20101ee131A_1_midterm-2009-winter-soln

20101ee131A_1_midterm-2009-winter-soln - Problem 1 4 4 13...

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Unformatted text preview: Problem 1. 4 4 13 12 3 2 ( ) 52 5 P full house × × = 13 is the choices of the number (2, 3, 4, …, Q, K, A) of the “three of a kind”. 12 is the choices of the number of the pair, which cannot be the same as the “three of a kind”. 4 3 is the number of ways of choosing a “three of a kind” once its number is specified. 4 2 is the number of ways of choosing a pair once its number is specified. 52 5 is the number of ways of choosing 5 cards out of 52, without any constraint. Problem 2. 1) Each die roll of the three rolls is an independent Bernoulli experiment, with 1/6 probability of “success” --- having the same number as the player’s bet. Thus, the number of matches among the three rolls is a Binomial random variable: 3 3 1 5 0, ( ) (# ) ( ) ( ) 6 6 i i For i P win i P matches i i- ≥ = = = = . We have 75 15 1 ( 1) , ( 2) , ( 3) 216 216 216 P win P win P win = = = = = = , and 3 1 125 ( 1) (# 0) 1 ( ) 216 i...
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This note was uploaded on 01/19/2012 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

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20101ee131A_1_midterm-2009-winter-soln - Problem 1 4 4 13...

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