Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.