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: Solutions to Homework 4, ISyE 2027 Problem 1 (a) P ( A B ) = P ( A ) + P ( B ) P ( A B ) = . 4 + . 3 . 5 = . 2, so P ( B  A ) = P ( B A ) P ( A ) = . 2 . 4 = 1 2 . (b) Similarly, P ( A  B ) = P ( A B ) P ( B ) = . 2 . 3 = 2 3 . (c) P ( A  B ) = 1 P ( A  B ) = 1 3 . (d) P ( A B ) = P ( A ) P ( A B ) = . 4 . 2 = . 2, so P ( A  B ) = . 2 . 7 = 2 7 . (e) P ( A ) P ( B ) = . 12, but P ( A B ) = . 2, so A and B are not independent. Problem 2 (a) The probability of having 3 pairs is just ( 13 3 )( 4 2 )( 4 2 )( 4 2 ) ( 52 6 ) . (b) The probability of having 3 pairs, with one pair being aces is ( 12 2 )( 4 2 )( 4 2 )( 4 2 ) ( 52 6 ) . (c) The probability of having 3 aces and 1 pair is just ( 4 3 ) (12) ( 4 2 ) (44) ( 52 6 ) . (d) The probability of having a six card straight assuming that ace can only be high is just (8)(4) 6 ( 52 6 ) . (e) Now we need to compute the probability of being dealt a six card flush in hearts given that all six cards are red. Letgiven that all six cards are red....
View
Full
Document
This note was uploaded on 02/01/2010 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
 Spring '08
 Zahrn

Click to edit the document details