# answersbookchapter4 - Chapter 4(1 P(4 = 4 2 8 4 2 2 = 6 91...

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Unformatted text preview: Chapter 4 (1) P (4) = 4 2 8 + 4 + 2 2 = 6 / 91, , P (2) = 4 1 Â· 2 1 8 + 4 + 2 2 = 8 / 91, P (1) = 4 1 Â· 8 1 8 + 4 + 2 2 = 32 / 91, P (0) = 2 2 8 + 4 + 2 2 = 1 / 91, P (- 1) = 8 1 Â· 2 1 8 + 4 + 2 2 = 16 / 91, P (- 2) = 8 2 8 + 4 + 2 2 = 28 / 91 (2) P { X = 3 } = P ( { (1 , 3) , (3 , 1) } ) = 2 / 36, P { X = 4 } = P ( { (1 , 4) , (2 , 2) , (4 , 1) } ) = 3 / 36, P { X = 7 } = 0, etc. (4) P { X = 1 } = 1 / 2 (this can be seen directly from symmetry, or calculated as 5 Â· 9! 10! ), P { X = 2 } = 5 Â· 5 Â· 8! 10! = 5 18 , P { X = 3 } = 5 Â· 4 Â· 5 Â· 7! 10! = 5 36 , P { X = 4 } = 5 Â· 4 Â· 3 Â· 5 Â· 6! 10! = 5 84 , P { X = 5 } = 5 Â· 4 Â· 3 Â· 2 Â· 5 Â· 5! 10! = 2 252 , P { X = 6 } = 5! Â· 5! 10! = 1 252 , P { X = n } = 0 for n > 6 P19 p (0) = 1 / 2, p (1) = 3 / 5- 1 / 2 = 1 / 10, p (2) = 1 / 5, p (3) = 1 / 10, p (3 . 5) = 1 / 10 P22 Let the teams be A and B, and let, say, AB denotes the event that A wins the first game and B wins the second. Let N be the expected number of games. (a) N = 2 Â· P ( AA ) + 3 Â· P ( ABA ) + 3 Â· P ( ABB ) + 3 Â· P ( BAA ) + 3 Â· P ( BAB ) + 2 Â· P ( BB ) = 2 p 2 + 3( p (1- p ) p + p (1- p ) 2 + (1- p ) p 2 + (1- p ) p (1- p )) + 2(1- p ) 2 = 2(1 + p- p 2 ); dN dp = 2(1- 2 p ) = 0 when p = 1 / 2....
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answersbookchapter4 - Chapter 4(1 P(4 = 4 2 8 4 2 2 = 6 91...

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