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Math Help - Probabilities...

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Probabilities...

    I thought this was gonna be easy...but I'm stumbling a little bit:

    While dressing in the dark, Joe selects 8 socks at random from a drawer containing 10 different pairs. What is the probability that at least one pair of socks match?
    I don't want anyone to solve this, but rather, I'd like to see if I'm doing this right.

    Let X represent the number of matching pairs. We are looking for P(X\geq1). I figured that P(X\geq1)=1-P(X=0), where P(X=0) is the probability of none matching.

    Since there are 20 different socks, and 2 are needed to make a pair (also noting that the pairs are different), I think that P(X=0)=\frac{\displaystyle\binom{2}{1}^8\binom{2}{  0}^2}{\displaystyle\binom{20}{8}}

    Thus, I get P(X\geq1)=1-\frac{\displaystyle\binom{2}{1}^8\binom{2}{0}^2}{\  displaystyle\binom{20}{8}}\approx\boxed{.998}

    Does this look right?

    I'd appreciate any help!!
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  2. #2
    MHF Contributor

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    Quote Originally Posted by Chris L T521 View Post
    Since there are 20 different socks, and 2 are needed to make a pair (also noting that the pairs are different), I think that P(X=0)=\frac{\displaystyle\binom{2}{1}^8\binom{2}{  0}^2}{\displaystyle\binom{20}{8}}

    Thus, I get P(X\geq1)=1-\frac{\displaystyle\binom{2}{1}^8\binom{2}{0}^2}{\  displaystyle\binom{20}{8}}\approx\boxed{.998}

    Does this look right?
    Not quite: on the numerator, you count the number of ways to choose socks in a given sequence of 8 pairs. As a consequence, you should multiply it by the number of ways to choose 8 pairs among 10 (like, first we choose the (unordered) pairs, and then the left/right sock in each pair).
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