Question

A poker hand consisting of 9 cards is dealt from a standard deck of 52 cards. Find the probability that the hand contains exactly 6 face cards.

Answer 1

Answer: $\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}$ or  $\dfrac{228}{91885}$ .

Step-by-step explanation:

Given : A poker hand consisting of 9 cards is dealt from a standard deck of 52 cards.

The total number of cards in a deck 52

Number of faces cards in a deck = 12

Number of cards not face cards = 40

The total number of combinations of drawing 9 cards out of 52 cards = $^{52}C_9$

Now , the combination of 9 cards such that exactly 6 of them are  face cards =  $^{12}C_{6}\times^{40}C_3$

Now , the probability that the hand contains exactly 6 face cards will be :-

$\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}$

$=\dfrac{\dfrac{12!}{6!6!}\times\dfrac{40!}{3!37!}}{\dfrac{52!}{9!\times43!}}\ \ [\because\ ^nC_r=\dfrac{n!}{r!(n-r)!}]\\\\=\dfrac{228}{91885}$

Hence, the probability that the hand contains exactly 6 face cards. is  $\dfrac{228}{91885}$ .