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If 2 different representatives are to be selected at random

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by rakeshd347 » Sat Oct 05, 2013 6:10 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.

OA is E
Last edited by rakeshd347 on Sat Oct 05, 2013 10:52 pm, edited 1 time in total.
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Source: — Data Sufficiency |
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by vinay1983 » Sat Oct 05, 2013 6:44 pm
rakeshd347 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.
Is it A
You can, for example never foretell what any one man will do, but you can say with precision what an average number will be up to!
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by rakeshd347 » Sat Oct 05, 2013 6:56 pm
vinay1983 wrote:
rakeshd347 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.
Is it A
No
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by theCodeToGMAT » Sat Oct 05, 2013 8:53 pm
10 employees --> 2 selected
P --> P(WW) > 1/2?

Statement 1:
Women's count : 6,7,8,9,10
P = .6 * .5 = .30 NO
p = .9 * .8 = .72 YES
INSUFFICIENT

Statement 2:
Men's probability < 1/10
M*(M-1)/90 < 1/10
M * (M-1) < 9
M = 0, 1, 2, 3
sO, W = ANY NUMBER
INSUFFICIENT

COMBINING...

W = 6,7,8,9,10
INSUFFICIENT

Answer [spoiler]{E}[/spoiler]

What is the OA?
Last edited by theCodeToGMAT on Mon Oct 07, 2013 7:30 am, edited 1 time in total.
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by [email protected] » Sun Oct 06, 2013 1:16 am
Hi rakeshd347,

White Code has used TESTing Values to do some of his work, and his answer is correct, I want to point out WHY it's correct.

We're told that there are 10 employees (men and women). We're asked if the probability of selected 2 women is > 1/2.

Fact 1: More than half are women. So W COULD BE 6, 7, 8, 9 or 10. Code's explanation provides proof of what happens when there are 6 women (the answer is NO) vs what happens when there are 9 women (the answer is YES).
Fact 1 is INSUFFICIENT

Fact 2: The probability of getting 2 men is < 1/10.

This will only happen when M = 0, 1, 2, or 3 (you'd have to do a bit of "probability math" to prove it)
So, the number of women COULD BE 7, 8, 9 or 10 (NOT any number)

If W = 7, then the odds of getting 2 women is (7/10)(6/9) = 42/90 and the answer to the question is NO.
If W = 9, then the odds of getting 2 women is (9/10)(8/9) = 72/90 and the answer to the question is YES.
Fact 2 is INSUFFICIENT.

Combining Facts, we have an "overlap" or 7, 8, 9 or 10
If W = 7, then the answer is NO.
If W = 9, then the answer is YES.
Combined, INSUFFICIENT.

Final Answer: E

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by ceilidh.erickson » Mon Oct 07, 2013 6:46 am
theCodeToGMAT wrote: Statement 1:
Women's count : 6,7,8,9,10
P = .6 * .6 = .36 NO
p = .9 * .9 = .81 YES
INSUFFICIENT
BE CAREFUL HERE! If we have 6 women, the probability of selecting 2 women is NOT (6/10)*(6/10). We've already chosen one woman, so we only have 5 women left, and a pool of 9 people left. The actual probabilities are:
6 women: (6/10)(5/9) = 1/3 NO
9 women: (9/10)(8/9) = 4/5 YES

So statement 1 is INSUFFICIENT, but not for the reason you listed.

Statement 2:
As Rich said, test numbers to prove that this only works when the number of men is 3, 2, 1, or 0, and thus 7-10 women.
7 women: (7/10)(6/9) = 7/15 NO (it's easiest to reduce the fraction)
10 women: 1*1 = 1 YES
Test the easiest number here - it's a lot less math! Easy to see that we'd have a 100% chance if we had all women.
INSUFFICIENT

As Rich said, when we combine the statements, we could still have a "no" answer with 7 or 8, and a "yes" answer with 9 or 10 women.

But remember - when you're drawing from the same pool without replacement, you have a diminishing number of choices!
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by theCodeToGMAT » Mon Oct 07, 2013 7:30 am
ceilidh.erickson wrote:
theCodeToGMAT wrote: Statement 1:
Women's count : 6,7,8,9,10
P = .6 * .6 = .36 NO
p = .9 * .9 = .81 YES
INSUFFICIENT
BE CAREFUL HERE! If we have 6 women, the probability of selecting 2 women is NOT (6/10)*(6/10). We've already chosen one woman, so we only have 5 women left, and a pool of 9 people left. The actual probabilities are:
6 women: (6/10)(5/9) = 1/3 NO
9 women: (9/10)(8/9) = 4/5 YES

So statement 1 is INSUFFICIENT, but not for the reason you listed.

Statement 2:
As Rich said, test numbers to prove that this only works when the number of men is 3, 2, 1, or 0, and thus 7-10 women.
7 women: (7/10)(6/9) = 7/15 NO (it's easiest to reduce the fraction)
10 women: 1*1 = 1 YES
Test the easiest number here - it's a lot less math! Easy to see that we'd have a 100% chance if we had all women.
INSUFFICIENT

As Rich said, when we combine the statements, we could still have a "no" answer with 7 or 8, and a "yes" answer with 9 or 10 women.

But remember - when you're drawing from the same pool without replacement, you have a diminishing number of choices!
Yeah.. correct..
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by GMATGuruNY » Mon Oct 07, 2013 6:03 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.
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.
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