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MGMAT - probabilty

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MGMAT - probabilty

by sarahw_gmat » Fri Jul 22, 2011 3:04 pm
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?

OA : 1/7

Can you please explain what am I doing wrong here?

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

= probability to select first woman x P of second woman X first man X second man

Thanks.
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by vineeshp » Fri Jul 22, 2011 7:33 pm
I dont think this is the way to solve it. You have assumed that they are selected in a certain order.
What about the other cases where there are other orders such as MWWM MWMW etc.

Infact there are six such orders.

So multiplying by 6 gives you 3/7 and that is the answer.

The link below should give you an answer to your specific question.
https://www.beatthegmat.com/a-small-comp ... 14052.html
Vineesh,
Just telling you what I know and think. I am not the expert. :)
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by top_business_2011 » Sat Jul 23, 2011 1:23 am
sarahw_gmat wrote: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?

OA : 1/7

Can you please explain what am I doing wrong here?

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

= probability to select first woman x P of second woman X first man X second man

Thanks.
This is to help you pin down a flaw in your reasoning.
In your solution, you have assumed that only the last spots shall be taken by men employees-an unwarranted assumption. 1/14 is an answer should the question be framed as: What's the probability that the first two employees in the team be women?

Draw a tree diagram to help you figure out the different possible combinations for two women and two men.
WWMM
So there are (4!)/(2!*2!) = 6 such arrangements. Therefore you've to multiply your answer by 6
The result shall be 3/7.

But the best way of doing it is maybe using the combination format:

(3C2 * 5C2)/ (8C4)..........[ Here, you can read it as 2 men from the 3 and(usually represented by '*' in mathematics) 2 women from the 5 divided the whole by 4 employees from the 8]
= 3/7.

Hope that helps.
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by GMATGuruNY » Sat Jul 23, 2011 1:50 am
sarahw_gmat wrote: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?

OA : 1/7

Can you please explain what am I doing wrong here?

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

= probability to select first woman x P of second woman X first man X second man

Thanks.

P(exactly n times) = P(one way) * total possible ways

One way to choose exactly 2 women is to select women on the first 2 picks and men on the last 2 picks:
P(WWMM) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14.

But WWMM is only ONE WAY to choose 2 women. We need to account for ALL THE WAYS in which 2 women could be chosen.
Any arrangement of the letters WWMM will result in choosing 2 women and 2 men:
WMWM = choosing women on the 1st and 3rd picks
MMWW = choosing women on the last 2 picks
etc.

Thus, we multiply the result above by the number of ways to arrange the letters WWMM.
Number of ways to arrange WWMM = 4!/2!2! = 6.

Putting it all together, we get:
P(exactly 2 women) = 6 * 1/14 = 3/7.
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