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655-705 Level|   Probability|      
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tom09b
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A committee consists of n women and k men. In addition there Updated on: 17 Feb 2012, 05:19
Originally posted by tom09b on 15 Feb 2012, 11:07.
Last edited by tom09b on 17 Feb 2012, 05:19, edited 2 times in total.
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A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probability that the number of women on the committee will increase?

(1) n + k = 12
(2) k/n = 1/3
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A committee consists of n women and k men. In addition there 15 Feb 2012, 12:20
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A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

Answer: B.
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A committee consists of n women and k men. In addition there 07 Jun 2013, 09:17
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tom09b
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

(1) n + k = 12
(2) k/n = 1/3
Show more

Rewording of the question:
What is the probability that a man is chosen to be replaced and the alternate to replace him is a woman.

What you need is (probability of man chosen) x (probability of woman alternate)

\(\frac{k}{k+n} * \frac{2}{4} = \frac{k}{2(k+n)}\)

(1) n+k = 12
if n=1 and k=11, \(\frac{11}{2(12)} = \frac{11}{24}\)
if n=2 and k=10, \(\frac{10}{2(12)} = \frac{10}{24}\)
insufficient

(2) \(\frac{k}{n} = \frac{1}{3}\)
if k=1 and n=3, \(\frac{1}{2(4)} = \frac{1}{8}\)
if k=2 and n=6, \(\frac{2}{2(8)} = \frac{1}{8}\); etc...
sufficient

Answer is B
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A committee consists of n women and k men. In addition there 15 Feb 2012, 12:31
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Bunuel
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

Answer: B.
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My intuition drove me to B, as well but.. I couldn't find the way! Thank you!!
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A committee consists of n women and k men. In addition there 03 Jul 2012, 09:40
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Here is how I analyzed it if it helps:

The probability of selecting a woman from the alternates as given is - (2/4) = (1/2)
The probability of selecting a woman from the committee is - n/(n+k)

Now, we need to figure out the probability of pick a woman from the committee AND from the alternates [P(W&W)]. Therefore this is an AND problem.

1. n/12 Insufficient
2. k/n=1/3. Therefore n/(n+k)=3/4
Sufficient because P(W&W)=(3/4)*(1/2)
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A committee consists of n women and k men. In addition there 07 Jun 2013, 06:12
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Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

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A committee consists of n women and k men. In addition there 20 Mar 2014, 05:34
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Bunuel
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

Answer: B.
Show more

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.
Please elaborate on this ..
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A committee consists of n women and k men. In addition there 20 Mar 2014, 05:57
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tusharGupta1
Bunuel
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

Answer: B.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.
Please elaborate on this ..
Show more

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.
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A committee consists of n women and k men. In addition there 20 Mar 2014, 06:33
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Bunuel
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

Answer: B.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.
Please elaborate on this ..

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.
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Thankx a ton :-D ............................................................................................................................................................
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A committee consists of n women and k men. In addition there 12 May 2014, 05:12
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Bunuel
A committee consists of n women and k men. In addition there are 4 alternates, 2 of whom are women. If one of the committee members is selected at random to be replaced by one of the alternates, also selected at random, what is the probabilty that the number of women on the committee will increase?

Notice that the probability of selecting a woman from 4 alternates is 1/2.

Now, the number of women on the committee will increase if we replace a man from the committee with a woman from 4 alternates. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.

(1) n + k = 12. Not sufficient as we can not get the probability of selecting a man from the committee.
(2) k/n = 1/3 --> k/(k+n)=1/4. Sufficient.

Answer: B.

. Hence, the probabilty that the number of women on the committee will increase is k/(k+n)*1/2.
Please elaborate on this ..

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.
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Dear Bunnel

Statement A: was clearly insufficient as we didnt know n & k individually.
statement B:
ratio of m:w = 1:3

so suppose, 1 m and 3 women r there in the committee

and if we replace the man with women, then the no. of women will increase.

now we have no man just 4 women..

probability of a new alternate as women in the committee: 3/4*1/2 = 3/8... lesser than 3/4


but if we replace the man with a man... there should be no change right as there will be 1 man and 3 women still
ans is diff therefore insufficient.. :(

Please explain.
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A committee consists of n women and k men. In addition there 12 May 2014, 08:53
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nandinigaur


Dear Bunnel

Statement A: was clearly insufficient as we didnt know n & k individually.
statement B:
ratio of m:w = 1:3

so suppose, 1 m and 3 women r there in the committee

and if we replace the man with women, then the no. of women will increase.

now we have no man just 4 women..

probability of a new alternate as women in the committee: 3/4*1/2 = 3/8... lesser than 3/4


but if we replace the man with a man... there should be no change right as there will be 1 man and 3 women still
ans is diff therefore insufficient.. :(

Please explain.
Show more

Don't understand what is your question...

The question asks what is the probability that the number of women on the committee will increase? The probability that the number of women on the committee will increase is k/(k+n)*1/2.

From (2) we get that k/(k+n)*1/2 = 1/4*1/2 = 1/8.
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A committee consists of n women and k men. In addition there 01 Nov 2015, 13:56
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unnecessary to get a number:

in order to increase the number of women, the only way is to replace a man with a woman

1. if a woman is replaced by a woman, the number will remain;

2. if a woman is replaced by a man, the number will decrease;

3. if a man is replaced by a man, the number will remain

so the possibility is:

(k/n+k)*(2/4), the key is the ratio of woman to man

(1) n+k=12, insufficient

(2)k/n=1/3, so k/n+k=1/4, sufficient

B
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A committee consists of n women and k men. In addition there 17 Oct 2016, 08:01
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Hi Bunuel,
Why does k/(k+n)*1/2 will represent the number of woman increase? How it relates to increment. Will you explain it?
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A committee consists of n women and k men. In addition there 17 Oct 2016, 08:05
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NaeemHasan
Hi Bunuel,
Why does k/(k+n)*1/2 will represent the number of woman increase? How it relates to increment. Will you explain it?
Show more

For the number of women on the committee to increase we must replace a man from the committee with a woman from 4 alternates.

The probability of selecting a man from the committee is (# of men in the committee)/(total # of people in the committee) = k/(k+n);
The probability of selecting a woman from 4 alternates is (# of women)/(total # of people) = 2/4 = 1/2.

Therefore, the probability that we select a man from the committee AND a woman from 4 alternates is k/(k+n)*1/2.

Hope it's clear.
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A committee consists of n women and k men. In addition there 02 Jun 2018, 22:56
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Given,
A committee of
n - women
k - men

Alternate choices, in case of replacement, available are 2 women & 2 men.

The probability of increasing the # of women, is by replacement of 1 man in the committee with 1 woman from the alternates.

Consider this as a selection of 2 people, one man from k men & one woman from the 2 alternates.

#of ways selecting 1 man from k men = k
#of way of selecting 1 woman from 2 women = 2

Required Probability = Probability of selecting 1 man from (n+k) members * Probability of selecting 1 woman from 4 alternates

Therefore required Probability is \({k/(n+k)}*{2/4}\)

Ok now lets check

Statement 1 : \(n+k = 12\), clearly not sufficient, as n=10, k= 2 or n=7, k=5, or many other combinations.

Statement 2: \(k/n =1/3\)

which is \(n/k = 3/1\)

By adding 1 to each side can be converted to

\((n+k)/k = 4/1\)

\(k/(n+k) = 1/4\)

Hence statement 2 is sufficient to find the required probability.
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A committee consists of n women and k men. In addition there 30 Oct 2024, 05:34
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the problems is basically asking what is the probability of picking one man out of the comitee of n+k people (n woman and k man) AND one woman from the other group of 4.

picking one man from the first group, we have k ways, picking a woman from the second group we have 2 ways --> in total 2K ways

the sample space is given by the total number of groups of 2 we can create by extracting ANY person from the first group and ANY person from the second group.
from the first group we can extract 1 person in (n+k) ways
from the second group we can extract 1 person in 4 ways

probability of picking 1 man from the first group and 1 woman from the second group =
2k / 4(n+k) = k/2(n+k)

(1) not sufficient
(2) clearly sufficient

probability of picking 1 man from the first group and 1 woman from the second group =
2k / 4(n+k) = k/2(n+k)
n=3k
k/2(n+k) = k/(3k+k)*2 = 1/8
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