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Thread: counting problem?

  1. #1
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    counting problem?




    A closet contains 6 different pairs of shoes. Five shoes are drawn at
    random. What is the probability that at least one pair of shoes is
    obtained?

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    Re: counting problem?

    6 pairs => 12 shoes , we are drawing 5 shoes. So the number of total unordered combination of draws is C(12,5) = 792
    For no pairs to occur selection of 5 shoes must be 5 different =>C(6,5) and mult by 2^5 since there are two pairs for each shoe.

    P(no pairs) = (2^5)C(6,5) / C(12,5) = 8/33

    P(at least 1 pair) = 1- (8/33) =25/33

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    Re: counting problem?


    Another way to do it is just think about it sequentially.

    P(no pairs): P(1st shoe doesn't make a match) * P(2nd shoe doesn't make a match | 1st shoe didn't make a match) * P(3rd shoe doesn't make a match | 1st and 2nd shoes didn't make a match) * ... * P(5th shoe doesn't make a match | 1st, 2nd, 3rd, 4th shoes didn't match) = 1 * (10/11) * (8/10) * (6/9) * (4/8)

    which leads to the same answer as darkdream gave.
    I don't have emotions and sometimes that makes me very sad.

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