# hw3sln - IEOR 161 Operations Research II University of...

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IEOR 161 Operations Research II University of California, Berkeley Spring 2008 Homework 3 Suggested Solution Chapter 3. 49. (a) Let A be the event that A wins. Let Y be the number of wins in the first two games. P ( A ) = P ( A | Y = 0) P ( Y = 0) + P ( A | Y = 1) P ( Y = 1) + P ( A | Y = 2) P ( Y = 2) = P ( A )2 p (1 - p ) + p 2 P ( A ) = p 2 1 - 2 p (1 - p ) (b) Let N be the number of games played. E ( N ) = E ( N | Y = 0) P ( Y = 0) + E ( N | Y = 1) P ( Y = 1) + E ( N | Y = 2) P ( Y = 2) = 2(1 - p ) 2 (2 + E ( N ))2 p (1 - p ) + 2 p 2 E ( N ) = 2 1 - 2 p (1 - p ) 54. Let X be the coin picked. P { X = n } = n 10 . P { N = k } = 10 n =1 P { N = k | X = n } P { X = n } = 10 n =1 (1 - n 10 ) k - 1 n 10 1 10 N is not a geometric. If the coin was reselected after each flip, N will be geometric. 63. (a) Let I i = 1 there is only one type i in the final selection; 0 more than one type i in the final selection. P { T = j } = n - 1 n n - 2 n - 1 . . . n - j - 1 n - j 1 n - j - 1 = 1 n P { I i = 1 | T = j } = n - j - 1 n - j n - j - 2 n - j - 1 . . . 1 2 = 1 n - j P { I i = 1 } = n - 1 j =0 P { S i = 1 | T = j } P { T = j } = 1 n n - 1 j =0 1 n - j (b) E [ n i =1 I i ] = nE [ I i ] = nP { I i = 1 } = n - 1 j =0 1 n - j 1

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Chapter 5. 6. Let X 1 be the service time of server 1 and X 2 be the service time of server 2. If server 1(2) finishes first, then X 1 < X 2 (X 2 < X 1 ). For Smith not to be the last, server 1(2) needs to finish first again. P { Smith is not last } = P { Smith is not last | server 1 finishes first } P { X
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• Spring '08
• Lim
• Operations Research, Tier One, Scaled Composites, Time t, Operations Research II University of California

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