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Math Help - probability exercise with dependent events

  1. #1
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    probability exercise with dependent events

    I have some difficulties in solving the following problem:

    A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events E1, E2, E3, E4 as follows:

    E1 = {the first pile has exactly 1 ace},
    E2 = {the second pile has exactly 1 ace},
    E3 = {the third pile has exactly 1 ace},
    E4 = {the fourth pile has exactly 1 ace},

    Use the formula
    P(E_{1}\cap E_{2}...E_{n})=P(E_{1})P(E_{2}\mid E_{1})P(E_{3}\mid E_{1}\cap E_{2})...P(E_{n}\mid E_{1}...E{n-1})

    to find
     P(E_{1}\cap E_{2}\cap E_{3}\cap E_{4})

    the probability that each pile has an ace.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Re: probability exercise with dependent events

    Looks like the product of four hypergeometrics

    Quote Originally Posted by achacy View Post
    I have some difficulties in solving the following problem:

    A deck of 52 playing cards, containing all 4 aces, is randomly divided into 4 piles of 13 cards each. Define events E1, E2, E3, E4 as follows:

    E1 = {the first pile has exactly 1 ace},
    E2 = {the second pile has exactly 1 ace},
    E3 = {the third pile has exactly 1 ace},
    E4 = {the fourth pile has exactly 1 ace},

    Use the formula
    P(E_{1}\cap E_{2}...E_{4})=P(E_{1})P(E_{2}\mid E_{1})P(E_{3}\mid E_{1}E_{2})P(E_{4}\mid E_{1}E_2E_3)

    to find
     P(E_{1}\cap E_{2}\cap E_{3}\cap E_{4})

    the probability that each pile has an ace.
    P(E_{1})= {{4\choose 1}{48\choose 12}\over {52\choose 13}}

    That's how the first pile has exactly one ace and the rest has three.
    So, now we have 3 aces and 39 total cards, so

    P(E_{2}\mid E_{1})= {{3\choose 1}{36\choose 12}\over {39\choose 13}}

    That leaves us with 2 aces and 26 cards...

    P(E_{3}\mid E_{1}E_2)= {{2\choose 1}{24\choose 12}\over {26\choose 13}}

    Finally we are stuck with a fourth pile that MUST have one ace...

    P(E_{4}\mid E_{1}E_2E_3)= {{1\choose 1}{12\choose 12}\over {13\choose 13}}

    The product of these probabilities should be the answer.
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  3. #3
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    Re: probability exercise with dependent events

    Thanks a lot for this help. It seems that I have to refresh my knowledge of combinatorics...
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  4. #4
    MHF Contributor matheagle's Avatar
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    Re: probability exercise with dependent events

    Thats just the basic hypergeometric, which is my next lecture in Probability on Tuesday.
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