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liq [111]
3 years ago
14

I need help asap. Math Writing Expressions.

Mathematics
1 answer:
erica [24]3 years ago
3 0

Answer:

1:

6y-3+y

<u>D =7y-3</u>

2.

<u>50C</u>

3.

<u>6÷K</u>

4.

<u>7-x</u>

5.

<u>8p-4p</u>

6.

<u>2g-3</u>

7.

1-7/p

<u>D.(p-7)/p</u>

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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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