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Thread: Random variable distribution

  1. #1
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    Random variable distribution

    2 cards are drawn from a 52 card deck. The random variable X represents the number of aces drawn. The random variable distribution:
    X = 0 : P(0) = 0.849
    X = 1 : P(1) = 0.145
    X = 2 : P(2) = 0.005

    Is this how you do it?


    Probability of the draws:

    0 aces out of 2 cards:

    C(48,2) * C(4,0) / C(52,2)

    1128 / 1328 = 0.849

    1 ace out of 2 cards:

    C(48,1) * C(4,1) / C(52,2)

    48 * 4 / 1328 = 0.145

    2 aces out of 2 cards:

    C(48,0) * C(4,2) / C(52,2)

    6 / 1328 = 0.005
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  2. #2
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    Re: Random variable distribution

    100%
    Thanks from topsquark
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  3. #3
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    Re: Random variable distribution

    Your "formula" is correct, your arithmetic is not: _{50}C_{2}= \frac{52!}{50!2!}= \frac{52(51)}{2}= 26(52)= 1326, not 1328.

    How I would do this: there are 52 cards in the deck of which 4 are aces and 48 are not. The probability the first card drawn is NOT an ace is 48/52= 12/13. Then there are 51 cards left of which 47 are not aces. The probability the second card drawn is also not an ace is 47/51. The probability neither card is an ace is P(0)= \frac{12}{13}\frac{47}{51}= 0.8507. P(0)= 0.8507.

    The probability the first card is an ace is 4/52= 1/13. Then there are 51 cards left of which the 48 are not aces. The probability the second card is not an ace is 48/51. The probability the first card drawn is an ace and the second is not is \frac{1}{13}\frac{48}{51}= 0.072.
    The probability the first card is not an ace is 48/52= 12/13. Then there are 51 cards left of which the 4 are aces. The probability the second card is an ace is 4/51. The probability the first card drawn is an ace and the second is not is \frac{12}{13}\frac{4}{51}= 0.072. (It is always true that these are the same- we are just changing the order of numbers in the numerator and denominator.)
    The probability of one ace and one non-ace in either order is P(1)= 0.072+ 0.072= 0.144.

    There are 52 cards of which 4 are aces. The probability the first card drawn is 4/52= 1/13. There at then 51 cards left of which 3 are aces. The probability the second card drawn is also an ace is 3/51. The probability both cards drawn are aces is [tex]P(2)= \frac{1}{13}\frac{3}{51}= P(2)= 0.0045
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