If 2 different representatives are to be selected at random from 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 ithink the answer should be D but GWD answer said it is E why?

(1) more than 1/2 out of 10 are woman. Therefore, p> 5/10 x 4/9 . So can dertermin p<1/2

(2) can use man+man<1/10 to find how many man tther eare/ so can find if p>1/2

1)P=c(2,n)/c(2,10)={n(n-1)/2}/{10*9/2]=n(n-1)/90, since n>5, so P=n(n-1)/90>5*4/90=2/9

(2)men=m, P(m)=c(2,m)/c(2,10)=m(m-1)/90<1/10, m(m-1)<9, m<4, n>=6, if n=6, P=n(n-1)/90=1/3