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DS- Difficult Probability

This topic has 4 expert replies and 3 member replies
appu.keshav@gmail.com Newbie | Next Rank: 10 Posts Default Avatar
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DS- Difficult Probability

Post Wed Jan 28, 2015 7:31 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 > 21 ?.
(1) More than! of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 11.

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Post Wed Jan 28, 2015 8:12 pm
The problem should read as follows:

Quote:
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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Marty Murray Legendary Member
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Post Wed Jan 28, 2015 9:50 pm
Quote:
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.
This problem could also be done by calculating numbers of combinations.

Start by finding the total number of possible sets of two employees out of ten, regardless of gender.

(10*9)/2! = 45 possible sets of two employees

So for the probability, p, of a random set being just two women to be greater than 1/2, we need the possible sets of two women to be more than 1/2 of 45.

Statement 1 says that more than half of the 10 employees are women.

If 6 employees are women we can create 6*5/2! = 15 sets of women only. This is less than half. So more than half of the 10 employees can be women and the probability of creating sets of only women could be less than half.

If 10 employees are women we can create (10*9)/2! = 45 sets of women only. This is more than half.

So Statement 1 is insufficient.

Statement 2 can be translated to saying that fewer than ten percent of the sets are composed of men only.

If there are four men we can create (4*3)/2 = 6 sets of two men, which is more than 10 percent of 45.

So we know that there are fewer than 4 men.

3 men can be grouped into 3*2/2! = 3 sets of two men, which is less than 10 percent of 45. So 3 works and any fewer than 3 works.

3 men means 7 women. 7 women can be used to form (7*6)/2! = 21 sets of only women. This is less than half.

There could also be 0 men, meaning there would be 10 women, who could compose all the sets, which is obviously more than half.

So Statement 2 is insufficient.

Combining the statement doesn't help. We could still have between 3 and 0 men and between 7 and 10 women.

So that's still insufficient.

Choose E.

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Post Wed Jan 28, 2015 9:51 pm
Quote:
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

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binit Master | Next Rank: 500 Posts Default Avatar
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Post Sun May 03, 2015 3:33 am
Hi Brent,
I solved this Q considering cases: 6W, 4M; 7W, 3M and so on and that took me 3.26 minutes. Anyone pls suggest the level of this Question. Is it okay to take 3.26 min.??

~Binit.

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Post Sun May 03, 2015 6:00 am
binit wrote:
Hi Brent,
I solved this Q considering cases: 6W, 4M; 7W, 3M and so on and that took me 3.26 minutes. Anyone pls suggest the level of this Question. Is it okay to take 3.26 min.??

~Binit.
I'd put this question in the 700-800 range. Tough!
3.26 minutes is fine just as long as you don't spend that long on every question. Rather than look at timing on a question by question basis, I like to look at it in terms of batches of 5 questions. To this end, I suggest that students use the following Milestone Charts to stay on track:


This (and more) is covered in our free GMAT time management video at http://www.gmatprepnow.com/module/general-gmat-strategies?id=1244

NOTE: early in one's prep, there's a danger in focusing too much on timing. In my view, this focus has the potential to actually hinder your preparation. I feel so strongly about the whole timing issue that I wrote two articles about it:
- Making Friends with Time on the GMAT - Part I (http://www.gmatprepnow.com/articles/making-friends-time-gmat-%E2%80%93-part-i)
- Making Friends with Time on the GMAT - Part II (http://www.gmatprepnow.com/articles/making-friends-time-gmat-%E2%80%93-part-ii)
Executive Summary: Use a timer, but only to get an idea of what 2 minutes feels like. Later on (once you've covered all of the content), you can start working on your speed.

Executive Summary of the Executive Summary: Content First - Speed Second

I hope that helps.

Cheers,
Brent

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Post Sun May 03, 2015 11:22 pm
How about an easier way?

Suppose there are w women in the group.

The probability of selecting two women = (w/10) * (w-1)/9, or w*(w-1)/90

We want to know if this is greater than 1/2. Simplifying that inequality, the question becomes

is w*(w-1) / 90 > 1/2 ?

or

is w * (w - 1) > 45 ?

or

is w ≥ 8 ?

S1 tells us w > 5, which isn't sufficient.

S2 gives us (10 - w)/10 * (9 - w)/9 < 1/10, or (10 - w)(9 - w) < 9, or w ≥ 7. Not sufficient.

Together, obviously, we still don't know, so it's E.

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binit Master | Next Rank: 500 Posts Default Avatar
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Post Mon May 04, 2015 11:16 pm
Hi Brent,
Thanks a lot for your valuable suggestion.

~Binit.

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