Question

Imagine that after washing 5 distinct pairs of socks, you discover that two socks are missing. of course, you would like to have the largest number of complete pairs remaining. thus, you are left with 4 complete pairs in the best-case scenario and with 3 complete pairs in the worst case. assuming that the probability of disappearance for each of the 10 socks is the same, find the probability of the best-case scenario; the probability of the worst-case scenario; the number of pairs you should expect in the average case.

Answer 1

solution:

You have only two qualitatively different outcomes possible. Count

the number of ways to get each of the two.

There are just two possible outcomes here: the two missing socks

make a pair (the best case) and the two missing stocks do not make a

pair (the worst case). The total number of different outcomes (the ways

to choose the missing socks) is 10 C 2 = 45.

The number of best-case ones is 5; hence its probability is 5 /45 = 1/9

The number of worst-case ones is 45 − 5 = 40; hence its probability is 40/45 = 8/9.

On average, you should expect 4 • 1/ 9 + 3 • 8 /9 = 28/ 9 = 3 1/ 9 matching pairs.

Answer 2

You have only two qualitatively different outcomes possible. And hence we will count  the number of ways to get each of the two outcome.

There are just two outcomes that are possible here: the two missing socks

make a pair (the best-case) and the two missing stocks do not make a

pair (the worst-case).

The total number of different outcomes (that is the number ways  to choose the missing socks) = $\\ C^{10}_{2}= \frac{10!}{2!\times 8!}=\frac{10\times 9\times \not{8!}}{2\times \not{8!}}=5\times 9=45   \text{ways}$

The number of best-case ones are 5; hence their probability is  $\frac{5}{45} =\frac{1}{9}$

The number of worst-case ones are 45 − 5 = 40;

Hence, their probability is $\frac{40}{45} =\frac{8}{9}$.

In average case, one should expect = $4\times\frac{1}{9} + 3\times\frac{8}{9} =\frac{28}{9} = 3.11 (approx.)$ matching pairs.