Explanation Given: 13 cards are dealt to each player from a deck of 52 cards in the card game. Explanation: Consider the experiment of dealing 13 cards to each player from a deck of 52 cards. The number of elements in the sample space will be equal to the number of ways of selecting 13 cards from 52 cards. That is, n ( S ) = C ( 52 , 13 ) = 52 ! 13 ! × ( 52 − 13 ) ! = 635 , 013 , 559 , 600 Assume that, E represents the event that the bridge hand contains 6 of one suit, 4 of another and 3 of another. There are 4 suits in a deck of cards and each deck has 13 cards. The number of ways of selecting 1 suit from 4 suits is C ( 4 , 1 ) and the number of ways of selecting 6 cards from 13 cards of a suit is C ( 13 , 6 ) . After selection of 6 cards, the number of left suits is 3. The number of ways of selecting 1 suit from 3 suits is C ( 3 , 1 ) and the number of ways of selecting 4 cards from 13 cards of a suit is C ( 13 , 4 ) . After selection of 10 cards, the number of left suits is 2. The number of ways of selecting 1 suit from 2 suits is C ( 2 , 1 ) and the number of ways of selecting 3 cards from 13 cards of a suit is C ( 13 , 3 )

Finite Mathematics, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

11th Edition

ISBN: 9780133886818

Author: Margaret L. Lial, Raymond N. Greenwell, Nathan P. Ritchey

Publisher: PEARSON

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Chapter 8.3, Problem 58E

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The probability that a bridge hand contains 6 of one suit, 4 of another and 3 of another when 13 cards are dealt.

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Chapter 8.3, Problem 57E

Chapter 8.3, Problem 59E

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Finite Mathematics, Books a la Carte Plus MyLab Math Access Card Package (11th Edition)

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The probability that a bridge hand contains 6 of one suit, 4 of another and 3 of another when 13 cards are dealt.

Solution Summary: The author analyzes the probability that a bridge hand contains 6 of one suit, 4 of another, and 3 from another when 13 cards are dealt to each player.

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The probability that a bridge hand contains 6 of one suit, 4 of another and 3 of another when 13 cards are dealt.

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