# Probability that committee chosen from 8 men and 7 women has more men

A board of trustees of a university consists of 8 men and 7 women. A committee of 3 must be selected at random and without replacement. The role of the committee is to select a new president for the university. Calculate the probability that the number of men selected exceeds the number of women selected.

My try:

Given that the number of men should be greater, so I'll find the probability that 2 out of the 3 are men.

Probability that the first three selected are men : $\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}$

Probability that the first two selected are men with the third a woman: $\frac{8}{15}\times\frac{7}{14}$

Probability that the first selected is a woman and the other two are men : $\frac{8}{14}\times\frac{7}{13}$

Total Probability: $\frac{8}{15}\times\frac{7}{14}\times\frac{6}{13}+\frac{8}{15}\times\frac{7}{14}+\frac{8}{14}\times\frac{7}{13} = .697$

The correct answer is: $\frac{36}{65}=0.5538$, and thanks in advance

Number of committees consisting of two men and one woman: $\binom{8}{2}\binom{7}{1}$. Number of committees consisting of only men: $\binom{8}{3}$. The number of all possible committees is $\binom{15}{3}$. I think you can take it from there :)