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
ds
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- hemant_rajput
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Target question: Is P(2 men) greater than P(1 man and 1 woman)?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
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
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Brent
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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
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
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The number of ways to choose 2 men from 6 options = 6C2 = (6*5)/(2*1) = 15.hemant_rajput 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
Question rephrased:
Is the number of ways to choose 1 woman and 1 man less than 15?
Statement 1:
Case 1: w=3
Number of ways to choose 1 woman from 3 options = 3.
Number of ways to choose 1 man from 6 options = 6.
To combine these options, we multiply:
3*6 = 18.
In this case, the number of ways to choose 1 woman and 1 man is NOT less than 15.
If the value of w INCREASES, then the number of ways to choose 1 woman and 1 man will also INCREASE, since the pool of women will be GREATER.
Thus, the number of ways to choose 1 man and 1 woman CANNOT be less than 15.
SUFFICIENT.
Statement 2:
Case 1 also satisfies statement 2.
In Case 1, the number of ways to choose 1 man and 1 woman is NOT less than 15.
Case 2: w=0.
Here, the number of ways to choose 1 woman and 1 man is 0, since there are no women.
In this case, the number of ways to choose 1 woman and 1 man IS less than 15.
Since the answer is NO in Case 1 but YES in Case 2, INSUFFICIENT.
The correct answer is A.
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hemant_rajput 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
Given :
Number of Men = 6
Number of Women = w
Number of Individuals to be selected = 2
Question : Is P(2 Men) > P(1 Man and 1 Woman) ?
Statement 1) w≥ 3
Take extreme cases
i.e. Case 1: w = 3,
P(2 Men)= (6/9)x(5/8) = 30/72, AND P(1 Man and 1 Woman)= (6/9)x(3/8)x2! = 36/72
Inference : P(2 Men) < P(1 Man and 1 Woman)
and Case 2: w = 30,
P(2 Men)= (6/36)x(5/35) = 30/(35x36), AND P(1 Man and 1 Woman)= (6/36)x(30/35)x2! = 360/(35x36)
Inference : P(2 Men) < P(1 Man and 1 Woman)
Consistent answer
SUFFICIENT
Statement 2) w < 6
Take extreme cases
i.e. Case 1: w = 0,
P(2 Men)= 1, AND P(1 Man and 1 Woman)= 0 [No Women]
Inference : P(2 Men) > P(1 Man and 1 Woman)
and Case 2: w = 5,
P(2 Men)= (6/11)x(5/10) = 30/(11x10), AND P(1 Man and 1 Woman)= (6/11)x(5/10)x2! = 60/(10x11)
Inference : P(2 Men) < P(1 Man and 1 Woman)
Inconsistent answer
NOT SUFFICIENT
Answer: Option A
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