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Old 03-10-2017, 08:00 PM   #1
cicakman
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Simple question I think

You draw two cards and neither is an ace. Your friend is feeling generous and let's you draw 10 more. Still no ace. You know there are 40 cards left, 4 of which are aces.

If your friend now draws 2 cards, what's the probability she gets exactly 1 ace?
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Old 03-10-2017, 10:23 PM   #2
snark
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Re: Simple question I think

P(ace) x P(not ace) = (4/40) x (36/39) = 9.23%
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Old 03-11-2017, 02:38 AM   #3
nickthegeek
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Re: Simple question I think

Quote:
Originally Posted by snark View Post
P(ace) x P(not ace) = (4/40) x (36/39) = 9.23%
You have to double that. The above is the probability to draw an Ace and a non-Ace in that specific order. You have to consider the opposite order also.

- Two non-Ace cards: C(36,2)/C(40,2) = 0.807692308
- One Ace and one non Ace: C(4,1)*C(35,2)/C(40,2) = 0.184615385
- Two Aces: C(4,2)/C(40,2) = 0.007692308
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