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VP
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If 2 different representatives are to be selected at random
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Updated on: 31 Jul 2012, 04:51
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52% (01:16) correct 48% (01:02) wrong based on 95 sessions
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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. OPEN DISCUSSION OF THIS QUESTION IS HERE: http://gmatclub.com/forum/if2differen ... 28233.html== Message from the GMAT Club Team == THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION. This discussion does not meet community quality standards. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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Originally posted by chunjuwu on 22 Aug 2004, 03:27.
Last edited by Bunuel on 31 Jul 2012, 04:51, edited 1 time in total.
OA added.



VP
Joined: 26 Apr 2004
Posts: 1168
Location: Taiwan

Excuse me,
have any ideas?
thank you



CIO
Joined: 09 Mar 2003
Posts: 460

I choose E.
In the first one, we just know a majority is women. That could be 6 women or more. If there were 6 women, the probability for both choices to be women would be 1/3, not above 1/2. If there were 8 women, the prob would be 28/45, which is above 1/2. If there were 7 women, the prob would be 7/15, also not 1/2. So there has to be 8 women at least to ensure larger prob than 1/2.
The second one tells us that the probability of two men is less than 1/10. If there were ONLY 2 men, then the probability would be 1/45, which is definately less than 1/10, but if there were 3 men, the probability would be 1/15, also less than 1/10. Only at 4 men is the probability greater than 1/10. So there could up to 3 men, and that means as few as 7 women.
So even together, we can't say conclusively that there are 8 women at least, so it must be E.



VP
Joined: 26 Apr 2004
Posts: 1168
Location: Taiwan

ian7777
thanks for your explanation.
I found when we come across such questions, we must draw a random
number to verify. But, it takes a few risks, because it is too hard to think carefully.
Do you have any idea?



CIO
Joined: 09 Mar 2003
Posts: 460

I think the question posted here is a good one, and it could potentially be on the test.
I think most of the questions posted here in probability and permutations are too hard, and are above the test, but there is value to working those out and learning all the ins and outs of the topic.
In general, you should do as many as you can, and get a feel for what the range of questions are that are out there. That way, when you see a question, it'll look familiar and you'll know what to do.
I don't see most problems as just putting a couple of numbers in and hoping for the best. Even with this problem, we had to use numbers to figure it out, but not haphazardly. The numbers were chosen to find out what happened when women are the majority. The experience of doing multiple problems should tell you that just because they're the majority it doesn't mean that picking 2 in a row is 50% or more. So we try a bunch to see what DOES make 50% or more.
I guess what I'm saying is that given enough practice and the right frame of mind, you can think carefully on the test and not just try random numbers to verify.
I also think that it's fine to take 3 or even 4 minutes on a given problem, as long as you know that doing the work will get you the answer. There will always be some problems that take only 20 or 30 seconds, so there should be ballance. Just don't do 4 minutes for a problem you'll get wrong anyway. Those you should do in 30 seconds, decide you won't get it, and save the time for something else.
Hope that was helpful?



Director
Joined: 05 May 2004
Posts: 566
Location: San Jose, CA

E nice Q
1 is not sufficient ... if # of women=6 then prob is 30/90<0.5
if # of women=9 prob is 72/90>0.5
2 is not sufficient
prob of men selected can be <.1 only if number of men are 2 and 3
if # of men = 2, # of fem = 8, p=56/90 >.5
if # of men = 3, # of fem = 7, p=42/90 <.5



Senior Manager
Joined: 22 Feb 2004
Posts: 347

Ian:
Do you think that this kind of question can be called a tough question by GMAT standards.. I mean a kind of question which might come if you are approaching 5051? And do you feel that it is normal to spend around 34 minutes in this question in the real exam. I solved it in approximately 3:15 minutes but was wondering if I took more time than required.
Thanks



CIO
Joined: 09 Mar 2003
Posts: 460

crackgmat750 wrote: Ian: Do you think that this kind of question can be called a tough question by GMAT standards.. I mean a kind of question which might come if you are approaching 5051? And do you feel that it is normal to spend around 34 minutes in this question in the real exam. I solved it in approximately 3:15 minutes but was wondering if I took more time than required. Thanks
yeah, I think all those things are true. This question is definately fair and in line with what I'd expect from the GMAT. I also have no problem spending more than 2 minutes on a question, as long as I know I'm going to have a problem down the road that won't take as long. Remember, you're distributing your time around the test, and it has to average 2 min/question. If you've been studying and really know your stuff, then at some point you'll get an easy algebra data sufficiency or some fraction problem that will only take about a minute, and you'll make up time.
That's how I take the test. When I want to be sure I'm not getting suckered into a trap when I see a hard question, I'll spend extra time on it.



Director
Joined: 08 Jul 2004
Posts: 586

DS Hard Probability
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22 Sep 2004, 04:21
Q2:
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 half 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.



Senior Manager
Joined: 19 May 2004
Posts: 291

E ?
for the possibility being > 0.5 we need more than 7 women in the group.
(7/10) *(6/9) < 0.5
(8/10)* (7/9) > 0.5
Statement 1: Insufficient.
Statement 2: Insufficient.
We can learn that there should be 3 or less men in the group,
which means 7 or more women.
Still not enough.
Even together we still can't be sure that there are more than 7 women.
So i pick E.
Not easy indeed!



Director
Joined: 08 Jul 2004
Posts: 586

How have you decided stat2 is not sufficeint?
S



Director
Joined: 16 Jun 2004
Posts: 885

1. Insufficient. We only know that the count of women is greater than 5. However, for the p>1/2, the number of women have to necesarily be 8 and above. Why? because 8C2/10C2 would be the first number that will be greater than 1/2. So, we do not know if its 6 women or 8 women, hence insufficient.
2. Insufficient. Because, the number of Men should be 3 or less. Why? only 3C2/10C2, is the first number that will make the probability of choosing men less than 1/10. This would mean the number of women can still be 7 or 8 or 9 whatever. If the number of women is 7 then we dont meet the stimulus criteria. If number of women is 8 then we are good to meet the critieria.
Both combined we still will be unsure if the number of women are 7(and not meet the criteria) or 8 & above (meeting the criteria).
SO E.



Director
Joined: 08 Jul 2004
Posts: 586

venksune wrote: 1. Insufficient. We only know that the count of women is greater than 5. However, for the p>1/2, the number of women have to necesarily be 8 and above. Why? because 8C2/10C2 would be the first number that will be greater than 1/2. So, we do not know if its 6 women or 8 women, hence insufficient.
2. Insufficient. Because, the number of Men should be 3 or less. Why? only 3C2/10C2, is the first number that will make the probability of choosing men less than 1/10. This would mean the number of women can still be 7 or 8 or 9 whatever. If the number of women is 7 then we dont meet the stimulus criteria. If number of women is 8 then we are good to meet the critieria.
Both combined we still will be unsure if the number of women are 7(and not meet the criteria) or 8 & above (meeting the criteria).
SO E.
Is it 3 or less men OR 2 or less men? from statement 2?



Director
Joined: 16 Jun 2004
Posts: 885

Its 3 or less  In which case the women will be 7 or more.
Let me see if I am doing some fundamental mistake or not. Lets assume 4 men in the group and 2 are selected. We would have 4C2/10C2, which will be equal to 2/15 or 1/(7.5) and is not less than 1/10. However, if we assume 3 men, it would be 3C2/10C2 which is 1/15 and is < than 1/10. Yep its 3 or les..



Intern
Joined: 08 May 2006
Posts: 18

probability :Men and women
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Updated on: 13 Aug 2006, 05:28
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.
Answers ??
Originally posted by Achilless on 13 Aug 2006, 05:02.
Last edited by Achilless on 13 Aug 2006, 05:28, edited 1 time in total.



Retired Moderator
Joined: 05 Jul 2006
Posts: 1731

I think you should edit the question????
How come 21 women of the 10 employees
I think it is 21% of the 10 employees are women.



Senior Manager
Joined: 09 Aug 2005
Posts: 273

Re: probability :Men and women
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13 Aug 2006, 06:35
[quote="Achilless"]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.
Answers ??[/quote]
Is it E ?
for p > 1/2 atleast 8 should be women
for men (p)< 1/10 men should be >5



Senior Manager
Joined: 05 Mar 2006
Posts: 346

E
S1: Lets say WOMEN = 6 OR 9
Probability 1 = (6/10)*(5/9) = 30/90 = 1/3 (therefore less then 0.5)
Probability 2 = (9/10)*(8/9) = 72/90 (greater then 0.5)
Insuff
S2: If prob 2 Men < 1/10
Therefore, (n/10)*(n1)/9 <1/10
n(n1) < 9 Since no fractions can exist, Men must be less then or equal to 3.
which also means Women must be greater then or equal to 7
So, (7/10)*(6/9) = 42/90 (less then 0.5)
If we try 8 Women: (8/10)*(7/9) = 56/90 (greater then 0.5)
Insuff
Both S1&S2, since S2 already satisfies S1 and still Insuff, both S1&S2 Insuff.
E, Pw can be greater then or less then 0.5



Senior Manager
Joined: 19 Jul 2006
Posts: 339

should be E
nicely explained by agsfaltex



Manager
Joined: 30 Jan 2006
Posts: 62

employees
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21 Aug 2007, 15:09
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.







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