Among $9$ people there are $7$ females and $2$ males. You want to make a committee of $3$ people. What is the probability the committee has at least two females?
I wanted to go about this problem $2$ ways to check my answer...however I'm getting different answers and I'm not sure which to trust.
So ideally there should be ${9\choose3}$ different total combinations of committees which is equal to $84$. However if I break this into cases I don't get the same answer.
The first case would be a committee of $3$ females: ${7\choose3} =35$, then $2$ females and $1$ male: ${7\choose2}{2\choose1}=42$ and finally $1$ female and $2$ males ${7\choose1}{2\choose2}= 14$ but this sums to $91$. So which total amount of arrangements is correct, and why aren't I getting the same result?