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10 Jun 2021, 08:00
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Difficulty:
45%
(medium)
Question Stats:
63% (01:26) correct
37%
(01:50)
wrong
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Three cards are drawn from a standard deck of 52 cards pack at random and witohut replacement. Find the probability that they consist of both colours.
(A) \(\frac{26C1*26C2}{52C3}\)
(B) \(\frac{2*26C1*26C2}{52C3}\)
(C) \(\frac{26C3}{52C3}\)
(D) \(\frac{26C3}{52C3}\)
(E) None of these
(A) \(\frac{26C1*26C2}{52C3}\)
(B) \(\frac{2*26C1*26C2}{52C3}\)
(C) \(\frac{26C3}{52C3}\)
(D) \(\frac{26C3}{52C3}\)
(E) None of these
10 Jun 2021, 20:56
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Both the colors cards can be drawn if there is one black and two red or one red and two black cards
So favorable outcomes = 26C1 x 26C2 + 26C2 x 26C1 = 2 x 26C1 x 26C2
Total Outcomes = 52C3
So the answer is B. 2 x 26C1 x 26C2/ 52C3
So favorable outcomes = 26C1 x 26C2 + 26C2 x 26C1 = 2 x 26C1 x 26C2
Total Outcomes = 52C3
So the answer is B. 2 x 26C1 x 26C2/ 52C3
16 Jun 2025, 14:29
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Can someone please expand on the explanation of why for the favourable outcomes the pool from where to choose becomes 26 (and not 52).
Thanks in advance.
Regards,
Juan C. Avellan
Thanks in advance.
Regards,
Juan C. Avellan
LaveenaPanchal
Both the colors cards can be drawn if there is one black and two red or one red and two black cards
So favorable outcomes = 26C1 x 26C2 + 26C2 x 26C1 = 2 x 26C1 x 26C2
Total Outcomes = 52C3
So the answer is B. 2 x 26C1 x 26C2/ 52C3
So favorable outcomes = 26C1 x 26C2 + 26C2 x 26C1 = 2 x 26C1 x 26C2
Total Outcomes = 52C3
So the answer is B. 2 x 26C1 x 26C2/ 52C3
