HW5Solutions (dragged)

# HW5Solutions (dragged) - P ABAA P BAAA P BBAB P BABB P ABBB...

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Home Work 5 Solutions IEOR 172, Fall 2015 Problem 4.20 a. Let p = 18 38 = 9 19 and q = 20 38 = 10 19 P ( X > 0) = p + p 2 q = 9 19 + 10 19 9 19 2 = 0 . 5917 Problem 4.20 b. No, there are two paths where we win but three where we loose. One of these paths has a loss of -3 which is relatively large given the problem. Problem 4.20 c. E ( X ) = 1 p + 1 p 2 q - 1 pq 2 - 1 pq 2 - 3 q 3 = - 0 . 108 Problem 4.22 a. The expected number of games that are played when i = 2 can be found using P ( AA ) · 2 + P ( BB ) · 2 + P ( ABA ) · 3 + P ( ABB ) · 3 + P ( BAA ) · 3 + P ( BAB ) · 3 = 2 p 2 + 2(1 - p ) 2 + 3 p 2 (1 - p ) + 3 p (1 - p ) 2 + 3(1 - p ) p 2 + 3(1 - p ) 2 p = 2 + 2 p - 2 p 2 We solve the maximization problem by taking the first derivative and setting it equal to zero: 2 - 4 p = 0 ) p = 1 2 . Problem 4.22 b. The expected number of games that are played when i = 3 can be found using 3 [ P ( AAA ) + P ( BBB )] + 4 [
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Unformatted text preview: ) + P ( ABAA ) + P ( BAAA ) + P ( BBAB ) + P ( BABB ) + P ( ABBB )] + 5[ P ( AABBA ) + P ( ABABA ) + P ( BAABA ) + P ( ABBAA ) + P ( BBAAA ) + P ( BABAA ) + P ( BBAAB ) + P ( BABAB ) + P ( ABBAB ) + P ( BAABB ) + P ( AABBB ) + P ( ABABB )] = 3[ p 3 + (1-p ) 3 ] + 4[3 p 3 (1-p ) + 3 p (1-p ) 3 ] + 5[6 p 3 (1-p ) 2 + 6 p 2 (1-p ) 3 ] Problem 4.23 a. Lets assume one buys x of the commodity at the start of the week, then in cash one has C = 1000-2 x . Here we have x ounces of our commodity with 0 x 500. Then at the end of the week our total value is given by V = 1000-2 x + Y x 1...
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