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 (duh! ) 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:

Part A:

The probability on the best case scenario

The best case scenario results when the two missing socks are of the same pair.
Thus, there are 4 complete pairs of socks not missing.

Now, the number of ways of selecting 2 socks from a total of 5 x 2 = 10 socks is given by 10C2 = 45

The number of ways of selecting 2 socks from 10 sucks such that the two socks are of the same pair is the same as the probability of selecting 1 pair of socks from 5 pairs and is given by 5C1 = 5

Therefore, the probability of the best case scenario is given by 5 / 45 = 1 / 9

Part B:

The probability on the worst case scenario

The worst case scenario results when the two missing socks are not of the same pair.
Thus, there are 3 complete pairs of socks not missing.

Now, the number of ways of selecting 2 socks from a total of 5 x 2 = 10 socks is given by 10C2 = 45

The number of ways of selecting 2 socks from 10 sucks such that the two socks are not of the same pair is given by 10C2 - 5C1 = 45 - 5 = 40

Therefore, the probability of the worst case scenario is given by 40 / 45 = 8 / 9

Where’s the table? be more specific so we can you