Answer: [tex]\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}[/tex] or [tex]\dfrac{228}{91885}[/tex] .
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 = [tex]^{52}C_9[/tex]
Now , the combination of 9 cards such that exactly 6 of them are face cards = [tex]^{12}C_{6}\times^{40}C_3[/tex]
Now , the probability that the hand contains exactly 6 face cards will be :-
[tex]\dfrac{^{12}C_{6}\times^{40}C_3}{^{52}C_9}[/tex]
[tex]=\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}[/tex]
Hence, the probability that the hand contains exactly 6 face cards. is [tex]\dfrac{228}{91885}[/tex] .
