Probability 5 9. The required probability is Pr( A ∩ B C ) + Pr( A C B ) = [Pr( A ) − Pr( A ∩ B )] + [Pr( B ) − Pr( A ∩ B )] = Pr( A ) + Pr( B ) − 2 Pr( A ∩ B ) . 10. Theorem 1.4.11 says that A = ( A ∩ B ) ∪ ( A ∩ B c ). Clearly the two events A ∩ B and A ∩ B c are disjoint. It follows from Theorem 1.5.6 that Pr( A ) = Pr( A ∩ B ) + Pr( A ∩ B c ). 11. (a) The set of points for which ( x − 1 / 2) 2 + ( y − 1 / 2) 2 < 1 / 4 is the interior of a circle that is contained in the unit square. (Its center is (1 / 2 , 1 / 2) and its radius is 1/2.) The area of this circle is π/ 4, so the area of the remaining region (what we want) is 1 − π/ 4. (b) We need the area of the region between the two lines y = 1 / 2 − x and y = 3 / 2 − x . The remaining area is the union of two right triangles with base and height both equal to 1/2. Each triangle has area 1/8, so the region between the two lines has area 1 − 2 / 8 = 3 / 4. (c) We can use calculus to do this. We want the area under the curve y = 1 − x 2 between x = 0 and x = 1. This equals ∫ 1 0 (1 − x 2 ) dx = x − x 3 3 ∣ ∣ ∣ ∣ ∣ 1 x =0 = 2 3 . (d) The area of a line is 0, so the probability of a line segment is 0. 12. The events B 1 , B 2 , . . . are disjoint, because the event B 1 contains the points in A 1 , the event B 2 contains the points in A 2 but not in A 1 , the event B 3 contains the points in A 3 but not in A 1 or A 2 , etc. By this same reasoning, it is seen that ∪ n i =1 A i = ∪ n i =1 B i and ∪ ∞ i =1 A i = ∪ ∞ i =1 B i . Therefore, Pr ( n ⋃ i =1 A i ) = Pr ( n ⋃ i =1 B i ) and Pr ( ∞ ⋃ i =1 A i ) = Pr ( ∞ ⋃ i =1 B i ) . However, since the events B 1 , B 2 , . . . are disjoint, Pr ( n ⋃ i =1 B i ) = n ∑ i =1 P r ( B i ) and Pr ( ∞ ⋃ i =1 B i ) = ∞ ∑ i =1 Pr( B i ) . 13. We know from Exercise 12 that Pr ( n ⋃ i =1 A i ) = n ∑ i =1 Pr( B i ) .

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    6 Chapter 1. Introduction to Probability Furthermore, from the definition of the events B 1 , . . . , B n it is seen that B i ⊂ A i for i = 1 , . . . , n . Therefore, by Theorem 1.5.4, Pr( B i ) ≤ Pr( A i ) for i = 1 , . . . , n . It now follows that Pr ( n ⋃ i =1 A i ) ≤ n ∑ i =1 Pr( A i ) . (Of course, if the events A 1 , . . . , A n are disjoint, there is equality in this relation.) For the second part, apply the first part with A i replaced by A c i for i = 1 , . . . , n . We get Pr (⋃ A c i ) ≤ n ∑ i =1 Pr( A c i ) . (S.1.1) Exercise 5 in Sec. 1.4 says that the left side of (S.1.1) is Pr ([ ⋂ A i ] c ). Theorem 1.5.3 says that this last probability is 1 − Pr ( ⋂ A i ). Hence, we can rewrite (S.1.1) as 1 − Pr (⋂ A i ) ≤ n ∑ i =1 Pr( A c i ) . Finally take one minus both sides of the above inequality (which reverses the inequality) and produces the desired result. 14. First, note that the probability of type AB blood is 1 − (0 . 5+0 . 34+0 . 12) = 0 . 04 by using Theorems 1.5.2 and 1.5.3. (a) The probability of blood reacting to anti-A is the probability that the blood is either type A or type AB. Since these are disjoint events, the probability is the sum of the two probabilities, namely 0 . 34 + 0 . 04 = 0 . 38. Similarly, the probability of reacting with anti-B is the probability of being either type B or type AB, 0 . 12 + 0 . 04 = 0 . 16. (b) The probability that both antigens react is the probability of type AB blood, namely 0.04. 1.6 Finite Sample Spaces Solutions to Exercises 1. The safe way to obtain the answer at this stage of our development is to count that 18 of the 36 outcomes in the sample space yield an odd sum. Another way to solve the problem is to note that regardless of what number appears on the first die, there are three numbers on the second die that will yield an odd sum and three numbers that will yield an even sum. Either way the probability is 1/2. 2. The event whose probability we want is the complement of the event in Exercise 1, so the probability is also 1/2. 3. The only differences greater than or equal to 3 that are available are 3, 4 and 5. These large difference only occur for the six outcomes in the upper right and the six outcomes in the lower left of the array in Example 1.6.5 of the text. So the probability we want is 1 − 12 / 36 = 2 / 3. 4. Let x be the proportion of the school in grade 3 (the same as grades 2–6). Then 2 x is the proportion in grade 1 and 1 = 2 x + 5 x = 7 x . So x = 1 / 7, which is the probability that a randomly selected student will be in grade 3.

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    Section 1.7. Counting Methods 7 5. The probability of being in an odd-numbered grade is 2 x + x + x = 4 x = 4 / 7. 6. Assume that all eight possible combinations of faces are equally likely. Only two of those combinations have all three faces the same, so the probability is 1 / 4. 7. The possible genotypes of the offspring are aa and Aa , since one parent will definitely contribute an a , while the other can contribute either A or a . Since the parent who is Aa contributes each possible allele with probability 1/2 each, the probabilities of the two possible offspring are each 1/2 as well. 8. (a) The sample space contains 12 outcomes: (Head, 1), (Tail, 1), (Head, 2), (Tail, 2), etc. (b) Assume that all 12 outcomes are equally likely. Three of the outcomes have Head and an odd number, so the probability is 1/4. 1.7 Counting Methods Commentary If you wish to stress computer evaluation of probabilities, then there are programs for computing factorials and log-factorials. For example, in the statistical software R , there are functions factorial and lfactorial that compute these. If you cover Stirling’s formula (Theorem 1.7.5), you can use these functions to illustrate the closeness of the approximation. Solutions to Exercises 1. Each pair of starting day and leap year/no leap year designation determines a calendar, and each calendar correspond to exactly one such pair. Since there are seven days and two designations, there are a total of 7 × 2 = 14 different calendars. 2. There are 20 ways to choose the student from the first class, and no matter which is chosen, there are 18 ways to choose the student from the second class. No matter which two students are chosen from the first two classes, there are 25 ways to choose the student from the third class. The multiplication rule can be applied to conclude that the total number of ways to choose the three members is 20 × 18 × 25 = 9000. 3. This is a simple matter of permutations of five distinct items, so there are 5! = 120 ways. 4. There are six different possible shirts, and no matter what shirt is picked, there are four different slacks. So there are 24 different combinations. 5. Let the sample space consist of all four-tuples of dice rolls. There are 6 4 = 1296 possible outcomes. The outcomes with all four rolls different consist of all of the permutations of six items taken four at a time. There are P 6 , 4 = 360 of these outcomes. So the probability we want is 360 / 1296 = 5 / 18. 6. With six rolls, there are 6 6 = 46656 possible outcomes. The outcomes with all different rolls are the permutations of six distinct items. There are 6! = 720 outcomes in the event of interest, so the probability is 720 / 46656 = 0 . 01543. 7. There are 20 12 possible outcomes in the sample space. If the 12 balls are to be thrown into different boxes, the first ball can be thrown into any one of the 20 boxes, the second ball can then be thrown into any one of the other 19 boxes, etc. Thus, there are 20 · 19 · 18 · · · 9 possible outcomes in the event. So the probability is 20! / [8!20 12 ].

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    8 Chapter 1. Introduction to Probability 8. There are 7 5 possible outcomes in the sample space. If the five passengers are to get off at different floors, the first passenger can get off at any one of the seven floors, the second passenger can then get off at any one of the other six floors, etc. Thus, the probability is 7 · 6 · 5 · 4 · 3 7 5 = 360 2401 . 9. There are 6! possible arrangements in which the six runners can finish the race. If the three runners from team A finish in the first three positions, there are 3! arrangements of these three runners among these three positions and there are also 3! arrangements of the three runners from team B among the last three positions. Therefore, there are 3! × 3! arrangements in which the runners from team A finish in the first three positions and the runners from team B finish in the last three positions. Thus, the probability is (3!3!) / 6! = 1 / 20. 10. We can imagine that the 100 balls are randomly ordered in a list, and then drawn in that order. Thus, the required probability in part (a), (b), or (c) of this exercise is simply the probability that the first, fiftieth, or last ball in the list is red. Each of these probabilities is the same r 100 , because of the random order of the list. 11. In terms of factorials, P n,k = n ! / [ k !( n − k )!]. Since we are assuming that n and n = k are large, we can use Stirling’s formula to approximate both of them. The approximation to n ! is (2 π ) 1 / 2 n n +1 / 2 e − n , and the approximation to ( n − k )! is (2 π ) 1 / 2 ( n − k ) n − k +1 / 2 e − n + k . The approximation to the ratio is the ratio of the approximations because the ratio of each approximation to its corresponding factorial converges to 1. That is, n ! k !( n − k )! ≈ (2 π ) 1 / 2 n n +1 / 2 e − n k !(2 π ) 1 / 2 ( n − k ) n − k +1 / 2 e − n + k = e − k n k k ! ( 1 − k n ) − n − k − 1 / 2 . Further simplification is available if one assumes that k is small compared to n , that is k/n ≈ 0. In this case, the last factor is approximately e k , and the whole approximation simplifies to n k /k !. This makes sense because, if n/ ( n − k ) is essentially 1, then the product of the k largest factors in n ! is essentially n k . 1.8 Combinatorial Methods Commentary This section ends with an extended example called “The Tennis Tournament”. This is an application of combinatorics that uses a slightly subtle line of reasoning. Solutions to Exercises 1. We have to assign 10 houses to one pollster, and the other pollster will get to canvas the other 10 houses. Hence, the number of assignments is the number of combinations of 20 items taken 10 at a time, ( 20 10 ) = 184,756 . 2. The ratio of ( 93 30 ) to ( 93 31 ) is 31 / 63 < 1, so ( 93 31 ) is larger.

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    Section 1.8. Combinatorial Methods 9 3. Since 93 = 63 + 30, the two numbers are the same. 4. Let the sample space consist of all subsets (not ordered tuples) of the 24 bulbs in the box. There are ( 24 4 ) = 10626 such subsets. There is only one subset that has all four defectives, so the probability we want is 1 / 10626. 5. The number is 4251! (97!4154!) = ( 4251 97 ) , an integer. 6. There are ( n 2 ) possible pairs of seats that A and B can occupy. Of these pairs, n − 1 pairs comprise two adjacent seats. Therefore, the probability is n − 1 ( n 2 ) = 2 n . 7. There are ( n k ) possible sets of k seats to be occupied, and they are all equally likely. There are n − k +1 sets of k adjacent seats, so the probability we want is n − k + 1 ( n k ) = ( n − k + 1)! k ! n ! . 8. There are ( n k ) possible sets of k seats to be occupied, and they are all equally likely. Because the circle has no start or end, there are n sets of k adjacent seats, so the probability we want is n ( n k ) = ( n − k )! k ! ( n − 1)! . 9. This problem is slightly tricky. The total number of ways of choosing the n seats that will be occupied by the n people is ( 2 n n ) . Offhand, it would seem that there are only two ways of choosing these seats so that no two adjacent seats are occupied, namely: X0X0 . . . 0 and 0X0X . . . 0X Upon further consideration, however, n − 1 more ways can be found, namely: X00X0X . . . 0X, X0X00X0X . . . 0X, etc. Therefore, the total number of ways of choosing the seats so that no two adjacent seats are occupied is n + 1. The probability is ( n + 1) / ( 2 n n ) .

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