Five men and four women are waiting to be interviewed for jobs. If they are all selected in random​ order, find the probability that no man will be interviewed until at least two women have been interviewed.

Answer 1

We must exclude the cases where the
first two are MW and MM
with no restriction, the number of ways to arrange the interviews is
9!/(4!5!) = 126
case of MM = 7!/(3!4!) = 35
case of MF = 7!/(4!3!) = 35
Prob(your stated event) = 1 - 70/126 = 4/9

Answer 2

answer ' =55