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I'm tutoring a math student and I'm having trouble with this question

Q : A sub-committee of 7 members must be formed from 15 students. There are 9 male students and 6 female students. What is the probability that there are EXACTLY 3 females on the sub committee.

A : The official answer is : 56/143

But according to my calculations :

(6C3 + 9C4) / 15C7 = (20 + 126) / 6435 = 146/6435

I'm not sure where I'm going wrong?

Please help?

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up vote 5 down vote accepted

Multiply instead of adding. How many ways are there to choose a committee of $3$ females and $4$ males?

The females can be chosen in $\binom{6}{3}$ ways. For every way of choosing the females, there are $\binom{9}{4}$ ways to choose the females.

It follows that there are $\binom{6}{3}\binom{9}{4}$ possible committees of the desired type.

Remark: In calculating the probability, we make the assumption that the $\binom{15}{7}$ possible committees are all equally likely. This assumption should probably have been made explicit in the statement of the problem. In real-world committee formation, it is quite common to pay some attention to female-male balance. Thus the "equally likely" assumption is real-world unreasonable.

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Thank you so much, Mr Nicolas, this one was really bugging me because I knew I was using the right numbers. – Andy Jan 3 '13 at 19:27
I could not have explained better. +1 – mick Jan 3 '13 at 21:13

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