A committee of three is to be selected from a pool of candidates consisting of five men and four women. If all the candidates are equally likely to be chosen, what is the probability that the committee will have an odd number of female members?

Here's my solution: $\frac{_5\rm C_2\times_4\rm C_1+_5\rm C_0\times_4\rm C_3}{_9\rm C_3} \Rightarrow \frac{11}{21}$. Is this correct?

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    $\begingroup$ Looks fine to me $\endgroup$ – Shubham Johri Sep 2 '19 at 10:13

Looks good to me. In your numerator you have the number of potential committees that contains $1$ woman (and two men), added to the number of potential committees that contain $3$ women (and no men). In the denominator you have the total number of potential committees. Since each potential committee is equally likely, this gives the probability that there are either exactly one, or exactly three women in the committee.

The actual calculation is correct too, which always helps.

As a teacher, I have two things I would've liked to see. One is a sentence or two (like my paragraph above, but you don't have to be that thorough) explaining where all the numbers come from. The other is that I would've preferred a $=$ rather than $\Rightarrow$, since that's what you have: two fractions with the same value.

Exactly how important these are is something only your teacher can tell you, though. I am not the one correcting your tests, hand-ins and exams.


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