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Math Help - Probability problem: random selection of shoes

  1. #1
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    Probability problem: random selection of shoes

    I'm sure a lot of people have seen questions like this one. I've gotten half of it done, but I could use a hand with the other half.

    A large box contains 10 pairs of shoes. If 6 shoes are selected at random, what is the probability of:

    a) no complete pairs
    My answer: P=\frac{\left(\begin{array}{cc}10\\6\end{array}\ri  ght) 2^6}{\left(\begin{array}{cc}20\\6\end{array}\right  )}

    b) exactly one complete pair
    My answer: P=\frac{10 \left(\begin{array}{cc}9\\4\end{array}\right) 2^4}{\left(\begin{array}{cc}20\\6\end{array}\right  )}

    c) exactly two complete pairs

    d) exactly three complete pairs

    I could use a hand with the latter two.
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  2. #2
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    I might interest you to know that the number of ways to choose at least one pair is  \displaystyle \sum\limits_{k = 1}^3 {\binom{10-k}{6-2k}\binom{10}{k}2^{6 - 2k} }

    If you see the pattern, you can finish the question.
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  3. #3
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    Quote Originally Posted by Plato View Post
    I might interest you to know that the number of ways to choose at least one pair is  \displaystyle \sum\limits_{k = 1}^3 {\binom{10-k}{6-2k}\binom{10}{k}2^{6 - 2k} }

    If you see the pattern, you can finish the question.
    Just so I'm sure, for the last one, it would be the following?
    P=\frac{\left(\begin{array}{cc}10\\3\end{array}\ri  ght)}{\left(\begin{array}{cc}20\\6\end{array}\righ  t)}
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  4. #4
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    That is three pair. Is it not?
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