Data Sufficiency

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 > 21 ?.
(1) More than! of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 11.

The problem should read as follows:

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.

The following cases satisfy both statements.

7 women, 3 men:
Statement 1: more than 1/2 the employees are women.
Statement 2: P(MM) = 3/10 * 2/9 = 1/15, which is less than 1/10.
Here, P(WW) = 7/10 * 6/9 = 7/15, so p<1/2.

10 women, 0 men:
Statement 1: more than 1/2 the employees are women.
Statement 2: P(MM) = 0.
Here, P(WW) = 1, so p>1/2.

Since p< 1/2 in the first case and p>1/2 in the second case, the two statements combined are INSUFFICIENT.

The correct answer is E.

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.

This problem could also be done by calculating numbers of combinations.

Start by finding the total number of possible sets of two employees out of ten, regardless of gender.

(10*9)/2! = 45 possible sets of two employees

So for the probability, p, of a random set being just two women to be greater than 1/2, we need the possible sets of two women to be more than 1/2 of 45.

Statement 1 says that more than half of the 10 employees are women.

If 6 employees are women we can create 6*5/2! = 15 sets of women only. This is less than half. So more than half of the 10 employees can be women and the probability of creating sets of only women could be less than half.

If 10 employees are women we can create (10*9)/2! = 45 sets of women only. This is more than half.

So Statement 1 is insufficient.

Statement 2 can be translated to saying that fewer than ten percent of the sets are composed of men only.

If there are four men we can create (4*3)/2 = 6 sets of two men, which is more than 10 percent of 45.

So we know that there are fewer than 4 men.

3 men can be grouped into 3*2/2! = 3 sets of two men, which is less than 10 percent of 45. So 3 works and any fewer than 3 works.

3 men means 7 women. 7 women can be used to form (7*6)/2! = 21 sets of only women. This is less than half.

There could also be 0 men, meaning there would be 10 women, who could compose all the sets, which is obviously more than half.

So Statement 2 is insufficient.

Combining the statement doesn't help. We could still have between 3 and 0 men and between 7 and 10 women.

So that's still insufficient.

Choose E.

Hi Brent,
I solved this Q considering cases: 6W, 4M; 7W, 3M and so on and that took me 3.26 minutes. Anyone pls suggest the level of this Question. Is it okay to take 3.26 min.??

~Binit.

How about an easier way?

Suppose there are w women in the group.

The probability of selecting two women = (w/10) * (w-1)/9, or w*(w-1)/90

We want to know if this is greater than 1/2. Simplifying that inequality, the question becomes

is w*(w-1) / 90 > 1/2 ?

or

is w * (w - 1) > 45 ?

or

is w ≥ 8 ?

S1 tells us w > 5, which isn't sufficient.

S2 gives us (10 - w)/10 * (9 - w)/9 < 1/10, or (10 - w)(9 - w) < 9, or w ≥ 7. Not sufficient.

Together, obviously, we still don't know, so it's E.

Hi Brent,
Thanks a lot for your valuable suggestion.

~Binit.