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Probability- Whats wrong with how my method??

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Probability- Whats wrong with how my method??

by MBAsa » Sat Jun 29, 2013 1:38 am
My answer to the following question was wrong. Can someone please explain whats wrong with my method.

Q:
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

A. 1/14 B.1/7 C.2/7 D. 3/7 E. 1/2

My solution:

To have exactly 2 women, there will also be 2 men, so my group will look like WWMM

Probablity of selecting W first = 5/8
Probablity of selecting W next = 4/7
Probablity of selecting M next= 3/6
Probablity of selecting M first = 2/5

Total probability = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

I realize the order of picking does not matter. Even if it was MWMW, WMWM or MMWW my answer would be the same.

Answer = D[/list][/list]
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by GMATGuruNY » Sat Jun 29, 2013 2:40 am
MBAsa wrote:My answer to the following question was wrong. Can someone please explain whats wrong with my method.

Q:
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

A. 1/14 B.1/7 C.2/7 D. 3/7 E. 1/2

My solution:

To have exactly 2 women, there will also be 2 men, so my group will look like WWMM

Probablity of selecting W first = 5/8
Probablity of selecting W next = 4/7
Probablity of selecting M next= 3/6
Probablity of selecting M first = 2/5

Total probability = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

I realize the order of picking does not matter. Even if it was MWMW, WMWM or MMWW my answer would be the same.

Answer = D[/list][/list]
You have correctly determined the following:
P(WWMM) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14.
This result represents the probability that the first and second people selected are women and that the third and fourth people selected are men.

But WWMM is only ONE WAY to select exactly two women.
To determine the total probability, we must account for ALL OF THE WAYS to select two women and one man.

Any arrangement of WWMM will yield exactly 2 W and 2 M.
Thus, the resulting probability above must be multiplied by the number of ways to arrange the 4 elements WWMM.
Number ways to arrange WWMM = 4!/(2!2!) = 6.
Thus:
P(exactly 2 women) = 1/14 * 6 = 3/7.

The correct answer is D.

I posted a complete solution for this problem in my two posts here:

https://www.beatthegmat.com/select-exact ... 88786.html

Similar problems:

https://www.beatthegmat.com/probability- ... 14250.html
https://www.beatthegmat.com/a-single-par ... 28342.html
https://www.beatthegmat.com/at-a-blind-t ... 20058.html
https://www.beatthegmat.com/rain-check-t79099.html
https://www.beatthegmat.com/probability-t227448.html
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by ganeshrkamath » Thu Jul 04, 2013 3:13 am
MBAsa wrote:My answer to the following question was wrong. Can someone please explain whats wrong with my method.

Q:
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?

A. 1/14 B.1/7 C.2/7 D. 3/7 E. 1/2

My solution:

To have exactly 2 women, there will also be 2 men, so my group will look like WWMM

Probablity of selecting W first = 5/8
Probablity of selecting W next = 4/7
Probablity of selecting M next= 3/6
Probablity of selecting M first = 2/5

Total probability = 5/8 * 4/7 * 3/6 * 2/5 = 1/14

I realize the order of picking does not matter. Even if it was MWMW, WMWM or MMWW my answer would be the same.

Answer = D[/list][/list]
Select any 2 women = 5C2 = 10
Select any 2 men = 3C2 = 3
Favourable cases = 10 * 3 = 30

Sample space = (3+5)C4 = 70

Probability = 30/70 = 3/7
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by vipulgoyal » Thu Jul 04, 2013 11:15 pm
P of selecting excatly 2 women = 1 - P of selecting (1 women + 3 women + 4 women )

1 - ( 5/8*3/7*2/6*1/5*4 + 5/8*4/7*3/6*3/5*4 + 5/8*4/7*3/6*2/5)
1 - (1/14 + 3/7 + 1/14)
1 - 4/7 = 3/7
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