Homework2Solution

Homework2Solution - Homework 2 Solutions July 23, 2007 p111...

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Unformatted text preview: Homework 2 Solutions July 23, 2007 p111 10,13,28,59,76,79 p125 7a, 12 1. #10 Three cards are selected without replacement from an ordinary deck of 52 playing cards. Compute the conditional probability that the first card selected is a spade given that the second and third cards are spades. Denote the i th card drawn by C i . P ( C 1 = s | C 2 = s,C 3 = s ) = P ( C 1 = s,C 2 = s,C 3 = s ) P ( C 2 = s,C 3 = s ) . P ( C 1 = s,C 2 = s,C 3 = s ) = ( 13 3 ) ( 52 3 ) . P ( C 2 = s,C 3 = s ) = ( 13 2 ) ( 52 2 ) . We get 13!3!49! 3!10!52! * 13!2!50! 2!11!52! = 11 50 . 2. #13 Suppose that an ordinary deck of 52 cards is randomly divided into 4 hands of 13 cards each. Determine p , the probability that each hand has an ace. Let E i be the event that the i th hand has exactly one ace. Determine p = P ( E 1 E 2 E 3 E 4 ) using the multiplication rule. By the multiplication rule, P ( E 1 E 2 E 3 E 4 ) = P ( E 4 | E 1 E 2 E 3 ) * P ( E 3 | E 2 E 1 ) * P ( E 2 | E 1 ) * P ( E 1 ). P ( E 1 ) = ( 4 1 )( 48 12 ) ( 52 13 ) . P ( E 2 | E 1 ) = ( 3 1 )( 36 12 ) ( 39 13 ) . P ( E 3 | E 1 E 2 ) = ( 2 1 )( 24 12 ) ( 26 13 ) . 1 P ( E 4 | E 1 E 2 E 3 ) = 1. Then P ( E 1 E 2 E 3 E 4 ) = ( 4 1 )( 48 12 ) ( 52 13 ) * ( 3 1 )( 36 12 ) ( 39 13 ) * ( 2 1 )( 24 12 ) ( 26 13 ) ....
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This note was uploaded on 11/15/2008 for the course STATS 116 taught by Professor Staff during the Summer '07 term at Stanford.

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Homework2Solution - Homework 2 Solutions July 23, 2007 p111...

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