# chap-6 - Chapter 6 Problems 2(a p(0 0 = 87 = 14/39 13 12 85...

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Chapter 6 77 Chapter 6 Problems 2. (a) p (0, 0) = 12 13 7 8 = 14/39, p (0, 1) = p (1, 0) = 12 13 5 8 = 10/39 p (1, 1) = 12 13 4 5 = 5/39 (b) p (0, 0, 0) = 11 12 13 6 7 8 = 28/143 p (0, 0, 1) = p (0, 1, 0) = p (1, 0, 0) = 11 12 13 5 7 8 = 70/429 p (0, 1, 1) = p (1, 0, 1) = p (1, 1, 0) = 11 12 13 4 5 8 = 40/429 p (1, 1, 1) = 11 12 13 3 4 5 = 5/143 3. (a) p (0, 0) = (10/13)(9/12) = 15/26 p (0, 1) = p (1, 0) = (10/13)(3/12) = 5/26 p (1, 1) = (3/13)(2/12) = 1/26 (b) p (0, 0, 0) = (10/13)(9/12)(8/11) = 60/143 p (0, 0, 1) = p (0, 1, 0) = p (1, 0, 0) = (10/13)(9/12)(3/11) = 45/286 p ( i , j , k ) = (3/13)(2/12)(10/11) = 5/143 if i + j + k = 2 p (1, 1, 1) = (3/13)(2/12)(1/11) = 1/286 4. (a) p (0, 0) = (8/13) 2 , p (0, 1) = p (1, 0) = (5/13)(8/13), p (1, 1) = (5/13) 2 (b) p (0, 0, 0) = (8/13) 3 p ( i , j , k ) = (8/13) 2 (5/13) if i + j + k = 1 p ( i , j , k ) = (8/13)(5/13) 2 if i + j + k = 2 5. p (0, 0) = (12/13) 3 (11/12) 3 p (0, 1) = p (1, 0) = (12/13) 3 [1 (11/12) 3 ] p (1, 1) = (2/13)[(1/13) + (12.13)(1/13)] + (11/13)(2/13)(1/13)

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78 Chapter 6 8. f Y ( y ) = y y y dx e x y c ) ( 2 2 = y e cy 3 3 4 , 0 < y < 0 ) ( dy y f Y = 1 c = 1/8 and so f Y ( y ) = 6 3 y e y , 0 < y < f X ( x ) = x y dy e x y ) ( 8 1 2 2 = ) 1 ( 4 1 x e x + upon using + + = y y y y e ye e y e y 2 2 2 2 9. (b) f X ( x ) = ) 2 ( 7 6 2 7 6 2 2 0 2 x x dy xy x + = + (c) P { X > Y } = 56 15 2 7 6 1 00 2 = + ∫∫ x dydx xy x (d) P { Y > 1/2 X < 1/2} = P { Y > 1/2, X < 1/2}/ P { X < 1/2} = + + 2 / 1 0 2 2 2 / 1 2 / 1 0 2 ) 2 ( 2 dx x x dxdy xy x 10. (a) f X ( x ) = e x , f Y ( y ) = e y , 0 < x < , 0 < y < P { X < Y } = 1/2 (b) P { X < a } = 1 e a 11. ! 2 ! 1 ! 2 ! 5 (.45) 2 (.15)(.40) 2 12. e 5 + 5 e 5 + 5 3 5 2 ! 3 5 ! 2 5 + e e
Chapter 6 79 14. Let X and Y denoted respectively the locations of the ambulance and the accident of the moment the accident occurs. P { Y X < a } = P { Y < X < Y + a } + P { X < Y < X + a } = ∫∫ + L L a y y dxdy L 0 ) , min( 2 2 = + + a L a y y L a L L y dxdy dxdy L 0 2 2 = 1 = + L a L a a L L a L a L 2 ) ( 2 , 0 < a < L 15. (a) 1 = ∫ ∫ = ) , ( ) , ( y xR dydx c dydx y x f = cA ( R ) where A ( R ) is the area of the region R . (b) f ( x , y ) = 1/4, 1 x , y 1 = f ( x ) f ( y ) where f ( v ) = 1/2, 1 v 1. (c) P { X 2 + Y 2 1} = c dydx 4 1 = (area of circle)/4 = π /4. 16. (a) A = A i , (b) yes (c) P ( A ) = ) ( i A P = n (1/2) n 1 17. 3 1 since each of the 3 points is equally likely to be the middle one. 18. P { Y X > L /3} = > x yL dydx L 3 / 2 4 L y L < < 2 0 < x < 2 L = + + 2 / 6 /3 / 6 / 02 / 2 4 L L L L x LL L dydx dydx L = + 72 7 24 5 12 4 2 2 2 2 L L L L = 7/9

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80 Chapter 6 19. 11 00 0 1 x dydx dx x = ∫∫ = 1 (a) 1 1 ln( ) y dx y x =− , 0 < y < 1 (b) 0 1 x dy x = 1, 0 < y < 1 (c) 1 2 (d) Integrating by parts gives that l n () 1 (l n ) yy d y y y y d y =− − yielding the result E [ Y ] = 1 0 ln( ) d y = 1/4 20. (a) yes: f X ( x ) = xe x , f Y ( y ) = e y , 0 < x < , 0 < y < (b) no: f X ( x ) = ) 1 ( 2 ) , ( 1 x dy y x f x = , 0 < x < 1 f Y ( y ) = y dx y x f y 2 ) , ( 0 = , 0 < y < 1 21. (a) We must show that dxdy y x f ) , ( = 1. Now, dxdy y x f ) , ( = 1 0 1 0 24 y dxdy xy = 1 0 2 ) 1 ( 12 dy y y = + 1 0 3 2 ) 2 ( 12 dy y y y = 12(1/2 2/3 + 1/4) = 1 (b) E [ X ] = 1 0 ) ( dx x xf X = x dydx xy x 1 0 1 0 24 = 1 0 2 2 ) 1 ( 12 dx x x = 2/5 (c) 2/5
Chapter 6 81 22. (a) No, since the joint density does not factor.

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chap-6 - Chapter 6 Problems 2(a p(0 0 = 87 = 14/39 13 12 85...

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