4 ms·
You need to divide by 3! and 2!. One way to look at it is that every selection of five from the 4M and 10F is equally likely. There are \binom{14}{5} (i.e. 14 c
by ohmygodel 16y ago
You need to divide by 3! and 2!. One way to look at it is that every selection of five from the 4M and 10F is equally likely. There are \binom{14}{5} (i.e. 14 choose 5) such selections. Of these, \binom{4}{3} * \binom{10}{2} have exactly 3M and 2F. Therefore, the probability is \binom{4}{3} * \binom{10}{2} / \binom{14}{5}, or (4 * 3 * 2 * 10 * 9) 5! / (3! 2! (14 * 13 * 12 * 11 * 10)) = 10 * 9/(13 * 11 * 7) = 0.08991...
- dcosson 16y agoThis is how I'd solve it too. Plus, with a username like that and a LaTex formatted answer, I'm already inclined to trust your math skills :)
- light3 16y agoAhh indeed, I thought there were 5!=120 ways to order 3M and 2F, but apparently there are only 5!/(2!3!) = 10 ways: {MMMFF,MMFFM,MFFMM,FFMMM,FMMMF,MFMMF,MMFMF,FMMFM,MFMFM,FMFMM}. I made the mistake of thinking you can distinguish between individual males/females.