hmwk4sol2027

# hmwk4sol2027 - Solutions to Homework 4, ISyE 2027 Problem 1...

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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....
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## 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.

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hmwk4sol2027 - Solutions to Homework 4, ISyE 2027 Problem 1...

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