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:
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
this is because you can simply do the butterfly method and multiply 4 times 2 which is 8 for 2/7 and 7 times 1 for 1/4 which is 7 so 2/7 is greater than 1/4
