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?

  • 3
    $\begingroup$ $\binom 22=1$ not $2$. Your approach is sound otherwise. $\endgroup$ – lulu Sep 23 '18 at 13:48
  • $\begingroup$ silly mistake! thank you! $\endgroup$ – Lil Sep 23 '18 at 14:04
  • $\begingroup$ Oh, I make worse mistakes than that all the time. Points to you for checking the sum. $\endgroup$ – lulu Sep 23 '18 at 14:07
  • $\begingroup$ The word arrange is misleading since it implies order. Using the word select or choose would be more appropriate in this context. $\endgroup$ – N. F. Taussig Sep 23 '18 at 14:13

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