This is a great question! What it really boils down to is finding a way to count the number of ways each of these things can happen. For instance, for Question a. we need to compute
# of ways to get an all-female committee
# of ways to get a committee
This is pretty straight forward when we think about it. We need 3 teachers from the 20 available teachers, and 3 parents of the 100 available parents. These are independent groups of people, so the combined number of ways to choose both teachers and parents is given by
# of ways to get a committee = 20C3 * 100C3 = 184338000
where C is the choose operator (20 choose 3). Since only half the teachers are female, we have 10 female teachers, and we already know 80 of the 100 parents are female. To select an all female committee means selecting from only possible committees selected from these groups. That gives
# of ways to get an all-female committee = 10C3 * 80C3 = 9859200
It follows that we have the following probability:
P(all-female committee) = 9859200 / 184338000 = 0.05348436025 (or around 5%)
For part b. we will have to be a little more carefully because, if we select 1 female teacher, that means we select 2 male teachers, and we have to account for both of the different ways to do that! (There are only 1 way to select 0 male teachers from the 10 male teachers, so we did not have to worry about this before.) At any rate, we need 1 female teacher from 10, 2 male teachers from 10, 1 female parent from 80, and 2 male parents from 20. This gives
# of ways to get this committee = 10C1*10C2*80C1*20C2 = 6840000.
It follows that we have
P(this committee) = 6840000 / 184338000 = 0.03710575139 (or around 3.7%)
So it seems that we are more likely to get an all-female committee than this configuration. Cool!