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A small company employs 3 men and 5 women. If a team of 4

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A small company employs 3 men and 5 women. If a team of 4 [#permalink] New post 22 Nov 2007, 18:46
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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. 1/14
B. 1/7
C. 2/7
D. 3/7
E. 1/2
[Reveal] Spoiler: OA

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Re: Probability [#permalink] New post 26 Aug 2008, 21:02
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P=\frac{3C2*5C2}{8C4}
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Re: Probability [#permalink] New post 27 Sep 2009, 22:24
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?


1/14
1/7
2/7
3/7
1/2

Soln:
= 3C2 * 5C2/8C4
= 3/7
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Re: Probability [#permalink] New post 07 Jul 2010, 17:24
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For this question answer is: 3/7

Solution:
There are 5 women and 3 men. Total number of employees are 8. So, out of 8 employees we have to select 4 employees = 8C4

Out of 8 employees, we must select exactly 2 women, this implies that we must select exactly 2 men.

(5C2 X 3C2) / 8C4

Ans: 3/7

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Re: Probability [#permalink] New post 11 Jul 2010, 02:05
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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
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Re: Probability [#permalink] New post 10 Aug 2010, 20:22
is probablity the same if
rather than "2 women" it says "jena and monica" must be in (ie 2 in particular among 4)?
bit confused in that
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Re: Probability [#permalink] New post 10 Aug 2010, 23:17
frank1 wrote:
is probablity the same if
rather than "2 women" it says "jena and monica" must be in (ie 2 in particular among 4)?
bit confused in that


Then there is only C^2_2 or 1 way to pick women instead of C^4_2.So probability will change.
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Re: Probability [#permalink] New post 24 Mar 2011, 21:15
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I used to have trouble with this but for those having trouble with probablility and combinatorics. I think you can have some help from ANAGRAM.

Example: Ways to rearrange LEVEL

5! = because we have 5 letters
2! = because we have 2 Ls
2! = because we have 2 Es

Formula is 5!/2!2! = 30

YOU CAN USE THIS WITH THE PROBLEM ABOVE.

SOLUTION:

How many ways to select 4 from 8 people?
(imagine this as rearranging YYYYNNNN) 8!/4!4! = 70

How many ways to select 2 women from 5 women?
(imagine this as rearranging YYNNN) 5!/2!3! = 10

How many ways to select 2 men from 3 men?
(imagine this as rearranging YYN) 3!/2! = 3

Probability = 3 x 10 / 70 = 3/7
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Probability Questn [#permalink] New post 24 Apr 2011, 14:52
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
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Re: Probability Questn [#permalink] New post 24 Apr 2011, 15:17
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coldteleporter wrote:
Hi all,

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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 [#permalink] New post 25 Apr 2011, 10:22
VeritasPrepKarishma wrote:
coldteleporter wrote:
Hi all,

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
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Re: [#permalink] New post 25 Apr 2011, 10:53
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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)
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Re: Probability Questn [#permalink] New post 25 Apr 2011, 12:55
VeritasPrepKarishma wrote:
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)


Thank you Karishma. You were very helpful.
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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink] New post 25 May 2013, 09:58
Can someone please explain to me why you can't solve this by multiplying:

5/8 * 5/8 * 3/8 * 3/8?

Is it because this method doesn't take into account different order possibilities?

In general how do you know whether to solve a probability by multiplying plain fractions like this or by using factorals (!s) ?
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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink] New post 25 May 2013, 10:16
Expert's post
dawip wrote:
Can someone please explain to me why you can't solve this by multiplying:

5/8 * 5/8 * 3/8 * 3/8?

Is it because this method doesn't take into account different order possibilities?

In general how do you know whether to solve a probability by multiplying plain fractions like this or by using factorals (!s) ?


Check here: a-small-company-employs-3-men-and-5-women-if-a-team-of-56037.html#p913073

Hope it helps.
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Re: Re: [#permalink] New post 10 May 2014, 14:24
VeritasPrepKarishma wrote:
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?

Thanks!
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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink] New post 15 Aug 2014, 11:12
Hi,

I have tried reverse approach & got it wrong :cry: please help me understand..

4W- 5C4
3W, 1M - 5C3*3C1
1W,3M - 5C1*3C3

Total - 8C4

1-( 4W+3W,1M +1W,3M) / TOTAL....

= WRONG ANSWER (PLEASE HELP)
Re: A small company employs 3 men and 5 women. If a team of 4   [#permalink] 15 Aug 2014, 11:12
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