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# In the card game bridge, each of 4 players is dealt a hand of 13 of the 52 cards. What is the probability that each player receives exactly

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In the card game bridge, each of 4 players is dealt a hand of 13 of the 52 cards. What is the probability that each player receives exactly one Ace? (You may use a calculator to compute the probability, but answer as an exact number. Entering a few decimal places of a nonterminating decimal is not considered exact; if you reach such an answer, enter it as a fraction.)

Guest Apr 9, 2015

#1
+17711
+5

For the first player:

The total number of possible hands of 13 cards from a deck of 52 is 52C13.

The total number of possible hands for this player that contains one ace and twelve other cards:

4C1·48C12  (which is choosing 1 ace from 4 aces and 12 other cards from the 48 remaining cards)

So, the probability for the first person getting one ace is:   ( 4C1·48C12 ) / 52C13

For the second player:

Probability  =  ( 3C1·36C12 ) / 39C13

because:  there are only 3 aces left, only 36 other cards left, and only a total of 39 cards left.

For the third player:

Probability  =  ( 2C1·24C12 ) / 26C13

because:  there are only 2 aces left, only 24 other cards left, and only a total of 26 cards left.

For the fourth player:

Probability  =  ( 1C1·12C12 ) / 13C13

because:  there is only 1 ace left, only 12 other cards left, and only a total of 13 cards left.

Multiply these four probabilities together to get the total probability. ...

geno3141  Apr 10, 2015
Sort:

#1
+17711
+5

For the first player:

The total number of possible hands of 13 cards from a deck of 52 is 52C13.

The total number of possible hands for this player that contains one ace and twelve other cards:

4C1·48C12  (which is choosing 1 ace from 4 aces and 12 other cards from the 48 remaining cards)

So, the probability for the first person getting one ace is:   ( 4C1·48C12 ) / 52C13

For the second player:

Probability  =  ( 3C1·36C12 ) / 39C13

because:  there are only 3 aces left, only 36 other cards left, and only a total of 39 cards left.

For the third player:

Probability  =  ( 2C1·24C12 ) / 26C13

because:  there are only 2 aces left, only 24 other cards left, and only a total of 26 cards left.

For the fourth player:

Probability  =  ( 1C1·12C12 ) / 13C13

because:  there is only 1 ace left, only 12 other cards left, and only a total of 13 cards left.

Multiply these four probabilities together to get the total probability. ...

geno3141  Apr 10, 2015

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