This problem is tricky in that, on top of being a hypergeometric selection-without-replacement question, it also invokes conditional probability. The best approach is to break it down to first principles and keep as close a link to careful logical expression as possible.
The first thing to deal with is the fact that we're not talking about a sample space of all possible ways to fill the subcommittee, because some ways of filling it involve only one gender. What we need to find out is how many ways this can happen.
The best approach is to first consider all possible outcomes and then subtract the ones that we don't want.
The number of all possible outcomes, when selecting 3 people from a total of 18, is 18C3, or 816.
However, some of those outcomes, as I've said before, are ones with only men or only women. So we need to find out how many of those outcomes exist and subtract them 816.
How many 3-person subcommittees consist of only men and no women? The 3 men would have to be selected from the 10 men, which is 10C3, or 120.
How many 3-person subcommittees consist of only women and no men? The 3 women would have to be selected from the 8 women, which is 8C3, or 56.
These 176 gender-exclusive outcomes must be removed from the total 816 outcomes. This is 816 - 176 = 640. This will be the denominator.
Now it remains to get the number of successful outcomes. This is is composed of a selection of 2 out of 10 men and 1 out of 8 women, which are 10C2 and 8C1, respectively. These numbers are 45 and 8. Since they happen together, there are 45 • 8 ways this can happen, or 360.
Put it all together, and the probability is 360/640, or 0.5625.
