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Thread: Counting

  1. #1
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    Counting


    What is the counting logic behind this question?
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  2. #2
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    Re: Counting

    There are 52 cards in the pack, 4 kings (K) and 48 non-kings (N). There are 3 ways to get "exactly one king" in three cards: "KNN", "NKN", and "NNK". The probability the first card dealt is a king is 4/52. There are then 51 cards left in the deck, 3 kings and 48 non-kings. The probability the second card dealt is a non-king is 48/51. There are then 50 cards, 3 kings and 47 non-kings. The probability the third card dealt is a non-king is 3/50. The probability of "king, non-king, non-king" (KNN) in that order is (4/52)(48/51)(47/50). I will leave it to you to show that the probabilities of "non-king, king, non-king" (NKN) and "non-king, non-king, king" (NNK) are also (4/52)(48/51)(47/50) so that the probability of "exactly one king" in any order is (4/52)(48/51)(47/50)+ (4/52)(48/51)(47/50)+ (4/52)(48/51)(47/50)= 3(4/52)(48/51)(47/50). You can, of course, reduce those fractions.
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  3. #3
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    Re: Counting

    Quote Originally Posted by lionraja View Post

    What is the counting logic behind this question?
    Look at this calculation, from which can you see the logic?
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