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I will assume that you mean a five card Poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408

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9y ago
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3y ago

An integer is chosen at random from the first 20,000 positive integers . What is the probability that the integer chosen is divisible by atleast one of the numbers 5,7,8? *

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Q: What is the probability that a random selected poker hand contains exactly 3 aces given that it contains at least 2 aces?
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