Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

B) is wrong because the probability is less than 1/10

so the most you can have is 3 men as if you have 4 the probability is 12/90 > 1/10. so 3 men, which means 7 women or more women. but 7 women means7/10*6/9 = 42/90 = ~46%

I. Ins., 'cause it doesn't give us actual number:
- it could be 6 women, then P = 6/10 * 5/9 = 30/90 = 1/3 <1> 1/2
II. Ins. We can say only that number of men is less then 3, 'cause: 1/10 = 9/90, so m/10 * (m-1)/9 should = 9/90 and this is possible only wnen number of men is less then 3.
But in this case:
- 7/10 * 6/9 = 42/90 <1> 1/2

Together: again - we can't figure out what is the number of men and women - ins.

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 >0.5?

(1) More than half of the 10 employees are women.
(2) The probability that both representatives selected will be men is less than 0.1

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.

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 >0.5?

(1) More than half of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 0.1

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.

I picked B...any thought????

S1: More than half are women.

Well we could have 6/10*5/9 = 30/90 ---> 1/3 So NO p is not greater than 1/2.

Or we could have 8/10*7/9 = 56/90 which is greater than 1/2. Insuff.

S2:

Let m equal the probability that 2 men will be selected.

m<.1 or 1/10. Since we have ten employees. 1/10 must have originall looked like this: 3/10*3/9 (we are always going to take away 1 from the bottom). Since this representation is illogical ( i.e. we can't have 3 men on first and 3 men on second). The next highest number is 3/10*2/9. =6/90 which is less than .1.

This means that there are 7 women. 7/10*6/9 =42/90 slightly less than half so NO.

However, there is no restraint on how many men there are.

So we can have 2/10*1/9 = 2/90 or m=1/45. this means that there are
8 women.. 8/10*7/9 =56/90 p>1/2.

Together, I still don't see how this can be suff. B already tells us that the number of men is less than half of the number of women.

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 >0.5?

(1) More than half of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 0.1

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.

I picked B...any thought????

S1: More than half are women.

Well we could have 6/10*5/9 = 30/90 ---> 1/3 So NO p is not greater than 1/2.

Or we could have 8/10*7/9 = 56/90 which is greater than 1/2. Insuff.

S2:

Let m equal the probability that 2 men will be selected.

m<.1 or 1/10. Since we have ten employees. 1/10 must have originall looked like this: 3/10*3/9 (we are always going to take away 1 from the bottom). Since this representation is illogical ( i.e. we can't have 3 men on first and 3 men on second). The next highest number is 3/10*2/9. =6/90 which is less than .1.

This means that there are 7 women. 7/10*6/9 =42/90 slightly less than half so NO.

However, there is no restraint on how many men there are.

So we can have 2/10*1/9 = 2/90 or m=1/45. this means that there are 8 women.. 8/10*7/9 =56/90 p>1/2.

Together, I still don't see how this can be suff. B already tells us that the number of men is less than half of the number of women.

I say E.

Yeah you are correct i realized it later! I didn't consider the probability of selection of a man & a woman in considering P = 1-(1/10)

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.

Last edited by marcodonzelli on 05 Jan 2008, 08:44, edited 1 time in total.

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 21 of the 10 employees are women. (2) The probability that both representatives selected will be men is less than 101.

Something is wrong with numbers.
_________________

from s1, no of women may be 6, 7 , 8, or 9 with 6, the prob of both women is (6/10) x (5/9) = 30/90 < 1/2 with 9, the prob of both women is (9/10) x (8/9) = 72/90 > 1/2 so S1 not sufficient.

from S2, prob of both men is less than 1/10 let x be no of men, then prob of both men is (x/10)(x-1/9) we are given that (x/10)(x-1/9) < 1/10 then, x(x-1)/90 < 1/10 ie x(x-1) < 9 now find integer values for x less than 9 which satisfies x(x-1) < 9 , so here x=3or x=2 when x=3, no of women = 7 and the prob of two women is (7/10)(6/9)= 42/90 < 1/2 when x=2, no of women =8 and the prob of two women is (8/10) x (7/9) = 56/90 > 1/2 so, S2 is insufficient answer is E

Last edited by Dellin on 05 Jan 2008, 09:51, edited 1 time in total.

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

1) From 1, more than 1/2 of 10 are women. So no of women can be 6,7,8,9, or 10.

If 6, then 6C2/10C2 = 15/45 = 1/3 . This is less than half If 7, then 7C2/10C2 = 21/45 = 7/15. This is less than half as 15 is greater than twice of 7. If 8, then 8C2/10C2 = 28/45. This is more than half as 45 is less than twice of 28.

So if 6,7 , the prob is less than half. If 8,9,10 as we go up, the prob is more than half. NOT SUFF.

2) From 2, The probability that both representatives selected will be men is less than 1/10.

We have seen from 1 as no. of women went up the probability went up. So for prob of 2 men getting selected less than 2, the no of men has to be really low.

Lets start off with no. of men = 4. 4C2/10C2 = 6/45 = 2/15. This is greater than 1/10. 3C2/10C2 = 3/45 = 1/15. This is smaller than 1/0. So no. of men could be 3. 2C2/10C2 is going to be still less. No need to calculate. So no of men could be 2 also. 1 man is NA as we calculating prob of 2 men getting selected.

So is no of men are 2 /3 , the no of women will be 7/8 respectively. But we have seen from 1 that when no of women is 7, the prob is less than 1/2. So NOT SUFF.

Taking together, we get that the result could be either. So answer is E

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

6/10*5/9 --> 30/90 1/3 So S1 is insuff

2: Take the most scenario: 4 men 4/10*3/9-- 12/90 6/30 --> 1/5

Wait correct. Change to E.

there are at least 7 women. Together offers nothing.

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

6/10*5/9 --> 30/90 1/3 So S1 is insuff

2: Take the most scenario: 4 men 4/10*3/9-- 12/90 6/30 --> 1/5

Wait correct. Change to E.

there are at least 7 women. Together offers nothing.

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
_________________

------------------------------------------------------------- When you come to the end of your rope, tie a knot and hang on.

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

x women, 10 - x men prob 2 women: xC2/10C2 = x*(x-1)/90 prob 2 men: (10-x)C2/10C2 = (10-x)*(9-x)/90

1: insufficient x >= 6 -> x*(x-1)/90 >= 6*5/90 = 1/3 - insuff (if x = 6, p < 1/2; x = 8, p=8*7/90 > 1/2) 2: insufficient (10-x)*(9-x) < 1/10 -> (10-x)*(9-x) < 9 -> 10 - x <= 3 -> x >=7 -> p = 7*6/90 = 42/90 < 1/2 (x = 7, p < 1/2; x = 8, p > 1/2)

In this case I applied combinations, though I wasn't able to finish it in 2 mins. Is there any other quicker approach? Thanks a lot

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.

Statement 1: Says More than 1/2 are women. Assume there is 6 women. So selecting 1st women is you have = 6C1 options Selecting 2nd women you have = 5C1 options Probability = 6C1 * 5C1/10C2 = 2/9 So it can answer the question.

If you have more women then probability for both women will increase. So it will be always greater than 1/2.

Statement 2: Says probability of of both men is 1/10, then probability of not getting both men is = 1 - 1/10 = 9/10 But keep in mind both not men might also mean even 1 of them can be men so second option cannont answer the question.

Answer A.

gmatclubot

Re: gmatprep combs and probs ds
[#permalink]
09 Mar 2008, 17:07