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Unformatted text preview: Math 361, Problem set 5 Due 10/04/10 1. (1.6.8) Let X have the pmf p ( x ) = ( 1 2 ) x , x = 1 , 2 , 3 , . . . , and zero else- where. Find the pmf of Y = X 3 . Answer: Y has pmf p ( x ) = ( 1 2 ) x/ 3 for x = 1 , 8 , 27 , . . . , and zero elsewhere. 2. (a) Pick a card from a standard deck. Let X denote the rank of the card(counting ace as one, J=11, Q=12, K=13.) Find the pmf of X . (b) Pick two cards from a deck, with replacement. Let Y denote the highest rank picked. Find the pmf of Y . Answer: Since all ranks are equally likely X has pmf p X ( x ) = 1 13 for x = 0 , . . . , 13 and p X ( x ) = 0 elsewhere. For Y ; p ( Y )( x ) = 1 13 2 + 2 n- 1 13 2 for x = 0 , . . . , 13 and p Y ( x ) = 0 elsewhere. Here the 1 13 2 represents both numbers being x , and the 2 x- 1 13 2 represents picking the x (with prob. 1 13 ), picking the number less than x (with prob. x- 1 13 ), and considering the two possible orders....
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This note was uploaded on 10/26/2010 for the course MATHCS Math 316 taught by Professor Dr.paulhorn during the Fall '10 term at Emory.
- Fall '10