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Vanyuwa [196]
4 years ago
13

A pair of fair dice is cast. what is the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one

of the numbers falling uppermost is a 3?
Mathematics
1 answer:
WITCHER [35]4 years ago
5 0
If a pair of fair dice is cast, the probability that the <span>sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is given by:

\frac{P(sum\, of\, 9\, \cup\,3)}{P(3)}

The number of outcomes of sum of 9 where at last one is 3 is (3, 6) and (6, 3) = 2.

</span>The number of outcomes of last one of the numbers falling uppermost is a 3 is (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3) and (6, 3) = 11.

Therefore, <span>the probabiliy that the sum of the numbers falling uppermost is 9, given that at least one of the numbers falling uppermost is a 3 is 2 / 11.</span>
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Andrew [12]

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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siniylev [52]
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Read 2 more answers
What's the answer I'm confused yeahhhhhhhhhh
Fantom [35]
-3 2/5 - 5/6 = -127/30

-2/3 + 6 1/8 = 107/24

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