If exactly 4 boys volunteer, what is the probability that the team is all boys or no boys?
There is not enough information to make conclusions. We assume in addition that there are 4 boys, m girls and the aim is to make a team with n persons, "n\\leq m+4" . In case we want to make a team from 4 boys, the total number of choices is: "C_{n-4}^{m}=\\frac{(n-4)!}{m!(n-4-m)!}" . I.e., we take 4 boys and for the other places take m girls. In case we wish to make a team with no boys, the number of choices is: "C_n^m=\\frac{n!}{(n-m)!m!}".
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