alltimeacheiver wrote: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

Statement 1: more than 1/2 of the 10 employees are women
Plug in w=7, m=3.
P(ww) = 7/10 * 6/9 = 7/15. Is 7/15 > 1/2? No.

Plug in w=10, m=0.
P(ww) = 10/10 * 9/9 = 1. Is 1 > 1/2? Yes.
Since the answer can be both No and Yes, insufficient.

Statement 2: The probability that both representatives selected will be men is less than 1/10
Plug in w=7, m=3. This works because P(mm) = 3/10 * 2/9 = 1/15, and 1/15 < 1/10.
P(ww) = 7/10 * 6/9 = 7/15. Is 7/15 > 1/2? No.

Plug in w=10, m=0. This works because P(mm) = 0*0 = 0, and 0 < 1/10.
P(ww) = 10/10 * 9/9 = 1. Is 1 > 1/2? Yes.
Since the answer can be both No and Yes, insufficient.

Statements 1 and 2 together:
Since w=7, m=3 and w=10, m=0 satisfy both statements, and the first combination yields an answer of No while the second yields an answer of Yes, insufficient.

The correct answer is E.

alltimeacheiver wrote: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

Let us call men=x, women=y; P(2 women selected) is y/10 *(y-1)/9 OR (y^2-y)/90, and our question is P(2 women selected)>1/2? OR (y^2-y) >45 ? ___important thing: x and y are integers, as men and women can't be divided.

st(1) more than 1/2 of 10 are women ---> at least 6 women, test Is 6^2-6<45 -Yes, But 9^2-9<45 -No, hence Not Sufficient
st(2) P(2 men selected)<1/10; x/10 * (x-1)/9 = (x^2-x)/90 <1/10 OR x^2-x<9; with x=4 we have 4^2-4=12 and 12 is more than 9, so we have at most 3 men and this is Not sufficient to state that Yes or No, P(2 women selected)<1/2 as with 2 men we have P(2 women selected)>1/2 and with 3 men, P(2 women selected)<1/2
Combined st(1&2) still no change in probabilities as we compare relative increases/decreases in men/women

[spoiler]After careful calc I make E (1 point less for women's probability luck )[/spoiler]