This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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....
View
Full
Document
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
 Dr.PaulHorn

Click to edit the document details