shubhamkumar wrote: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
