A company is hiring to fill four managerial vacancies. The candidates are five men and three women. If every candidate is equally likely to be chosen then the probability that at least one woman will be selected is _______(round off to $2$ decimal places).

It can be solved in both the ways. Answer is 0.92
Also ,in a naive way,

P(atleast 1 woman) = 1 – P(all men)
=1-(5/8).(4/7).(3/6).(2/5)

=1 – (0.625 * 0.571 * 0.5 * 0.4 )

= 1- 0.07137

= 0.9286

by
180 points 2
2