The problem here is that you are arranging the people. When you select a woman out of 8 as 5/8, you are saying that you are picking a woman first. You are arranging them in this way:
WWMM
Now, if you want to un-arrange them, multiply it by the total number of arrangements i.e. 4!/(2!*2!) (because there are 2 men and 2 women so you divide by 2!s to get the total number of arrangements)

When you do that, you get 1/14 * 4!/(2!*2!) = 3/7
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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

I find the explanation provided in MGMAT very complicated and would like to know what techniques other people would recommend to solve this problem in 2 mins.
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P=Favorable outcomes / Total # of outcomes

Favorable outcomes: 2 women and 2 men = \(C^2_5*C^2_3=30\);
Total # of outcomes: 4 people out of 5+3=8 = \(C^4_8=70\);

\(P=\frac{30}{70}=\frac{3}{7}\).

Answer: D.
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Ok I get it. The order is not given. its random.

So we have 4*3*2/2*2 = 6 ways to arrange wwmm

so final probability is 1/14 * 6 = 3/7

Hi all,

I was working on a practice test when i saw this question. Now, the problem is; i have solved it in two ways, one way(using combinations) leads to the correct answer, the other gives me something different. i would appreciate if you could point out what im doing wrong here.

Questn
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




first method:
5C2 * 3C2 -> it gives combination of exactly 2 women and 2 men.
8C4 -> gives total possibilities of 4 people from 5 women and 3 men.

Probability = 5C2*3C2 / 8C4 = 3/7


SECOND METHOD:
Probability of two women -> 5/8 * 4/7.

probability of two men -> 3/6 * 2/5.

Probability: (5/8 * 4/7) * (3/6 * 2/5) = 1/14.

I know something is wrong with the second method but i can't really figure out why its flawed.

Any pointers ?? thanks

Yes. What you have to do is select a group. Hence you do not have to arrange them. In your second method, you got 1/14 which is the probability of getting wwmm. Since there are other such arrangements too which are all acceptable to us e.g. Wmwm since we just need a group of 2 men and 2 women irrespective of their arrangement, we multiply 1/14 by 6 (since 6 such arrangements are possible as shown by you)
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The team should have 4 employees out of which 2 are women, thus the remaining 2 will be men.
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We need to consider which men are chosen since a committee with Ann, Bea, Joe and Ed is DIFFERENT FROM a committee with Ann, Bea, Joe and Kevin

Cheers,
Brent
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Could someone please explain the way how to solve this fraction mathematically? What does C2 and C4 stand for? What calculations are derived from this?

Thanks

Hi All,

You did the first 'step' of this question correctly. The number of groups of 2 men IS 3 and the number of groups of 2 women IS 10. Now, you have to do the rest of the math in a really specific way:

Since each "subgroup" of men can be paired with each "subgroup" of women, there are (3)(10) = 30 possible groups of four that include exactly 2 women. Since this is a probability question, we now have to determine the total number of groups of four that are possible...

That is 8C4 = (8)(7)(6)(5) / (4)(3)(2)(1) = (2)(7)(1)(5) = 70 possible groups of 4

30/70 = 3/7

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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