hmwk7sol2027 - Solutions to Homework 7, ISyE 2027 Fall 2006...

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Unformatted text preview: Solutions to Homework 7, ISyE 2027 Fall 2006 Problem 1 (a) For 0 ≤ i + j ≤ 5, we see that P ( X = i,Y = j ) = ( 13 i )( 13 j )( 26 5- ( i + j ) ) ( 52 5 ) . (b) The marginal pmf of Y is just P ( Y = k ) = ( 13 k )( 39 5- k ) ( 52 5 ) , ≤ k ≤ 5 . (c) The conditional probability that X = 0, given that Y = 4 is just P ( X = 0 | Y = 4) = P ( X = 0 ,Y = 4) P ( Y = 4) = ( 26 1 ) ( 39 1 ) = 2 / 3 . Problem 2 (a) To get a feel for how to compute the joint pmf of X and Y , let’s consider a few special cases. To make things easier to explain, label one die as die 1 and the other as die 2. Notice that P ( X = 1 ,Y = 1) = 1 / 16, since the event { X = 1 ,Y = 1 } corresponds to the event that both die 1 and die 2 are equal to 1, and the probability of this event is 1 / 16. Also, P ( X = 2 ,Y = 1) = 2 / 16, since the event { X = 2 ,Y = 1 } corresponds to either die 1 being 2 and die 2 being 1, or die 1 being 1 and die 2 being 2. Finally, we find that P ( X = 1 ,Y = 2) = 0, since it’s clear that X ≥ Y . It should be clear now that, for any arbitrary i , j : P ( X = i,Y = j ) = 1 / 16 , 1 ≤ i = j ≤ 4; 1 / 8 , 1 ≤ j < i ≤ 4; , otherwise. . (b) The marginal pmf of X is as follows: P ( X = 1) = P ( X = 1 ,Y = 1) = 1 / 16 P ( X = 2) = P ( X = 2 ,Y = 1) + P ( X = 2 ,Y = 2) = 3 / 16 P ( X = 3) = P...
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This note was uploaded on 02/01/2010 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Tech.

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hmwk7sol2027 - Solutions to Homework 7, ISyE 2027 Fall 2006...

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