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A committee consists of n women and k men. In addition there [#permalink]
15 Feb 2012, 10:07

2

This post received KUDOS

00:00

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B

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E

Difficulty:

65% (hard)

Question Stats:

63% (02:17) correct
37% (01:31) wrong based on 268 sessions

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?

Re: A committee consists of n women and k men. In addition there [#permalink]
15 Feb 2012, 11:20

1

This post received KUDOS

Expert's post

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.

Re: A committee consists of n women and k men. In addition there [#permalink]
15 Feb 2012, 11:31

Bunuel wrote:

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.

My intuition drove me to B, as well but.. I couldn't find the way! Thank you!!

Re: A committee consists of n women and k men. In addition there [#permalink]
03 Jul 2012, 08:40

1

This post received KUDOS

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.

Re: A committee consists of n women and k men. In addition there [#permalink]
07 Jun 2013, 08:17

1

This post received KUDOS

tom09b wrote:

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

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

Re: A committee consists of n women and k men. In addition there [#permalink]
20 Mar 2014, 04:34

Bunuel wrote:

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 ..

Re: A committee consists of n women and k men. In addition there [#permalink]
20 Mar 2014, 04:57

Expert's post

tusharGupta1 wrote:

Bunuel wrote:

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.

Re: A committee consists of n women and k men. In addition there [#permalink]
20 Mar 2014, 05:33

Bunuel wrote:

tusharGupta1 wrote:

Bunuel wrote:

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.

Thankx a ton ............................................................................................................................................................

Re: A committee consists of n women and k men. In addition there [#permalink]
12 May 2014, 04:12

Bunuel wrote:

tusharGupta1 wrote:

Bunuel wrote:

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.

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. _________________

Hope to clear it this time!! GMAT 1: 540 Preparing again

Re: A committee consists of n women and k men. In addition there [#permalink]
12 May 2014, 07:53

Expert's post

nandinigaur wrote:

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.

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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