I need help for a problem in my book probability A bridge hand is found by takin

Question

I need help for a problem in my book probability

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 hearts, 3 diamond, and 1 club.

b) 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.

I am having trouble setting this up (a) and having trouble understanding (b)

Explanation / Answer

a )

In standard deck there are 52 cards and 4 suits ( spades, hearts, diamonds, club )

Each suit contain 13 cards.

The number of possible ways that 13 cards are drawing each with 5 spades, 4 hearts, 3 diamonds, and 1 clubs is,

13C5*13C4*13C3*13C1

Total number of possible ways that 13 cards drawing from 52 cars is, 52C13

So probability that drawing each with 5 spades, 4 hearts, 3 diamonds, and 1 clubs is,

p = (13C5*13C4*13C3*13C1 ) / 52C13  = 0.0054

b )

Number of possible ways to select 5 cards from one suit is , 4C1*13C5

Number of ways to select 4 cards from another is , 3C1*13C4

Number of ways to to select 3 cards from remaining one suit and 1 from another suit is , 2C1*13C3*13C1

Total number of ways dealt 5 cards of one suit, 4 cards of another. other suits split 3 and 1 be

(4C1*13C5)*(3C1*13C4)*(2C1*13C3*13C1)

So , required probability is,

(4C1*13C5)*(3C1*13C4)*(2C1*13C3*13C1) / 52C13 = 0.1293

Similarly,

Probability that dealt 5 cards of one suit, 4 cards of another. other suits split 2 and 2 be

(4C1*13C5)*(3C1*13C4)*(13C2*13C2) / 52C13 = 0.1058

So the probability of having the other suits split 3 and 1 be greater than the probability of having them split 2 and 2.

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