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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. 1/14 B. 1/7 C. 2/7 D. 3/7 E. 1/2
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Re: Probability [#permalink]
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26 Aug 2008, 22:02
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\(P=\frac{3C2*5C2}{8C4}\)
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Re: Probability [#permalink]
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27 Sep 2009, 23:24
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]
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07 Jul 2010, 18:24
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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]
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11 Jul 2010, 03:05
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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]
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10 Aug 2010, 21:22
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]
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11 Aug 2010, 00:17
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]
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24 Mar 2011, 22:15
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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]
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24 Apr 2011, 15:52
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]
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24 Apr 2011, 16:17
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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 unarrange 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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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 unarrange 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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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]
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25 Apr 2011, 13:55
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]
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25 May 2013, 10:58
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]
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25 May 2013, 11:16



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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]
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15 Aug 2014, 12:12
Hi, I have tried reverse approach & got it wrong 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)



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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink]
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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink]
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30 Jan 2016, 17:15
Can someone explain to me why the answer is not 1/7? I did Desired Outcomes/Total Outcomes. Obviously total is 70 we all agree on that. We also all agree that 5 women choose 2 = 10. However, I don't see why we then do 3 men choose 2. Why does it matter which men are chosen? Can someone explain logically rather than mathematically?
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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink]
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30 Jan 2016, 18:54
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redfield wrote: Why does it matter which men are chosen? Can someone explain logically rather than mathematically? 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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Re: A small company employs 3 men and 5 women. If a team of 4
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