This is a little tricky. You could just try np = 2 x
½ = 1 which does give the correct answer, but this is
assuming the binomial distribution which does not apply in this case
because the sampling fraction is too small. To do this properly, you
have to compute the probabilities of finding 0, 1, or 2 men
on the committee.
The denominator will be 15, since there are 15 possible
committees. There are 3 committees consisting entirely of
men (each leaving out one of the three men), and thus the probability of
having a committee of two men is 3/15 = 1/5. Similarly, so
is the probability of two women (no men). The probability of having
exactly one man (and one woman) is what is left:
1 - 1/5 - 1/5 = 3/5.
Thus the expected number of men is:
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