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A small company employs 3 men and 5 women. If a team of 4 [#permalink]

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22 Nov 2007, 19:46

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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 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?

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.

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

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

Quote:

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

Hello karishma,

From what i understand, you have added all probabilities like so: 1) WWMM 2) WMMW 3) WMWM 4) MWMW 5) MMWW 6) MWWM

...such that you do not repeat arrangements that are the same, such as W1W2 and W2W1. ... right ? 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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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)

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)

Hi Karishma,

Very helpful answer. Maybe you could clarify something for me: When we use the combinatorics method in this problem - (2c5*2c3)/4c8 - why don't we add in the permutations part like we do for the probability method?

What I mean by that is -- If i use the probability approach (favorable outcomes/total outcomes) -- I get (5/8)(4/7)(3/6)(2/5) and then I multiply that by the permutations, which means that I multiply it 4!/2!2! -- why don't we do this last part when it comes to probability?

Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink]

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26 Sep 2015, 10:31

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