A closet contains $10$pairs of shoes. If $8$shoes are randomly selected, what is the probability that there will be

(a) no complete pair?

(b) exactly$1$complete pair?

A closet contains $10$pairs of shoes. If$8$shoes are randomly selected.

Since there shouldn't be a complete pair, we are allowed to choose atmost $1$shoes from each pair. Therefore, our choice can be made in the following way: first, choose $8$pairs from the$10$pairs in the closet; this can be done in$\left(\begin{array}{c}10\\ 8\end{array}\right)$ways. Then from the chosen$8$pairs, choose one shoe from each. This can be done in $2^8$ways as there are $2$choices for each pair. Therefore, the total number of choices is$\left(\begin{array}{c}10\\ 8\end{array}\right)2^8$. The number of ways to randomly choose $8$shoes out is$\left(\begin{array}{c}20\\ 8\end{array}\right)$. Therefore,

$P\left(\text{no complete pair}\right)=\frac{\left(\begin{array}{c}10\\ 8\end{array}\right)2^8}{\left(\begin{array}{c}20\\ 8\end{array}\right)}$

For getting exactly one pair, we can first choose the pair which will appear completely; there are $10$ways of doing it. Then we need to choose $6$shoes from the$9$remaining pairs such that there are no complete pairs. Proceeding as part$\left(a\right)$we see that there are $\left(\begin{array}{l}9\\ 6\end{array}\right)2^6$ways for this. Therefore, the total number is$\left(\begin{array}{l}9\\ 6\end{array}\right)2^6*10$. Thus,

$P\left(\text{exactly one pair}\right)=\frac{\left(\begin{array}{l}9\\ 6\end{array}\right)2^6*10}{\left(\begin{array}{c}20\\ 8\end{array}\right)}$