ChapterSolutions - 7

# ChapterSolutions - 7 - Chapter 7 Problems 1 Let X = 1 if...

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98 Chapter 7 Chapter 7 Problems 1. Let X = 1 if the coin toss lands heads, and let it equal 0 otherwise. Also, let Y denote the value that shows up on the die. Then, with p ( i , j ) = P { X = i , Y = j } E [return] = = = + 6 1 6 1 ) , 0 ( 2 ) , 1 ( 2 j j j p j j jp = ) 5 . 10 42 ( 12 1 + = 52.5/12 2. (a) 6 6 9 = 324 (b) X = (6 S )(6 W )(9 R ) (c) E [ X ] = 6(6)(6) P { S = 0, W = 0, R = 3} + 6(3)(9) P { S = 0, W = 3, R = 0} + 3(6)(9) P { S = 3, W = 0, R = 0} + 6(5)(7) P { S = 0, W = 1, R = 2} + 5(6)(7) P { S = 1, W = 0, R = 2} + 6(4)(8) P { S = 0, W = 2, R = 1} + 4(6)(8) P { S = 2, W = 0, R = 1} + 5(4)(9) P { S = 1, W = 2, R = 0} + 4(5)(9) P { S = 2, W = 1, R = 0} + 5(5)(8) P { S = 1, W = 1, R = 1} = + + + + + ) 9 )( 6 )( 6 ( 200 6 2 6 360 9 2 6 384 2 9 6 420 3 6 324 3 9 216 3 21 1 198.8 3. ] [ a Y X E = dydx y x a ∫∫ 1 0 1 0 . Now + = 1 0 1 0 ) ( ) ( x a x a a dy x y dy y x dy y x + = xx a a du u du u 0 1 0 ) 1 /( ] ) 1 ( [ 1 1 + + = + + a x x a a Hence, ] [ a Y X E = + + + + 1 0 1 1 ] ) 1 ( [ 1 1 dx x x a a a = ) 2 )( 1 ( 2 + + a a

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Chapter 7 99 4. ] [ Y X E = ∑∑ == m i m j j i m 11 2 1 . Now, + = = = + = m i j i j m j i j j i j i 1 1 1 ) ( ) ( = [ i ( i 1) + ( m i )( m i + 1)]/2 Hence, using the identity = m j j 1 2 = m ( m + 1)(2 m + 1)/6, we obtain that ] [ Y X E = m m m m m m m m m 3 ) 1 )( 1 ( 2 ) 1 ( 6 ) 1 2 )( 1 ( 1 2 + = + + + 5. The joint density of the point ( X , Y ) at which the accident occurs is f ( x , y ) = 9 1 , 3/2 < x , y < 3/2 = f ( x ) f ( y ) where f ( a ) = 1/3, 3/2 < a < 3/2. Hence we may conclude that X and Y are independent and uniformly distributed on ( 3/2, 3/2) Therefore, E [ X + Y ] = 2 / 3 3 4 3 1 2 2 / 3 0 2 / 3 2 / 3 = = xdx dx x . 6. = = = 10 1 10 1 ] [ i i i i X E X E = 10(7/2) = 35. 8. E [number of occupied tables] = = = = N i i N i i X E X E 1 1 ] [ Now, E [ X i ] = P { i th arrival is not friends with any of first i 1} = (1 p ) i 1 and so E [number of occupied tables] = = N i i p 1 1 ) 1
100 Chapter 7 7. Let X i equal 1 if both choose item i and let it be 0 otherwise; let Y i equal 1 if neither A nor B chooses item i and let it be 0 otherwise. Also, let W i equal 1 if exactly one of A and B choose item i and let it be 0 otherwise. Let X = = 10 1 i i X , Y = = 10 1 i i Y , W = = 10 1 i i W (a) E [ X ] = = 10 1 ] [ i i X E = 10(3/10) 2 = .9 (b) E [ Y ] = = 10 1 ] [ i i Y E = 10(7/10) 2 = 4.9 (c) Since X + Y + W = 10, we obtain from parts (a) and (b) that E [ W ] = 10 .9 4.9 = 4.2 Of course, we could have obtained E [ W ] from E [ W ] = = 10 1 ] [ i i W E = 10(2)(3/10)(7/10) = 4.2 9. Let X j equal 1 if urn j is empty and 0 otherwise. Then E [ X j ] = P {ball i is not in urn j , i j } = = n j i i ) / 1 1 ( Hence, (a) E [number of empty urns] = ∑∑ == n j n j i i 1 ) / 1 1 (b) P {none are empty} = P {ball j is in urn j , for all j } = = n j j 1 / 1 10. Let X i equal 1 if trial i is a success and 0 otherwise. (a) .6. This occurs when P { X 1 = X 2 = X 3 } = 1. It is the largest possible since 1.8 = = = = } 1 { 3 } 1 { i i X P X P . Hence, P { X i = 1} = .6 and so P { X = 3} = P { X 1 = X 2 = X 3 = 1} P { X i = 1} = .6.

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## This note was uploaded on 05/10/2011 for the course MATH 340L taught by Professor Seckin during the Spring '11 term at University of Texas.

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ChapterSolutions - 7 - Chapter 7 Problems 1 Let X = 1 if...

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