Asked by Zozina
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))
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))
Answers
Answered by
scott
10C3 committees with Anne and Billy out of 12C5 possible committees
what fraction (probability) of possible committees are they on?
what fraction (probability) of possible committees are they on?
Answered by
Zozina
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.
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.
Answered by
Zozina
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.
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