20101ee131A_1_midterm-2009-winter-soln

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

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
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 01/19/2012 for the course EE 131A taught by Professor Lorenzelli during the Spring '08 term at UCLA.

### Page1 / 3

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

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online