A jury pool consists of 6 men and w women. If 2 jurors are selected from the pool at random, is the probability that 2 men will be selected higher than the probability that 1 man and 1 woman will be selected?

(1) w> 3

(2) w < 6

Target question: Is P(2 men) greater than P(1 man and 1 woman)?

Statement 1: w> 3
Let's see what happens if w = 3 (note: this is the best chance that P(2 men) will be greater than P(1 man and 1 woman)

P(2 men) = P(man selected 1st and man selected 2nd)
= (6/9)(5/8)
= 30/72

P(1 man and 1 woman) = P(man selected 1st and woman selected 2nd OR woman selected 1st and man selected 2nd)
= P(man selected 1st and woman selected 2nd) + P(woman selected 1st and man selected 2nd)
= (6/9)(3/8) + (3/9)(6/8)
= 18/72 + 18/72
= 36/72

So, when w = 3, P(2 men) is not greater than P(1 man and 1 woman)

IMPORTANT: Now that we've shown that P(2 men) is not greater than P(1 man and 1 woman) when w = 3, we can see that, as the value of w increases, the answer to the target question will always remain the same.

As such, statement 1 is SUFFICIENT

Statement 2: w < 6
Consider these two conflicting cases:

Case a: w = 1
P(2 men) = (6/7)(5/6)
= 30/42

P(1 man and 1 woman) = P(man selected 1st and woman selected 2nd OR woman selected 1st and man selected 2nd)
= P(man selected 1st and woman selected 2nd) + P(woman selected 1st and man selected 2nd)
= (6/7)(1/6) + (1/6)(6/7)
= 6/42 + 6/42
= 12/42

So, when w = 1, P(2 men) is greater than P(1 man and 1 woman)

Case b: w = 3
In statement 1, we already showed that, when w = 3, P(2 men) is not greater than P(1 man and 1 woman)

Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent

Hi hemant_rajput,

This DS question can be solved by TESTing Values. It's also important to remember that Quant questions are almost always "pattern-based", so even if you don't immediately spot a pattern, you should continue to look for one.

We're told that we have 6 men and W women. We're asked if the probability of selected 2 men from the group is greater than the probability of selecting 1 man and 1 woman. This is a YES/NO question.

Fact 1: W ≥ 3

IF....
W = 3
M = 6
Total = 9
The probability of selecting 2 men is (6/9)(5/8) = 30/72
The probability of selecting 1 of each can be calculated in a couple of different ways. Here, I'll calculate the following:
1st = male, 2nd = female + 1st = female, 2nd = male.....
The probability is (6/9)(3(8) + (3/9)(6/8) = 18/72 + 18/72 = 36/72
The answer to the question is NO.

W = 4
M = 6
Total = 10
The probability of selecting 2 men is (6/10)(5/9) = 30/90 NOTICE how the numerator DOES NOT CHANGE
The probability of selecting 1 of each....
The probability is (6/10)(4/9) + (4/10)(6/9) = 24/72 + 24/72 = 48/72 NOTICE how the numerator GETS BIGGER
The answer to the question is NO.

We now have the pattern behind Fact 1: adding women makes the probability of selecting 2 men DECREASE and the probability of selecting 1 of each INCREASE. In this scenario, the answer to the question is ALWAYS NO.
Fact 1 is SUFFICIENT.

Fact 2: W < 6

Here, we can use W = 3 (from Fact 1 above) and the answer to the question is NO.

If W = 0, then the probability of selecting 2 men = 100% and the probability of selecting 1 of each = 0%
The answer to the question is YES.
Fact 2 is INSUFFICIENT

Final Answer: A

GMAT assassins aren't born, they're made,
Rich