DS question on probability

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DS question on probability

by spetznaz » Thu Apr 24, 2014 10:14 pm
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.

A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are not sufficient.

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by [email protected] » Thu Apr 24, 2014 10:40 pm
Hi spetznaz,

This DS question asks if the probability that 2 people selected from a group of 10 people will both be women will be greater than 1/2. This is a YES/NO question. It ultimately comes down to the number of women in the group of 10 employees; it's also perfect for TESTing Values.

Fact 1: More than 1/2 of the 10 people are women.

If all 10 people are women, then the probability that 2 women are chosen is: (10/10)(9/9) = 1/1 = 100% and the answer is YES.
If 6 of the 10 people are women, then the probability that 2 women are chosen is: (6/10)(5/9) = 30/90 = 33.333% and the answer is NO.
Fact 1 is INSUFFICIENT.

Fact 2 The probability that both people selected will be men is less than 1/10

There could be....
10 women and 0 men --- the probability of choosing 2 men is 0%; the probability of choosing 2 women is 100% and the answer is YES

9 women and 1 man --- the probability of choosing 2 men is 0%; the probability of choosing 2 women is (9/10)(8/9) = 72/90 and the answer is YES

8 women and 2 men --- the probability of choosing 2 men is (2/10)(1/9) = 2/90; the probability of choosing 2 women is (8/10)(7/9) = 56/90 and the answer is YES

7 women and 3 men --- the probability of choosing 2 men is (3/10)(2/9) = 6/90; the probability of choosing 2 women is (7/10)(6/9) = 42/90 and the answer is NO
Fact 2 is INSUFFICIENT

Combined, we could have...
7 women and 3 men and the answer is NO
10 women and 0 men and the answer is YES
Combined, INSUFFICIENT

Final Answer: E

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Rich
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by Brent@GMATPrepNow » Fri Apr 25, 2014 9:14 am
spetznaz 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.

Target question: Is the probability that both representatives selected will be women > 1/2?

Let's examine some scenarios and see which ones yield situation where the probability that both representatives selected will be women > 1/2

Scenario #1 - 5 women & 5 men: P(both selected people are women) = (5/10)(4/9) = 20/90 (NOT greater than 1/2)
Scenario #2 - 6 women & 4 men: P(both selected people are women) = (6/10)(5/9) = 30/90 (NOT greater than 1/2)
Scenario #3 - 7 women & 3 men: P(both selected people are women) = (7/10)(6/9) = 42/90 (NOT greater than 1/2)
Scenario #4 - 8 women & 2 men: P(both selected people are women) = (8/10)(7/9) = 56/90 (PERFECT - greater than 1/2)

IMPORTANT: So, if there are 8 or more women, the probability will be greater than 1/2
We can even rephrase the target question...
REPHRASED target question: Are there 8 or more women?

Statement 1: More than 1/2 of the 10 employees are women.
This is not enough information. Consider these two conflicting cases:
Case a: there are 7 women, in which case there are NOT 8 or more women
Case b: there are 8 women, in which case there ARE 8 or more women
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statement 2: The probability that both representatives selected will be men is less than 1/10.
This is not enough information. Consider these two conflicting cases:
Case a: there are 8 women & 2 men. Here P(both are men) = (2/10)(1/9) = 2/90, which is less than 1/2. In this case there ARE 8 or more women
Case b: there are 7 women & 3 men. Here P(both are men) = (3/10)(2/9) = 6/90, which is less than 1/2. In this case there are NOT 8 or more women
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Even when we combine the statements, we can see that it's possible to have 7 women in the group OR 8 women in the group.
Since we still cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT

Answer = E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by spetznaz » Sat Apr 26, 2014 8:06 am
Thank you Brent and Rich :)