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If 2 different representatives are to be selected at random

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 [#permalink] New post 21 Aug 2007, 15:57
E


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%

but 8 women is 8/10 * 7/9 = 56/90 > 50%

so INSUFFICIENT

E!!!!!!!
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 [#permalink] New post 21 Aug 2007, 17:41
Got E here:

1) From the problem we have:

w - women
m - men

w/10 * (w-1)/9 = probability of choosing 2 women.

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.

Ans. E
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DS - SET 23 - Q2 [#permalink] New post 26 Sep 2007, 12:46
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????
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Re: DS - SET 23 - Q2 [#permalink] New post 26 Sep 2007, 12:57
singh_amit19 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 >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.
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Re: DS - SET 23 - Q2 [#permalink] New post 26 Sep 2007, 13:05
GMATBLACKBELT wrote:
singh_amit19 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 >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)

Thanks though!
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 [#permalink] New post 26 Sep 2007, 19:18
St1:
Cases:
6 women, 4 men -> p = 6/10 * 5/9 = 0.33
7 women, 3 men -> p = 7/10 * 6/9 = 0.46
8 women, 2 men -> p = 8/10 * 7/9 = 0.62
Insufficient.

St2:
Let the number of men be x.

Then x/p + (x-1)/(p-1) = 0.1 --> can't solve.
Insufficient.

St1 and st2:
6 women, 4 men. Then p(2 men) = 4/10 * 3/9 = 0.13. Out.
7 women, 3 men. Then p(2 men) = 3/10 * 2/9 = 0.06. Possible. p(2 women) = 0.46
8 women, 2 men. Then p(2 men) = 2/10 * 1/9 = 0.02. Possible. p(2 women) = 0.62
Insufficient.


Ans E
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still prob [#permalink] New post 05 Jan 2008, 08:33
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.
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Re: still prob [#permalink] New post 05 Jan 2008, 08:42
Expert's post
marcodonzelli 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 > 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.
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Re: still prob [#permalink] New post 05 Jan 2008, 09:04
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.
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Re: still prob [#permalink] New post 05 Jan 2008, 09:04
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E

w - the number of women
m - the number of men.

w+m=10

p(w,w)=w/10*(w-1)/9=w*(w-1)/90
p(w,w)>1/2 --> w*(w-1)/90>1/2 --> w*(w-1)>45 --> w≥8, m≤2

1. w>5, m≤5 insuff.
2. p(m,m)<1/10 --> m*(m-1)/90<1/10 --> w≥7, m≤3. insuff.
1&2. w≥7, m≤3. insuff.
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Re: still prob [#permalink] New post 05 Jan 2008, 09:26
i think i was wrong since i considered only one value 3 for the eqn m(m-1)<9
correct answer should be E
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probab [#permalink] New post 19 Jan 2008, 10:29
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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Re: probab [#permalink] New post 19 Jan 2008, 10:41
1. 6C2/10C2 = 15/45 = 1/3 INSUFFICIENT

2. xC2/10C2 = x/45 = <1/10

which means that xC2 must be < 4.5

3C2 = 3
4C2 = 6

so there must be 3 or fewer men in the group. Which means there are AT LEAST 7 women in the group.

7C2/10C2 = 21/45 = 7/15 which is < 1/2

INSUFFICIENT

Together we don't learn anything new.

Answer E
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Re: probab [#permalink] New post 19 Jan 2008, 11:15
For me the answer is E. This is how is solved it.

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
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Re: probab [#permalink] New post 19 Jan 2008, 11:55
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marcodonzelli 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


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.

we can have 7/10*6/9 or 8/10*7/9
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Re: probab [#permalink] New post 20 Jan 2008, 04:36
GMATBLACKBELT wrote:
marcodonzelli 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


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.

we can have 7/10*6/9 or 8/10*7/9



OA is in fact E guys, good work!
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Probability [#permalink] New post 05 Feb 2008, 10:57
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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Re: Probability [#permalink] New post 05 Feb 2008, 11:12
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suntaurian 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


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)

2 -> 1 -> E
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gmatprep combs and probs ds [#permalink] New post 09 Mar 2008, 01:29
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.
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Re: gmatprep combs and probs ds [#permalink] New post 09 Mar 2008, 17:07
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.
Re: gmatprep combs and probs ds   [#permalink] 09 Mar 2008, 17:07
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