Final_practice_sol - ORIE 360/560 Practice Final Solutions Fall 2007 Problem 1(a 1 1 3 E[X|Y = 1 E[X|Y = 2 = 2 2 4 9 1 1 9 Var X = E[X 2 E 2[X =

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ORIE 360/560 Practice Final Solutions Fall, 2007 Problem 1. (a) EX = E [ E [ X | Y ]] = 1 2 E [ X | Y = 1] + 1 2 E [ X | Y = 2] = 3 4 . Var X = E [ X 2 ] - E 2 [ X ] = E [ E [ X 2 | Y ]] - 9 16 = 1 2 E [ X | Y = 1] + 1 2 E [ X | Y = 2] - 9 16 . Now, if X U (0 ,a ), then EX 2 = Z a 0 1 a x 2 dx = a 2 3 , so Var X = 1 2 ± 1 3 + 4 3 ² - 9 16 = 5 6 - 9 16 = 40 48 - 27 48 = 13 48 . (b) P [ X > 1] = P [ Y = 2 ,X > 1] = P [ Y = 2] P [ X > 1 | Y = 2] = 1 2 · 1 2 = 1 4 . Problem 2. (a) For a given suit, there are ( 13 k )( 39 13 - k ) ways to have k cards of that suit (and hence 13 - k cards of the other three). Therefore, the probability of getting a hand with one suit of at least nine cards is 4 13 X k =9 ( 13 k )( 39 13 - k ) ( 52 13 ) . (b) For a given pair of suits, there are ( 26 13 ) - 2 ways to choose 13 cards from those two suits (not including the two hands consisting of only cards from one suit or only cards from the other). There are ( 4 2 ) ways to choose two suits. Hence, the answer is ± 4 2 ² ( 26
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This note was uploaded on 11/12/2008 for the course ORIE 360 taught by Professor Ehrlichman during the Fall '07 term at Cornell University (Engineering School).

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Final_practice_sol - ORIE 360/560 Practice Final Solutions Fall 2007 Problem 1(a 1 1 3 E[X|Y = 1 E[X|Y = 2 = 2 2 4 9 1 1 9 Var X = E[X 2 E 2[X =

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