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

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    S OLUTIONS M ANUAL A N I NTRODUCTION TO M ATHEMATICAL S TATISTICS AND I TS A PPLICATIONS S IXTH E DITION Richard Larsen Vanderbilt University Morris Marx University of West Florida

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    iii Contents Chapter 2: Probability ..............................................................................................................1 2.2 Samples Spaces and the Algebra of Sets ............................................................................. 1 2.3 The Probability Function ..................................................................................................... 6 2.4 Conditional Probability ....................................................................................................... 7 2.5 Independence ..................................................................................................................... 13 2.6 Combinatorics ................................................................................................................... 17 2.7 Combinatorial Probability ................................................................................................. 24 Chapter 3: Random Variables ................................................................................................27 3.2 Binomial and Hypergeometric Probabilities ..................................................................... 27 3.3 Discrete Random Variables ............................................................................................... 36 3.4 Continuous Random Variables .......................................................................................... 41 3.5 Expected Values ................................................................................................................ 44 3.6 The Variance ..................................................................................................................... 52 3.7 Joint Densities ................................................................................................................... 57 3.8 Transforming and Combining Random Variables............................................................. 69 3.9 Further Properties of the Mean and Variance.................................................................... 73 3.10 Order Statistics .................................................................................................................. 79 3.11 Conditional Densities ........................................................................................................ 82 3.12 Moment-Generating Functions.......................................................................................... 88 Chapter 4: Special Distributions ............................................................................................93 4.2 The Poisson Distribution ................................................................................................... 93 4.3 The Normal Distribution ................................................................................................... 98 4.4 The Geometric Distribution............................................................................................. 105 4.5 The Negative Binomial Distribution ............................................................................... 107 4.6 The Gamma Distribution ................................................................................................. 109 Chapter 5: Estimation...........................................................................................................111 5.2 Estimating Parameters: The Method of Maximum Likelihood and Method of Moments ................................................................................................. 111 5.3 Interval Estimation .......................................................................................................... 118 5.4 Properties of Estimators .................................................................................................. 123 5.5 Minimum-Variance Estimators: The Cramér-Rao Lower Bound ................................... 128 5.6 Sufficient Estimators ....................................................................................................... 130 5.7 Consistency ..................................................................................................................... 133 5.8 Bayesian Estimation ........................................................................................................ 135

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    iv Chapter 6: Hypothesis Testing .............................................................................................137 6.2 The Decision Rule ........................................................................................................... 137 6.3 Testing Binomial Data ................................................................................................................................................................ 138 6.4 Type I and Type II Errors ................................................................................................ 139 6.5 A Notion of Optimality: The Generalized Likelihood Ratio ........................................... 144 Chapter 7: Inferences Based on the Normal Distribution ....................................................147 7.3 Deriving the Distribution of / Y S 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 0 : X Y H    ....................................................................................................... 163 9.3 Testing 2 2 0 : X Y H    —The F Test ................................................................................. 166 9.4 Binomial Data: Testing 0 : X Y H p p  ............................................................................. 168 9.5 Confidence Intervals for the Two-Sample Problem ........................................................ 170 Chapter 10: Goodness-of-Fit Tests ......................................................................................173 10.2 The Multinomial Distribution ......................................................................................... 173 10.3 Goodness-of-Fit Tests: All Parameters Known ............................................................... 175 10.4 Goodness-of-Fit Tests: Parameters Unknown ................................................................. 178 10.5 Contingency Tables ......................................................................................................... 185 Chapter 11: Regression ........................................................................................................189 11.2 The Method of Least Squares .......................................................................................... 189 11.3 The Linear Model ............................................................................................................ 199 11.4 Covariance and Correlation ............................................................................................. 204 11.5 The Bivariate Normal Distribution .................................................................................. 208 Chapter 12: The Analysis of Variance .................................................................................211 12.2 The F test ......................................................................................................................... 211 12.3 Multiple Comparisons: Tukey’s Method......................................................................... 214

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    v 12.4 Testing Subhypotheses with Constrasts .......................................................................... 216 12.5 Data Transformations ...................................................................................................... 218 Appendix 12.A.2 The Distribution of /( 1) /( ) SSTR k SSE n k   When 1 H Is True ............................................ 218 Chapter 13: Randomized Block Designs .............................................................................221 13.2 The F Test for a Randomized Block Design ................................................................... 221 13.3 The Paired t Test.............................................................................................................. 224 Chapter 14: Nonparametric Statistics...................................................................................229 14.2 The Sign Test .................................................................................................................. 229 14.3 Wilcoxon Tests ................................................................................................................ 232 14.4 The Kruskal-Wallis Test ................................................................................................. 236 14.5 The Friedman Test........................................................................................................... 239 14.6 Testing for Randomness .................................................................................................. 241

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    1 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 ): a 2 + b 2 = 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 p 1 and p 2 denote the two perpetrators and i 1 , i 2 , and i 3 , the three in the lineup who are innocent. Then S =   1 1 1 2 1 3 2 1 2 2 2 3 1 2 1 2 1 3 2 3 ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ), ( , ) p i p i p i p i p i p i p p i i i i i i . The event A contains every outcome in S except ( p 1 , p 2 ). 2.2.12 The quadratic equation will have complex roots—that is, the event A will occur—if b 2  4 ac < 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), …

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    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 =   ( , ) : 0,1,...,10, 0,1,...,10, and 1 19 x y x y x y      . Intuitively, the optimal strategies are (1, 0) and (9, 10). 2.2.15 Let A k be the set of chips put in the urn at 1/2 k minute until midnight . For example, A 1 = {11, 12, 13, 14, 15, 16, 17, 18, 19, 20}. Then the set of chips in the urn at midnight is 1 ( { 1}) k k A k        . 2.2.16 move arrow on first figure raise B by 1 2.2.17 If x 2 + 2 x  8, then ( x + 4)( x  2)  0 and A = { x :  4  x  2}. Similarly, if x 2 + 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 = ( A 11  A 21 )  ( A 12  A 22 ). 2.2.20 (a) (b) (c) empty set (d) 2.2.21 40 2.2.22 (a) { E 1, E 2} (b) { S 1, S 2, T 1, T 2} (c) { A , I } 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 ).

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    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 = A 1  A 2  …  A k . Then 1 2 ... C C C k A A A    = ( A 1  A 2  …  A k ) C = B C . Then the expression is simply B  B C = 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) A C  B C  C C (b) A  B  C (c) A  B C  C C (d) ( A  B C  C C )  ( A C  B  C C )  ( A C  B C  C ) (e) ( A  B  C C )  ( A  B C  C )  ( A C  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) A 1  A 2  A 3 (b) A 1  A 2  A 3 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 [( ) ] C N A B  = 4700 (implying that N ( A  B ) = 1300). Therefore, N ( A  B ) = number who saw movie twice = 850 + 690  1300 = 240.

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    4 Chapter 2: Probability 2.2.32 (a) (b) 2.2.33 (a) (b) 2.2.34 (a) A  ( B  C ) ( A  B )  C (b) A  ( B  C ) ( A  B )  C 2.2.35 A and B are subsets of A  B .

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    Section 2.2: Sample Spaces and the Algebra of Sets 5 2.2.36 (a) ( ) C C A B  = A C  B (b) ( ) C C B A B A B     (c) ( ) C C A A B A B     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 ( ) C C N A B  = [( ) ] C N A B  = 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 ( A C  B ) = 300. Then N ( A  B ) = N ( B )  ( ) C N A B  = 400  300 = 100. ( ) C N A B  = N ( A )  N ( A  B ) = 600  100 = 500. 2.2.40 [( ) ] C N A B  = 120  N ( A  B ) = 120  [ N ( C A  B ) + N ( A  C B ) + N ( A  B )] = 120  [50 + 15 + 2] = 53

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    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. 2.3.5 No. P(A1  A2  A3) = P(at least one “6” appears) = 1  P(no 6’s appear) = 3 5 1 1 6 2         . The Ai’s are not mutually exclusive, so P ( A 1  A 2  A 3 )  P ( A 1 ) + P ( A 2 ) + P ( A 3 ). 2.3.6 P(A or B but not both) = 0.5 – 0.2 = 0.3 2.3.7 By inspection, B = ( B  A 1 )  ( B  A 2 )  …  ( B  A n ). 2.3.8 (a) (b) (b)

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    Section 2.4: Conditional Probability 7 2.3.9 P (odd man out) = 1  P (no odd man out) = 1  P ( HHH or TTT ) = 1  2 3 8 4  2.3.10 A = {2, 4, 6, …, 24}; B = {3, 6, 9, …, 24); A  B = {6, 12, 18, 24}. Therefore, P ( A  B ) = P ( A ) + P ( B )  P ( A  B ) = 12 8 4 16 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 A 1 and A 2 are mutually exclusive and cover the entire sample space, p 1 + p 2 = 1. But 3 p 1  p 2 = 1 2 , so p 2 = 5 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 ( F C  T C ) = 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) X C  Y = {( H , T , T , H ), ( T , H , H , T )}, so P ( X C  Y ) = 2/16 (b) X  Y C = {( H , T , T , T ), ( T , T , T , H ), ( T , H , H , H ), ( H , H , H , T )} so P ( X  Y C ) = 4/16 2.3.16 A = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} A  B C = {(1, 5), (3, 3), (5, 1)}, so P ( A  B C ) = 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|sum exceeds 8) = (sum 10 and sum exceeds 8) (sum exceeds 8) P P  = (sum 10) 3 36 3 (sum 9,10,11, or 12) 4 36 3 36 2 36 1 36 10 P / P / / / /        .

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    8 Chapter 2: Probability 2.4.2 P ( A | B ) + P ( B | A ) = 0.75 = ( ) ( ) 10 ( ) 5 ( ) ( ) ( ) 4 P A B P A B P A B P A B P B P A        , which implies that P ( A  B ) = 0.1. 2.4.3 If P ( A | B ) = ( ) ( ) ( ) P A B P A P B   , then P ( A  B ) < P ( A )  P ( B ). It follows that P ( B | A ) = ( ) ( ) ( ) ( ) ( ) P A B P A P B P A P A    = P ( B ). 2.4.4 P ( E | A  B ) = ( ( )) ( ) ( ) ( ) 0.4 0.1 3 ( ) ( ) ( ) 0.4 4 P E A B P E P A B P A B P A B P A B P A B              . 2.4.5 The answer would remain the same. Distinguishing only three family types does not make them equally likely; (girl, boy) families will occur twice as often as either (boy, boy) or (girl, girl) families. 2.4.6 P ( A  B ) = 0.8 and P ( A  B )  P ( A  B ) = 0.6, so P ( A  B ) = 0.2. Also, P ( A | B ) = 0.6 = ( ) ( ) P A B P B  , so P ( B ) = 0.2 1 0.6 3  and P ( A ) = 0.8 + 0.2  1 2 3 3  . 2.4.7 Let R i be the event that a red chip is selected on the i th draw, i = 1, 2. Then P (both are red) = P ( R 1  R 2 ) = P ( R 2 | R 1 ) P ( R 1 ) = 3 1 3 4 2 8   . 2.4.8 P ( A | B ) = ( ) ( ) ( ) ( ) ( ) ( ) ( ) P A B P A P B P A B a b P A B P B P B b          . But P ( A  B )  1, so P ( A | B )  1 a b b   . 2.4.9 Let i W be the event that a white chip is selected on the i th draw, i = 1,2 . Then P ( W 2 | W 1 ) = 1 2 1 ( ) ( ) P W W P W  . If both chips in the urn are white, P ( W 1 ) = 1; if one is white and one is black, P ( W 1 ) = 1 2 . Since each chip distribution is equally likely, P ( W 1 ) = 1  1 1 1 3 2 2 2 4    . Similarly, P ( W 1  W 2 ) = 1  1 1 1 5 2 4 2 8    , so P ( W 2 | W 1 ) = 5 / 8 5 3 / 4 6  . 2.4.10 P [( A  B )| ( A  B ) C ] = [( ) ( ) ] ( ) 0 [( ) ] [( ) ] C C C P A B A B P P A B P A B         2.4.11 (a) P ( A C  B C ) = 1  P ( A  B ) = 1  [ P ( A ) + P ( B )  P ( A  B )] = 1  [0.65 + 0.55  0.25] = 0.05

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    Section 2.4: Conditional Probability 9 (b) P [( A C  B )  ( A  B C )] = P ( A C  B ) + P ( A  B C ) = [ P ( A )  P ( A  B )] + [ P ( B )  P ( A  B )] = [0.65  0.25] + [0.55  0.25] = 0.70 (c) P ( A  B ) = 0.95 (d) P [( A  B ) C ] = 1  P ( A  B ) = 1  0.25 = 0.75 (e) P {[( A C  B )  ( A  B C )]| A  B } = [( ) ( )] ( ) C C P A B A B P A B     = 0.70/0.95 = 70/95 (f) P ( A  B )| A  B ) = P ( A  B )/ P ( A  B ) = 0.25/0.95 = 25/95 (g) P ( B | A C ) = P ( A C  B )/ P ( A C ) ] = [ P ( B )  P ( A  B )]/[1  P ( A )] = [0.55  0.25]/[1  0.65] = 30/35 2.4.12 P (No. of heads  2| No. of heads  2) = P (No. of heads  2 and No. of heads  2)/ P (No. of heads  2) = P (No. of heads = 2)/ P (No. of heads  2) = (3/8)/(7/8) = 3/7 2.4.13 P (first die  4|sum = 8) = P (first die  4 and sum = 8)/ P (sum = 8) = P ({(4, 4), (5, 3), (6, 2)}/ P ({(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)}) = 3/5 2.4.14 There are 4 ways to choose three aces (count which one is left out). There are 48 ways to choose the card that is not an ace, so there are 4  48 = 192 sets of cards where exactly three are aces. That gives 193 sets where there are at least three aces. The conditional probability is (1/270,725)/(193/270,725) = 1/193. 2.4.15 First note that P ( A  B ) = 1  P [( A  B ) C ] = 1  0.2 = 0.8. Then P ( B ) = P ( A  B )  P ( A  B C )  P ( A  B ) = 0.8  0.3  0.1 = 0.5. Finally P ( A | B ) = P ( A  B )/ P ( B ) = 0.1/0.5 = 1/5 2.4.16 P ( A | B ) = 0.5 implies P ( A  B ) = 0.5 P ( B ). P ( B | A ) = 0.4 implies P ( A  B ) = (0.4) P ( A ). Thus, 0.5 P ( B ) = 0.4 P ( A ) or P ( B ) = 0.8 P ( A ). Then, 0.9 = P ( A ) + P ( B ) = P ( A ) + 0.8 P ( A ) or P ( A ) = 0.9/1.8 = 0.5. 2.4.17 P [( A  B ) C ] = P [( A  B ) C ] + P ( A  B C ) + P ( A C  B ) = 0.2 + 0.1 + 0.3 = 0.6 P ( A  B |( A  B ) C ) = P [( A  B C )  ( A C  B )]/ P (( A  B ) C ) = [0.1 + 0.3]/0.6 = 2/3 2.4.18 P (sum  8|at least one die shows 5) = P (sum  8 and at least one die shows 5)/ P (at least one die shows 5) = P ({(5, 3), (5, 4), (5, 6), (3, 5), (4, 5), (6, 5), (5, 5)})/(11/36) = 7/11 2.4.19 P (Outandout wins|Australian Doll and Dusty Stake don’t win) = P (Outandout wins and Australian Doll and Dusty Stake don’t win)/ P (Australian Doll and Dusty Stake don’t win) = 0.20/0.55 = 20/55 2.4.20 Suppose the guard will randomly choose to name Bob or Charley if they are the two to go free. Then the probability the guard will name Bob, for example, is P (Andy, Bob) + (1/2) P (Bob, Charley) = 1/3 + (1/2)(1/3) = 1/2. The probability Andy will go free given the guard names Bob is P (Andy, Bob)/ P (Guard names Bob) = (1/3)/(1/2) = 2/3. A similar argument holds for the guard naming Charley. Andy’s concern is not justified.

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    10 Chapter 2: Probability 2.4.21 P ( BBRWW ) = P ( B ) P ( B | B ) P ( R | BB ) P ( W | BBR ) P ( W | BBRW ) = 3 5 6 5 15 14 13 12 11      =0.0050 P (2, 6, 4, 9, 13) = 1 1 1 1 1 1 15 14 13 12 11 360,360      . 2.4.22 Let K i be the event that the i th key tried opens the door, i = 1, 2, …, n . Then P (door opens first time with 3rd key) = 1 2 3 1 2 1 3 1 2 ( ) ( ) ( ) ( ) C C C C C C C P K K K P K P K K P K K K       = 1 2 1 1 1 2 n n n n n n        . 2.4.23 (a) The complementary event is that the team loses three or four games. Assume the games are independent. The probability of the event is (0.6)(0.50)(0.6)(0.3) + (0.4)(0.50)(0.6)(0.3) + (0.4)(0.50)(0.4)(0.3) + (0.4)(0.50)(0.6)(0.7) + (0.4)(0.50)(0.6)(0.3) = 0.234. The probability of a bowl appearance is 1 – 0.234 = 0.766. (b) Let k A = probability team wins exactly k games, k = 3,4. Then 4 3 4 4 3 4 4 4 3 4 3 4 3 4 3 4 [ ( )] [( ) ] ( ) ( ) ( | ) ( ) ( ) ( ) P A A A P A A A P P A P A A A P A A P A A P A A              But this does not equal 4 ( ) P A . (c) Yes, the two events are independent. 2.4.24 (1/2)(1/2)(1/2)(2/3)(3/4) = 1/16 2.4.25 Let A i be the event “Bearing came from supplier i ”, i = 1, 2, 3. Let B be the event “Bearing in toy manufacturer’s inventory is defective.” Then P ( A 1 ) = 0.5, P ( A 2 ) = 0.3, P ( A 3 ) = 0.2 and P ( B | A 1 ) = 0.02, P ( B | A 2 ) = 0.03, P ( B | A 3 ) = 0.04 Combining these probabilities according to Theorem 2.4.1 gives P ( B ) = (0.02)(0.5) + (0.03)(0.3) + (0.04)(0.2) = 0.027 meaning that the manufacturer can expect 2.7% of her ball-bearing stock to be defective. 2.4.26 Let B be the event that the face (or sum of faces) equals 6. Let A 1 be the event that a Head comes up and A 2 , the event that a Tail comes up. Then P ( B ) = P ( B | A 1 ) P ( A 1 ) + P ( B | A 2 ) P ( A 2 ) = 1 1 5 1 6 2 36 2    = 0.15. 2.4.27 Let B be the event that the countries go to war. Let A be the event that terrorism increases. Then P ( B ) = P ( B | A ) P ( A ) + P ( B | A C ) P ( A C ) = (0.65)(0.30) + (0.05)(0.70) = 0.23. 2.4.28 Let B be the event that a donation is received; let A 1 , A 2 , and A 3 denote the events that the call is placed to Belle Meade, Oak Hill, and Antioch, respectively. Then P ( B ) = 3 1 1000 1000 2000 ( ) ( ) (0.60) (0.55) (0.35) 0.46 4000 4000 4000 i i i P B A P A          . 2.4.29 Let B denote the event that the person interviewed answers truthfully, and let A be the event that the person interviewed is a man. Then P ( B ) = P ( B | A ) P ( A ) + P ( B | A C ) P ( A C ) = (0.78)(0.47) + (0.63)(0.53) = 0.70.

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