If 2 different representatives are to be selected at random fro a group of 10 employees and if p is the probability that both representatives selected will be women, is p>1/2?

(1) More than 1/2 of the 10 employees are women.

(2) The probability that both representatives selected will be men is less than 1/10.

I chose B ?  The answer that I found is E ! Why?   Please, Help! 

(1) is obviously not sufficient.

(2) needs some computation.

Suppose that there are x men, then Prob(2 men selected)=(x/10)*((x-1)/9)<1/10. Solving the inequality involves an estimation of sqrt(37), which is close to 6. Therefore integer x must be <=3.

If x=3, then there are 7 women. Prob(2 women selected)=(7/10)*(6/9)=14/30<15/30=1/2. If x=1 or 2, Prob(2 women selected)>1/2. So (2) is not sufficient, either.

There are at least 7 women already, so (1) and (2) together are not sufficient.

Choose E.