Question

# 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?

Probability
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?

2021-08-22

(a)
Since there shouldn’t be a complete pair, we are allowed to choose at most 1 shoe 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 $$\displaystyle{\left(\begin{array}{c} {10}\\{8}\end{array}\right)}$$ ways.
Then from chosen 8 pairs, choose one shoe from each.
This can be done in $$\displaystyle{2}^{{8}}$$ ways as there are 2 choices for each pair.
Therefore, total number of choices is $$\displaystyle{\left(\begin{array}{c} {10}\\{8}\end{array}\right)}{2}^{{8}}.$$.
The number of ways to randomly choose 8 shoes out of 20 is $$\displaystyle{\left(\begin{array}{c} {20}\\{8}\end{array}\right)}$$. Therefore,
$$P(\text{no complete pair})=\frac{((10),(8))2^8}{((20),(8))}$$
(b)
For getting exactly one pair, we can first choose the pair which will appear completely. there are 10 ways of doin it.
Then we need to choose 6 shoes from 9 remaining pairs.
Proceeding as part(a) we see that there are ((9),(6))2^6 ways for this. Therefore, the total number is $$\displaystyle{\left(\begin{array}{c} {9}\\{6}\end{array}\right)}{2}^{{6}}\cdot{10}$$. Thus,
$$P(\text{exactly one pair})=\frac{((9),(6))2^6*10}{((20),(8))}$$
Result: $$P(\text{no complete pair})=\frac{((10),(8))2^8}{((20),(8))}$$
$$P(\text{exactly one pair})=\frac{((9),(6))2^6*10}{((20),(8))}$$