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Math Help - combinations/probability question please help!!

  1. #1
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    combinations/probability question please help!!

    Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, for each value from 1 to 6, there are two cards in the deck with that value. Bill likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for pairs of cards that have the same value. What is the chance that Bill finds at least one pair of cards that have the same value?
    8/33
    62/165
    17/33
    103/165
    25/33

    this is my solution but it is wrong:

    an Arbitrary deck of hands with at least one pair of similar cards = A, B, C, A
    the number of patterns that A,B,C,A hand appears = 4C2
    Therefore, number of ways A is chosen= 12
    when the first A is chosen, it locks in the value for the second A
    number of ways B is chosen = 10(because by choosing a value for A, two cards are taken off the choice list)
    number of ways C is chosen = 9
    P(at least two cards have the same value) = (12 x 10 x 9 x 4C2)/(12 x 11 x 10 x 9) = 18/33

    I hope that your explaination you will elaborate:
    1. why my method is wrong
    2. the actual solution.
    Thank you very much for even considering my post!!!
    Adhil
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  2. #2
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    Re: combinations/probability question please help!!

    P(at least 1) = 1 - P(None). This is probably more useful here.
    So there are  6C4 possible ways to select 4 distinct cards. Then there are  2^4 different suits for those cards. Finally there are  12C6 different cards.
    So you have
    P(at least one) = 1 - P(none) = 1 - \frac{6C4*2^4}{12C6} = 1 - \frac{16}{33} = \frac{17}{33}
    Last edited by Scopur; November 28th 2012 at 08:53 PM.
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  3. #3
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    Re: combinations/probability question please help!!

    Hello, warElephant!

    Another approach . . .


    Bill has a small deck of 12 playing cards made up of only 2 suits of 6 cards each.
    Each of the 6 cards within a suit has a different value from 1 to 6.
    Bill likes to play a game in which he shuffles the deck, turns over 4 cards,
    and looks for pairs of cards that have the same value.
    What is the probability that Bill finds at least one pair of cards that have the same value?

    . . (a)\;\tfrac{8}{33} \qquad(b)\;\tfrac{62}{165} \qquad (c)\;\tfrac{17}{33}\qquad (d)\;\tfrac{103}{165} \qquad (e)\;\tfrac{25}{33}

    We will find the probability that there are no matching pairs among the four cards.


    The first card can be any card (it doesn't matter): \tfrac{12}{12}

    The second card can be any of the other 10 non-matching cards: \tfrac{10}{11}

    The third card can be any of the other 8 non-matching cards: \tfrac{8}{10}

    The fourth card can be any of the other 6 non-matching cards: \ftrac{6}{9}

    Hence: . P(\text{no match}) \;=\;\frac{12}{12}\cdot\frac{10}{11}\cdot\frac{8}{  10}\cdot\frac{6}{9} \:=\:\frac{16}{33}


    Therefore: . P(\text{at least one match}) \;=\;1-\frac{16}{33} \;=\;\frac{17}{33}
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