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Please do at least two whole screenshots. Thx, I'm going to spam this now.

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Ket [755]3 years ago

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

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

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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 $\dfrac{228}{91885}$ .

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