hw6_418_10

# hw6_418_10 - 4 Let X denotes the highest ranking achieved...

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4. Let X denotes the highest ranking achieved by a woman. Then p ( X = 1) = ( 5 1 ) 9! 10! = 1 / 2, p ( X = 2) = ( 5 1 )( 5 1 ) 8! 10! = 5 / 18, p ( X = 3) = ( 5 2 )( 2 1 )( 5 1 ) 7! 10! = 5 / 34, p ( X = 4) = ( 5 3 )( 5 1 ) 3!6! 10! = 5 / 84, p ( X = 5) = ( 5 4 )( 5 1 ) 4!5! 10! = 5 / 252, p ( X = 6) = 5!5! 10! = 1 / 252. For i = 7 , 8 , 9 , 10, p ( X = i ) = 0 . 7. a) Let X denotes the max value to appear in the two rolls, then S x = { 1 , 2 , 3 , 4 , 5 , 6 } , b) Let Y denotes the min value to appear in the two rolls,, then S y = { 1 , 2 , 3 , 4 , 5 , 6 } , c) Let Z denotes the sum of the two rolls, then S z = { 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 11 , 12 } . d) Let A denotes the value of the first roll minus the value of the 2nd , then S A = {- 5 , - 4 , - 3 , - 2 , - 1 , 0 , 1 , 2 , 3 , 4 , 5 } . 8. Assume the die is fair, S x = { 1 , 2 , 3 , 4 , 5 , 6 } , then p ( X = 1) = 1 / 36 , p ( X = 2) = 3 / 36 , p ( X = 3) = 5 / 36 , p ( X = 4) = 7 / 36 , p ( X = 5) = 9 / 36 , p ( X = 6) = 11 / 36 . 13. Let Y 1 B (1 , 0 . 3), Y 2 B (1 , 0 . 6), X i = 1000 with conditional probability 1/2, X i = 500 with conditional probability 1/2. Let A1 denotes X1=1, X2=1; A2 denotes X1=1,X2=0; A3 denotes X1=0,X2=1; A4 denotes X1=0,X2=0. Then p ( X = 0) = p ( X = 0 | A 4) p ( A 4) = (1 - 0 . 3)(1 - 0 . 6) = 0 . 28 , p ( X = 500) = p ( X = 500 | A 2) p ( A 2) + p ( X = 500 | A 3) p ( A 3) = 1 / 2[0 . 3 * (1 - 0 . 6) + (1 - 0 . 3)0 . 6] = 0 . 27 .

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