Question

Hi everyone. I am struggling a bit with this one example question I am doing.
Problem:
There is a group of 12 people, 6 men and 6 women. A committee is to be formed consisting of 5 members from this group. Find the probability that Anne (one of the females) and her friend Billy (one of the males) are on the same committee.
My attempt:
I know that the total number of ways of selecting a committee is C(12,5).
I know that after Anne and Billy are chosen, there are 3 remaining places in the committee, and there are C(10,3) ways of doing this. But I am not sure what to do after this. At first, I thought this would work but it doesn't:
P = (C(6,1)*C(6,1)*C(10,3))/(C(12,5))

Answer 1

10C3 committees with Anne and Billy out of 12C5 possible committees

what fraction (probability) of possible committees are they on?

Answer 2

Hi Scott. Hmm, I am still unsure what you mean by that.
I guess the probability of choosing Anne, in particular, is 1/12, and then the probability of choosing Billy after that becomes 1/11. So maybe 1/12 * 1/11 * (10C3)/(12C5) ? I am quite confused.

Answer 3

Oh I simply divide those two results I mentioned, but I am not sure why that works. Because can't, Anne and Billy, be placed in the committee in different ways? 10C3 / 12C5 makes it seem like the ways in which Billy and Anne can be arranged in the committee doesn't matter.