Probability problem: random selection of shoes

**Runty** · September 28th 2010, 06:26 AM

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.

**Plato** · September 28th 2010, 07:12 AM

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.

**Runty** · September 28th 2010, 07:25 AM

Originally Posted by Plato

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?
$P=\frac{\left(\begin{array}{cc}10\\3\end{array}\ri ght)}{\left(\begin{array}{cc}20\\6\end{array}\righ t)}$

**Plato** · September 28th 2010, 07:27 AM

That is three pair. Is it not?

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