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

Question

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

Answer 1

In the line "Probability that the first two selected are men with the third a woman," the probability of choosing a woman third, after choosing two men, is 7/13. You need to multiply by that as well. Similarly, the probability of choosing the woman first is missing from the following line. Finally, you also need to add in the additional case where a man is chosen first, a woman second, and another man third.

Another approach:

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 :)