I'm working on probability in Calc & having problems w/ the following question I'd greatly appreciate any help

Question

A smoky hill IB committee is to consist of 3 teachers and 3 students who are to be chosen at random from 24 senior IB teachers and 145 IB students. If half of the teachers and 99 students are women, find the probability that the committee has:
a) Only female members
b) Only 1 teacher and only 1 student who are women

I'm not sure how to directly reply to answers so I'll just write it here. Cattbarf, your first answer is incorrect. The way you solved it would mean there has to be 3 female teachers AND 3 female students when actually there should be a total of 3 females, between the teachers and the students. I very much appreciate you replying though. Thanks for trying to help.

Answer 1

No of ways of choosing the committee
= C(24,3) * C(145,3)

No of ways of choosing only women
= C(12,3) * C(99,3)

So P(only women) = C(12,3) * C(99,3) / [C(24,3) * C(145,3)]
= 0.03426

No of ways of choosing 1 female teacher and 1 female student
= C(99,1)*C(46,2) * C(12,1) * C(12,2)
You also have to encorporate the no of ways of choosing 2 men out of the 12 and 46 etc etc.

P(2nd part) = C(99,1)*C(46,2) * C(12,1) * C(12,2) / [C(24,3) * C(145,3)]

You work that out.

Answer 2

The selection of teachers and students are assumed to be independent events, so we can consider each one separately.

For situation (a), p(female teacher)= 12/24=0.5. So the probability that we have 3 female teacher members is 1/8. For students p=(99/145), so the probability that we have 3 female students is
(99/145)^3. The probability that BOTH events happen is the product of these probabilities.

For situation (b), we find the binomial distribution for choices of 3 members from each group given the above probabilities. Then we evaluate the term corresponding to one female and two males for each group. Once we have those terms, they are multiplied together to find the desired probability.