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A bridge hand is found by taking 13 cards at random and without replacement from a deck of 52 playing cards. Find the probability of drawing each of the following hands"

(a) One in which there are 5 spades, 4 hears, 3 diamonds, and 1 club?
(b) One in which therea re 5 spades, 4 hears, 2 diamonds, and 2 clubs?
(c) One in which there are 5 spades, 4 hearts, 1 diamond, and 3 clubs?
(d) Suppose you are dealt 5 cards of one suit, 4 cards of another. Would the probability of having the other suits split 3 and 1 be greater than the probability of having them split 2 and 2?

1 Answer

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Final answer:

To find the probability of each given bridge hand, you need to consider the number of ways each hand can occur and divide it by the total number of possible bridge hands.

Step-by-step explanation:

To find the probability of drawing each of the given bridge hands, we need to consider the number of ways each hand can occur and divide it by the total number of possible bridge hands.

(a) One in which there are 5 spades, 4 hearts, 3 diamonds, and 1 club:

The number of ways to choose 5 spades from 13 is C(13, 5). Similarly, the number of ways to choose 4 hearts from 13, 3 diamonds from 13, and 1 club from 13 are C(13, 4), C(13, 3), and C(13, 1) respectively. The total number of bridge hands is C(52, 13). The probability is then (C(13, 5) * C(13, 4) * C(13, 3) * C(13, 1)) / C(52, 13).

(b) One in which there are 5 spades, 4 hearts, 2 diamonds, and 2 clubs:

Follow the same approach as in (a) to calculate the probability.

(c) One in which there are 5 spades, 4 hearts, 1 diamond, and 3 clubs:

Follow the same approach as in (a) to calculate the probability.

(d) The probability of having the other suits split 2 and 2 is greater than the probability of having them split 3 and 1. The reason is that the number of ways to choose 2 cards from each suit is greater than the number of ways to choose 3 cards from one suit and 1 card from another suit.

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