Solution Manual for An Introduction to Mathematical Statistics and Its Applications, 6th Edition

INSTRUCTOR’S SOLUTIONS MANUAL A N I NTRODUCTION TO M ATHEMATICAL S TATISTICS AND I TS A PPLICATIONS SIXTH EDITION Richard Larsen Vanderbilt University Morris Marx University of West Florida The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Reproduced by Pearson from electronic files supplied by the author. Copyright © 2018, 2012, 2006 Pearson Education, Inc. Publishing as Pearson, 330 Hudson Street, NY NY 10013 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher. Printed in the United States of America. ISBN-13: 978-0-13-411427-9 ISBN-10: 0-13-411427-2 Contents Chapter 2: Probability ………………………………………………………………………………………………..1 2.2 2.3 2.4 2.5 2.6 2.7 Samples Spaces and the Algebra of Sets ………………………………………………………………….. 1 The Probability Function ……………………………………………………………………………………….. 6 Conditional Probability …………………………………………………………………………………………. 7 Independence ……………………………………………………………………………………………………… 13 Combinatorics ……………………………………………………………………………………………………. 17 Combinatorial Probability ……………………………………………………………………………………. 24 Chapter 3: Random Variables ……………………………………………………………………………………27 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 Binomial and Hypergeometric Probabilities …………………………………………………………… 27 Discrete Random Variables ………………………………………………………………………………….. 36 Continuous Random Variables ……………………………………………………………………………… 41 Expected Values …………………………………………………………………………………………………. 44 The Variance ……………………………………………………………………………………………………… 52 Joint Densities ……………………………………………………………………………………………………. 57 Transforming and Combining Random Variables……………………………………………………. 69 Further Properties of the Mean and Variance………………………………………………………….. 73 Order Statistics …………………………………………………………………………………………………… 79 Conditional Densities ………………………………………………………………………………………….. 82 Moment-Generating Functions……………………………………………………………………………… 88 Chapter 4: Special Distributions ………………………………………………………………………………..93 4.2 4.3 4.4 4.5 4.6 The Poisson Distribution ……………………………………………………………………………………… 93 The Normal Distribution ……………………………………………………………………………………… 98 The Geometric Distribution………………………………………………………………………………… 105 The Negative Binomial Distribution ……………………………………………………………………. 107 The Gamma Distribution ……………………………………………………………………………………. 109 Chapter 5: Estimation……………………………………………………………………………………………..111 5.2 5.3 5.4 5.5 5.6 5.7 5.8 Estimating Parameters: The Method of Maximum Likelihood and Method of Moments ……………………………………………………………………………………. 111 Interval Estimation ……………………………………………………………………………………………. 118 Properties of Estimators …………………………………………………………………………………….. 123 Minimum-Variance Estimators: The Cramér-Rao Lower Bound …………………………….. 128 Sufficient Estimators …………………………………………………………………………………………. 130 Consistency ……………………………………………………………………………………………………… 133 Bayesian Estimation ………………………………………………………………………………………….. 135 Copyright © 2018 Pearson Education, Inc. iii Chapter 6: Hypothesis Testing …………………………………………………………………………………137 6.2 6.3 6.4 6.5 The Decision Rule …………………………………………………………………………………………….. 137 Testing Binomial Data……………………………………………………………………………………………………………………………………………. 138 Type I and Type II Errors …………………………………………………………………………………… 139 A Notion of Optimality: The Generalized Likelihood Ratio ……………………………………. 144 Chapter 7: Inferences Based on the Normal Distribution …………………………………………….147 7.3 Deriving the Distribution of SY/ n ………………………………………………………………………… 147 7.4 Drawing Inferences About  …………………………………………………………………………….. 150 7.5 Drawing Inferences About  2 ……………………………………………………………………………. 156 Chapter 8: Types of Data: A Brief Overview …………………………………………………………….161 8.2 Classifying Data ……………………………………………………………………………………………….. 161 Chapter 9: Two-Sample Inference ……………………………………………………………………………163 9.2 Testing H 0 :  X  Y …………………………………………………………………………………………. 163 9.3 9.4 9.5 Testing H 0 :  X2   Y2 —The F Test ……………………………………………………………………… 166 Binomial Data: Testing H 0 : p X  pY ………………………………………………………………….. 168 Confidence Intervals for the Two-Sample Problem ……………………………………………….. 170 Chapter 10: Goodness-of-Fit Tests …………………………………………………………………………..173 10.2 10.3 10.4 10.5 The Multinomial Distribution …………………………………………………………………………….. 173 Goodness-of-Fit Tests: All Parameters Known ……………………………………………………… 175 Goodness-of-Fit Tests: Parameters Unknown ……………………………………………………….. 178 Contingency Tables …………………………………………………………………………………………… 185 Chapter 11: Regression …………………………………………………………………………………………..189 11.2 11.3 11.4 11.5 The Method of Least Squares ……………………………………………………………………………… 189 The Linear Model ……………………………………………………………………………………………… 199 Covariance and Correlation ………………………………………………………………………………… 204 The Bivariate Normal Distribution ………………………………………………………………………. 208 Chapter 12: The Analysis of Variance ………………………………………………………………………211 12.2 12.3 The F test …………………………………………………………………………………………………………. 211 Multiple Comparisons: Tukey’s Method………………………………………………………………. 214 Copyright © 2018 Pearson Education, Inc. iv 12.4 Testing Subhypotheses with Constrasts ……………………………………………………………….. 216 12.5 Data Transformations ………………………………………………………………………………………… 218 /( k 1) Appendix 12.A.2 The Distribution of SSTR SSE /( n k ) When H1 Is True …………………………………….. 218 Chapter 13: Randomized Block Designs …………………………………………………………………..221 13.2 13.3 The F Test for a Randomized Block Design …………………………………………………………. 221 The Paired t Test……………………………………………………………………………………………….. 224 Chapter 14: Nonparametric Statistics………………………………………………………………………..229 14.2 14.3 14.4 14.5 14.6 The Sign Test …………………………………………………………………………………………………… 229 Wilcoxon Tests …………………………………………………………………………………………………. 232 The Kruskal-Wallis Test ……………………………………………………………………………………. 236 The Friedman Test…………………………………………………………………………………………….. 239 Testing for Randomness …………………………………………………………………………………….. 241 Copyright © 2018 Pearson Education, Inc. v Chapter 2: Probability Section 2.2: Sample Spaces and the Algebra of Sets 2.2.1 S = ( s, s, s ), ( s, s, f ), ( s, f , s ), ( f , s, s ), ( s, f , f ), ( f , s, f ), ( f , f , s ), ( f , f , f ) A =  ( s, f , s ), ( f , s, s ) ; B = ( f , f , f ) 2.2.2 Let (x, y, z) denote a red x, a blue y, and a green z. Then A = (2, 2,1), (2,1, 2), (1, 2, 2), (1,1,3), (1,3,1), (3,1,1) 2.2.3 (1,3,4), (1,3,5), (1,3,6), (2,3,4), (2,3,5), (2,3,6) 2.2.4 There are 16 ways to get an ace and a 7, 16 ways to get a 2 and a 6, 16 ways to get a 3 and a 5, and 6 ways to get two 4’s, giving 54 total. 2.2.5 The outcome sought is (4, 4). It is “harder” to obtain than the set {(5, 3), (3, 5), (6, 2), (2, 6)} of other outcomes making a total of 8. 2.2.6 The set N of five card hands in hearts that are not flushes are called straight flushes. These are five cards whose denominations are consecutive. Each one is characterized by the lowest value in the hand. The choices for the lowest value are A, 2, 3, …, 10. (Notice that an ace can be high or low). Thus, N has 10 elements. 2.2.7 P = {right triangles with sides (5, a, b): a2 + b2 = 25} 2.2.8 A = {SSBBBB, SBSBBB, SBBSBB, SBBBSB, BSSBBB, BSBSBB, BSBBSB, BBSSBB, BBSBSB, BBBSSB} 2.2.9 (a) S = {(0, 0, 0, 0) (0, 0, 0, 1), (0, 0, 1, 0), (0, 0, 1, 1), (0, 1, 0, 0), (0, 1, 0, 1), (0, 1, 1, 0), (0, 1, 1, 1), (1, 0, 0, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 0, 1, 1, ), (1, 1, 0, 0), (1, 1, 0, 1), (1, 1, 1, 0), (1, 1, 1, 1, )} (b) A = {(0, 0, 1, 1), (0, 1, 0, 1), (0, 1, 1, 0), (1, 0, 0, 1), (1, 0, 1, 0), (1, 1, 0, 0, )} (c) 1 + k 2.2.10 (a) S = {(1, 1), (1, 2), (1, 4), (2, 1), (2, 2), (2, 4), (4, 1), (4, 2), (4, 4)} (b) {2, 3, 4, 5, 6, 8} 2.2.11 Let p1 and p2 denote the two perpetrators and i1, i2, and i3, the three in the lineup who are innocent. Then S = ( p1 , i1 ), ( p1 , i2 ), ( p1 , i3 ), ( p2 , i1 ), ( p2 , i2 ), ( p2 , i3 ), ( p1 , p2 ), (i1 , i2 ), (i1 , i3 ), (i2 , i3 ) . The event A contains every outcome in S except (p1, p2). 2.2.12 The quadratic equation will have complex roots—that is, the event A will occur—if b2  4ac < 0. 2.2.13 In order for the shooter to win with a point of 9, one of the following (countably infinite) sequences of sums must be rolled: (9,9), (9, no 7 or no 9,9), (9, no 7 or no 9, no 7 or no 9,9), … Copyright © 2018 Pearson Education, Inc. 1 2 Chapter 2: Probability 2.2.14 Let (x, y) denote the strategy of putting x white chips and y red chips in the first urn (which results in 10  x white chips and 10  y red chips being in the second urn). Then S = ( x, y ) : x  0,1,…,10, y  0,1,…,10, and 1  x  y  19 . Intuitively, the optimal strategies are (1, 0) and (9, 10). 2.2.15 Let Ak be the set of chips put in the urn at 1/2k minute until midnight. For example, A1 = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. Then the set of chips in the urn at midnight is   ( A  {k  1})   . k k 1 2.2.16 move arrow on first figure raise B by 1 2.2.17 If x2 + 2x  8, then (x + 4)(x  2)  0 and A = {x: 4  x  2}. Similarly, if x2 + x  6, then (x + 3)(x  2)  0 and B = {x: 3  x  2). Therefore, A  B = {x: 3  x  2} and A  B = {x: 4  x  2}. 2.2.18 A  B  C = {x: x = 2, 3, 4} 2.2.19 The system fails if either the first pair fails or the second pair fails (or both pairs fail). For either pair to fail, though, both of its components must fail. Therefore, A = (A11  A21)  (A12  A22). 2.2.20 (a) (b) (c) empty set (d) 2.2.21 40 2.2.22 (a) {E1, E2} 2.2.23 (a) If s is a member of A  (B  C) then s belongs to A or to B  C. If it is a member of A or of B  C, then it belongs to A  B and to A  C. Thus, it is a member of (A  B)  (A  C). Conversely, choose s in (A  B)  (A  C). If it belongs to A, then it belongs to A  (B  C). If it does not belong to A, then it must be a member of B  C. In that case it also is a member of A  (B  C). (b) {S1, S2, T1, T2} (c) {A, I} Copyright © 2018 Pearson Education, Inc. Section 2.2: Sample Spaces and the Algebra of Sets 3 (b) If s is a member of A  (B  C) then s belongs to A and to B  C. If it is a member of B, then it belongs to A  B and, hence, (A  B)  (A  C). Similarly, if it belongs to C, it is a member of (A  B)  (A  C). Conversely, choose s in (A  B)  (A  C). Then it belongs to A. If it is a member of A  B then it belongs to A  (B  C). Similarly, if it belongs to A  C, then it must be a member of A  (B  C). 2.2.24 Let B = A1  A2  …  Ak. Then A1C  A2C  …  AkC = (A1  A2  … Ak)C = BC. Then the expression is simply B  BC = S. 2.2.25 (a) Let s be a member of A  (B  C). Then s belongs to either A or B  C (or both). If s belongs to A, it necessarily belongs to (A  B)  C. If s belongs to B  C, it belongs to B or C or both, so it must belong to (A  B)  C. Now, suppose s belongs to (A  B)  C. Then it belongs to either A  B or C or both. If it belongs to C, it must belong to A  (B  C). If it belongs to A  B, it must belong to either A or B or both, so it must belong to A  (B  C). (b) Suppose s belongs to A  (B  C), so it is a member of A and also B  C. Then it is amember of A and of B and C. That makes it a member of (A  B)  C. Conversely, if s is a member of (A  B)  C, a similar argument shows it belongs to A  (B  C). 2.2.26 (a) AC  BC  CC (b) A  B  C (c) A  BC  CC (d) (A  BC  CC)  (AC  B  CC)  (AC  BC  C) (e) (A  B  CC)  (A  BC  C)  (AC  B  C) 2.2.27 A is a subset of B. 2.2.28 (a) {0}  {x: 5  x  10} (b) {x: 3  x < 5} (c) {x: 0 < x  7} (d) {x: 0 < x < 3} (e) {x: 3  x  10} (f) {x: 7 < x  10} 2.2.29 (a) B and C (b) B is a subset of A. 2.2.30 (a) A1  A2  A3 (b) A1  A2  A3 The second protocol would be better if speed of approval matters. For very important issues, the first protocol is superior. 2.2.31 Let A and B denote the students who saw the movie the first time and the second time, respectively. Then N(A) = 850, N(B) = 690, and N [( A  B )C ] = 4700 (implying that N(A  B) = 1300). Therefore, N(A  B) = number who saw movie twice = 850 + 690  1300 = 240. Copyright © 2018 Pearson Education, Inc. 4 Chapter 2: Probability 2.2.32 (a) (b) 2.2.33 (a) (b) 2.2.34 (a) A  (B  C) (A  B)  C A  (B  C) (A  B)  C (b) 2.2.35 A and B are subsets of A  B. Copyright © 2018 Pearson Education, Inc. Section 2.2: Sample Spaces and the Algebra of Sets 2.2.36 5 (a) ( A  B C ) C = AC  B (b) B  ( A  B )C  AC  B (c) A  ( A  B )C  A  B C 2.2.37 Let A be the set of those with MCAT scores  27 and B be the set of those with GPAs  3.5. We are given that N(A) = 1000, N(B) = 400, and N(A  B) = 300. Then N ( AC  B C ) = N [( A  B )C ] = 1200  N(A  B) = 1200  [(N(A) + N(B)  N(A  B)] = 1200  [(1000 + 400  300] = 100. The requested proportion is 100/1200. 2.2.38 N(A  B  C) = N(A) + N(B) + N(C)  N(A  B)  N(A  C)  N(B  C) + N(A  B  C) 2.2.39 Let A be the set of those saying “yes” to the first question and B be the set of those saying “yes” to the second question. We are given that N(A) = 600, N(B) = 400, and N(AC  B) = 300. Then N(A  B) = N(B)  N ( AC  B ) = 400  300 = 100. N ( A  B C ) = N(A)  N(A  B) = 600  100 = 500. 2.2.40 N [( A  B )C ] = 120  N(A  B) = 120  [N( AC  B) + N(A  B C ) + N(A  B)] = 120  [50 + 15 + 2] = 53 Copyright © 2018 Pearson Education, Inc. 6 Chapter 2: Probability Section 2.3: The Probability Function 2.3.1 Let L and V denote the sets of programs with offensive language and too much violence, respectively. Then P(L) = 0.42, P(V) = 0.27, and P(L  V) = 0.10. Therefore, P(program complies) = P((L  V)C) = 1  [P(L) + P(V)  P(L  V)] = 0.41. 2.3.2 P(A or B but not both) = P(A  B)  P(A  B) = P(A) + P(B)  P (A  B)  P(A  B) = 0.4 + 0.5  0.1  0.1 = 0.7 2.3.3 (a) 1  P(A  B) (b) P(B)  P(A  B) 2.3.4 P(A  B) = P(A) + P(B)  P(A  B) = 0.3; P(A)  P(A  B) = 0.1. Therefore, P(B) = 0.2. 3 2.3.5 5 1 No. P(A1  A2  A3) = P(at least one “6” appears) = 1  P(no 6’s appear) = 1     . 6 2 The Ai’s are not mutually exclusive, so P(A1  A2  A3)  P(A1) + P(A2) + P(A3). 2.3.6 P(A or B but not both) = 0.5 – 0.2 = 0.3 2.3.7 By inspection, B = (B  A1)  (B  A2)  …  (B  An). 2.3.8 (a) (b) (b) Copyright © 2018 Pearson Education, Inc. Section 2.4: Conditional Probability 7 2 3  8 4 2.3.9 P(odd man out) = 1  P(no odd man out) = 1  P(HHH or TTT) = 1  2.3.10 A = {2, 4, 6, …, 24}; B = {3, 6, 9, …, 24); A  B = {6, 12, 18, 24}. 12 8 4 16    . Therefore, P(A  B) = P(A) + P(B)  P(A  B) = 24 24 24 24 2.3.11 Let A: State wins Saturday and B: State wins next Saturday. Then P(A) = 0.10, P(B) = 0.30, and P(lose both) = 0.65 = 1  P(A  B), which implies that P(A  B) = 0.35. Therefore, P(A  B) = 0.10 + 0.30  0.35 = 0.05, so P(State wins exactly once) = P(A  B)  P(A  B) = 0.35  0.05 = 0.30. 2.3.12 Since A1 and A2 are mutually exclusive and cover the entire sample space, p1 + p2 = 1. 1 5 But 3p1  p2 = , so p2 = . 2 8 2.3.13 Let F: female is hired and T: minority is hired. Then P(F) = 0.60, P(T) = 0.30, and P(FC  TC) = 0.25 = 1  P(F  T). Since P(F  T) = 0.75, P(F  T) = 0.60 + 0.30  0.75 = 0.15. 2.3.14 The smallest value of P[(A  B C)C] occurs when P(A  B  C) is as large as possible. This, in turn, occurs when A, B, and C are mutually disjoint. The largest value for P(A  B  C) is P(A) + P(B) + P(C) = 0.2 + 0.1 + 0.3 = 0.6. Thus, the smallest value for P[(A  B  C)C] is 0.4. 2.3.15 (a) XC  Y = {(H, T, T, H), (T, H, H, T)}, so P(XC  Y) = 2/16 (b) X  YC = {(H, T, T, T), (T, T, T, H), (T, H, H, H), (H, H, H, T)} so P(X  YC) = 4/16 2.3.16 A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} A  BC = {(1, 5), (3, 3), (5, 1)}, so P(A  BC) = 3/36 = 1/12. 2.3.17 A  B, (A  B)  (A  C), A, A  B, S 2.3.18 Let A be the event of getting arrested for the first scam; B, for the second. We are given P(A) = 1/10, P(B) = 1/30, and P(A  B) = 0.0025. Her chances of not getting arrested are P[(A  B)C] = 1  P(A  B) = 1  [P(A) + P(B)  P(A  B)] = 1  [1/10 + 1/30  0.0025] = 0.869 Section 2.4: Conditional Probability 2.4.1 P(sum  10 and sum exceeds 8) P (sum exceeds 8) P (sum  10) 3 / 36 3   . = P (sum  9, 10, 11, or 12) 4 / 36  3 / 36  2 / 36  1 / 36 10 P(sum = 10|sum exceeds 8) = Copyright © 2018 Pearson Education, Inc. 8 Chapter 2: Probability 2.4.2 P(A|B) + P(B|A) = 0.75 = P ( A  B) P ( A  B ) 10 P ( A  B )    5 P ( A  B) , which implies 4 P( B) P ( A) that P(A  B) = 0.1. P( A  B)  P ( A) , then P(A  B)

0. (b) No, because P(A  B) = 0.2  P(A)  P(B) = (0.6)(0.5) = 0.3 (c) P(AC  BC) = P((A  B)C) = 1  P(A  B) = 1  0.2 = 0.8. 2.5.2 Let C and M be the events that Spike passes chemistry and mathematics, respectively. Since P(C  M) = 0.12  P(C)  P(M) = (0.35)(0.40) = 0.14, C and M are not independent. Copyright © 2018 Pearson Education, Inc. 14 Chapter 2: Probability P(Spike fails both) = 1  P(Spike passes at least one) = 1  P(C  M) = 1  [P(C) + P(M)  P(C  M)] = 0.37. 6 . 36 2.5.3 P(one face is twice the other face) = P((1, 2), (2, 1), (2, 4), (4, 2), (3, 6), (6, 3)) = 2.5.4 Consider the probability of the complementary event that they have the same blood types: 0.42  0.012  0.052  0.452  0.375 . Then the probability they have different blood types is 1 – 0.375 = 0.625. 2.5.5 P(Dana wins at least 1 game out of 2) = 0.3, which implies that P(Dana loses 2 games out of 2) = 0.7. Therefore, P(Dana wins at least 1 game out of 4) = 1  P(Dana loses all 4 games) = 1  P(Dana loses first 2 games and Dana loses second 2 games) = 1  (0.7)(0.7) = 0.51. 2.5.6 Six equally-likely orderings are possible for any set of three distinct random numbers: x1 < x2 < x3, x1 < x3 < x2, x2 < x1 < x3, x2 < x3 < x1, x3 < x1 < x2, and x3 < x2 0, so A, B, and C are not mutually independent. However, P(A  B) = 36 3 3 9 3 18 9 = P(A)  P(B) =  , P(A  C) = = P(A)  P(C) =  , and P(B  C) = = P(B)  6 6 36 6 36 36 3 18 P(C) =  , so A, B, and C are pairwise independent. 6 36 2.5.16 Let Ri and Gi be the events that the ith light is red and green, respectively, i = 1, 2, 3, 4. Then 1 1 P(R1) = P(R2) = and P(R3) = P(R4) = . Because of the considerable distance between the 3 2 intersections, what happens from light to light can be considered independent events. P(driver stops at least 3 times) = P(driver stops exactly 3 times) + P(driver stops all 4 times) = P((R1  R2  R3  G4)  (R1  R2  G3  R4)  (R1  G2  R3  R4)  1  1  1  1   1  1  1  1   (G1  R2  R3  R4)  (R1  R2  R3  R4)) =             3  3  2  2   3  3  2  2   1  2  1  1   2  1  1  1   1  1  1  1  7 . +                    3  3  2  2   3  3  2  2   3  3  2  2  36 2.5.17 Let M, L, and G be the events that a student passes the mathematics, language, and general 6175 7600 8075 knowledge tests, respectively. Then P(M) = , P(L) = , and P(G) = . 9500 9500 9500 P(student fails to qualify) = P(student fails at least one exam) = 1  P(student passes all three exams) = 1  P(M  L  G) = 1  P(M)  P(L)  P(G) = 0.56. 2.5.18 Let Ai denote the event that switch Ai closes, i = 1, 2, 3, 4. Since the Ai’s are independent events, P(circuit is completed) = P((A1  A2)  (A3  A4)) = P(A1  A2) + P(A3  A4)  P((A1  A2)  (A3  A4)) = 2p2  p4. 2.5.19 Let p be the probability of having a winning game card. Then 0.32 = P(winning at least once in 5 tries) = 1  P(not winning in 5 tries) = 1  (1  p)5, so p = 0.074 Copyright © 2018 Pearson Education, Inc. 16 Chapter 2: Probability 2.5.20 Let AH, AT, BH, BT, CH, and CT denote the events that players A, B, and C throw heads and tails on individual tosses. Then P(A throws first head) = P(AH  (AT  BT  CT  AH)  ) 2  4 1 11 11 1 1 =           . 2 28 28 2 1  1/ 8  7 Similarly, P(B throws first head) = P((AT  BH)  (AT  BT  CT  AT  BH)  …) 2  2 1 11 11 1 1 =        …    . 4 48 48 4 1  1/ 8  7 P(C throws first head) = 1  4 2 1   . 7 7 7 2.5.21 P(at least one child becomes adult) = 1  P(no child becomes adult) = 1  0.8n . ln 0.25 Then 1  0.8n  0.75 implies n  or n ≥ 6.2 , so take n = 7. ln 0.8 2.5.22 P(at least one viewer can name actor) = 1  P(no viewer can name actor) = 1  (0.85)10 = 0.80. 2.5.23 Let B be the event that no heads appear, and let Ai be the event that i coins are tossed, 2 6 6 11 1 1  1   1  63 . i = 1, 2, …, 6. Then P(B) =  P ( B | Ai ) P( Ai )          …       26  2 6  2   6  384 i 1 2.5.24 P(at least one red chip is drawn from at least one urn) = 1  P(all chips drawn are white) r r r rm 4 4 4 4 = 1          1    . 7 7 7 7 n 2.5.25  35  P(at least one double six in n throws) = 1  P(no double sixes in n throws) = 1    . By  36  trial and error, the smallest n for which P(at least one double six in n throws) exceeds 0.50 is 24 25  35   35  25 [1    = 0.49; 1    = 0.51].  36   36  2.5.26 Let A be the event that a sum of 8 appears before a sum of 7. Let B be the event that a sum of 8 appears on a given roll and let C be the event that the sum appearing on a given roll is 5 25 , P(C) = , and P(A) = P(B) + P(C)P(B) + P(C)P(C)P(B) neither 7 nor 8. Then P(B) = 36 36 2 k  5 5 25 5  25  5 5   25  5 1 +  =        .     36 36 36  36  36 36 k  0  36  36  1  25 / 36  11 2.5.27 Let W, B, and R denote the events of getting a white, black and red chip, respectively, on a given draw. Then P(white appears before red) = P(W  (B  W)  (B  B  W)  ) 2 w b w b w   =       wbr wbr wbr  wbr  wbr =   w 1 w  .  w  b  r  1  b / (w  b  r )  w  r Copyright © 2018 Pearson Education, Inc. Section 2.6: Combinatorics 17 2.5.28 P(B|A1) = 1  P(all m I-teams fail) = 1  (1  r)m; similarly, P(B|A2) = 1  (1  r)nm. From Theorem 2.4.1, P(B) = [1  (1  r)m]p + [1  (1  r)nm](1  p). Treating m as a continuous variable and differentiating P(B) gives dP ( B ) dP ( B ) = p(1  r)mln(1  r) + (1  p)(1  r)nm ln(1  r). Setting = 0 implies that dm dm n ln[(1  p ) / p ] m=  . 2 2ln(1  r ) 2.5.29 P(at least one four) = 1  P(no fours) = 1  (0.9)n. 1  (0.9)n  0.7 implies n = 12 2.5.30 Let B be the event that all n tosses come up heads. Let A1 be the event that the coin has two heads, and let A2 be the event the coin is fair. Then (1 / 2) n (8 / 9) 8(1 / 2) n  1(1 / 9)  (1 / 2) n (8 / 9) 1  8(1 / 2) n By inspection, the limit of P ( A2 | B ) as n goes to infinity is 0. P ( A2 | B )  2.5.31 Assume the events that Stanley answers correctly from question to question are independent. (Is this a reasonable assumption?) The probability of answering at least one question on Final A is (0.45)(0.55) + (0.55)(0.45) + (0.45)(0.45) = 0.6975 The probability of answering at least one question on Final B is (0.30)(0.70)(0.70) + ((0.70)(0.30)(0.70) +(0.30)(0.30)(0.70) + (0.30)(0.30)(0.70) + (0.30)(0.70)(0.30) + (0.70)(0.30)(0.30) + (0.30)(0.30)(0.30) = 0.657 A simpler way to answer the question is to take the complement of the event that he answers none correctly. For Final A this is 1  (0.55) 2 = 0.6975. For Final B this is 1  (0.70)3 = 0.657. With either method of solution, Stanley will be a bit better off taking Final A. 2.5.32 Take the complementary event that none of n switches opens. This probability is 0.4n . Then 0.04n  0.02 implies nln(0.04) ≤ ln(0.02) or n ≥ ln(0.02)/ln(0.04) = 4.27. So the smallest n is 5. Section 2.6: Combinatorics 2.6.1 2  3  2  2 = 24 2.6.2 20  9  6  20 = 21,600 2.6.3 3  3  5 = 45. Included will be aeu and cdx. 2.6.4 (a) 262  104 = 6,760,000 (b) 262  10  9  9  8  7 = 3,407,040 (c) The total number of plates with four zeros is 26  26, so the total number not having four zeros must be 262  104  262 = 6,759,324. Copyright © 2018 Pearson Education, Inc. 18 Chapter 2: Probability 2.6.5 There are 9 choices for the first digit (1 through 9), 9 choices for the second digit (0 + whichever eight digits are not appearing in the hundreds place), and 8 choices for the last digit. The number of admissible integers, then, is 9  9  8 = 648. For the integer to be odd, the last digit must be either 1, 3, 5, 7, or 9. That leaves 8 choices for the first digit and 8 choices for the second digit, making a total of 320 (= 8  8  5) odd integers. 2.6.6 For each topping, the customer has 2 choices: “add” or “do not add.” The eight available toppings, then, can produce a total of 28 = 256 different hamburgers. 2.6.7 The bases can be occupied in any of 27 ways (each of the seven can be either “empty” or “occupied”). Moreover, the batter can come to the plate facing any of five possible “out” situations (0 through 4). It follows that the number of base-out configurations is 5  27, or 640. 2.6.8 (a) There are 3 choices for the leading digit—7, 8,9, There are 10 choices for each of the remaining eight places, for a total of 3  108 However, this count includes 7,000,000,000, so the answer is 3  108  1 . (b) Suppose the sequence starts with an even number. Counting 0 as even, there are 5 choices for each of the five even places and five choices for the four odd places, giving a total of 59 . But the same number occurs if the sequence starts with an odd number, so the answer is 2  59 (c) There are 5! choices for the first five digits. Then there are 6 choices for where the 2’s go, so the answer is 5! 6 . 2.6.9 4  14  6 + 4  6  5 + 14  6  5 + 4  14  5 = 1156 2.6.10 There are two mutually exclusive sets of ways for the black and white keys to alternate—the black keys can be 1st, 3rd, 5th, and 7th notes in the melody, or the 2nd, 4th 6th, and 8th. Since there are 5 black keys and 7 white keys, there are 5  7  5  7  5  7  5  7 variations in the first set and 7  5  7  5  7  5  7  5 in the second set. The total number of alternating melodies is the sum 54 7 4  7 4 54 = 3,001,250. 2.6.11 The number of usable garage codes is 28  1 = 255, because the “combination” where none of the buttons is pushed is inadmissible (recall Example 2.6.3). Five additional families can be added before the eight-button system becomes inadequate. 2.6.12 4, because 21 + 22 + 23 < 26 but 21 + 22 + 23 + 24  26. 2.6.13 In order to exceed 256, the binary sequence of coins must have a head in the ninth position and at least one head somewhere in the first eight tosses. The number of sequences satisfying those conditions is 28  1, or 255. (The “1” corresponds to the sequences TTTTTTTTH, whose value would not exceed 256.) 2.6.14 There are 3 choices for the vowel and 4 choices for the consonant, so there are 3  4 = 12 choices, if order doesn’t matter. If we are taking ordered arrangements, then there are 24 ways, since each unordered selection can be written vowel first or consonant first. 2.6.15 There are 1  3 ways if the ace of clubs is the first card and 12  4 ways if it is not. The total is then 3 + 12  4 = 51 Copyright © 2018 Pearson Education, Inc.