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A small company employs 3 men and 5 women. If a team of 4
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22 Nov 2007, 19:46
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66% (01:58) correct 34% (02:14) wrong based on 1058 sessions
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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 Questn
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24 Apr 2011, 16:17
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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Re: Probability
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26 Aug 2008, 22:02
\(P=\frac{3C2*5C2}{8C4}\)
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A small company employs 3 men and 5 women. If a team of 4
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23 May 2010, 12:11
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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Re: Proba and combinations (MGMAT)
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23 May 2010, 12:32
lets see Total amount of possible choices is 8! / 4! 4! = 8 x 6 x 7 x 5 / 4 x 3 x 2 = 70 the only way to have exactly 2 women is to have 2 men so the number of ways we can select 2men from 4 is 3!/ 2! = 3 the number of ways we can select 2 women from 5 is 5! /2! 3! = 5 x 4 / 2 = 10 the total different ways you can have 2 women and 2 men 3 x 10 = 30 desired outcome / total outcomes = probability > 30 / 70 = 3/7 so it should be D pS. Kudos are always welcomed
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Re: Proba and combinations (MGMAT)
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23 May 2010, 12:37
nifoui 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? 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. Will post OA later. 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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Re: Probability
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07 Jul 2010, 18:24
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
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11 Jul 2010, 03:05
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
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24 Mar 2011, 22:15
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
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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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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: A small company employs 3 men and 5 women. If a team of 4
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28 Nov 2014, 02:09
Bunuel,i did not understand why we took two men too?I did it like Favorable outcomes: selecting 2 women C^2_5 and Total # of outcomes: 4 people out of 5+3=8 = C^4_8=70;



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Re: A small company employs 3 men and 5 women. If a team of 4
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28 Nov 2014, 05:11
Ralphcuisak wrote: Bunuel,i did not understand why we took two men too?I did it like Favorable outcomes: selecting 2 women C^2_5 and Total # of outcomes: 4 people out of 5+3=8 = C^4_8=70; The team should have 4 employees out of which 2 are women, thus the remaining 2 will be men.
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Re: A small company employs 3 men and 5 women. If a team of 4
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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
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30 Jan 2016, 18:54
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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14 Dec 2017, 10:13
alimad 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?
A. 1/14 B. 1/7 C. 2/7 D. 3/7 E. 1/2 We need to determine the probability that when 4 employees are selected from 3 men and 5 women, 2 women and 2 men will be selected. Let’s determine the number of ways to select the 2 women and 2 men. Number of ways to select 2 women: 5C2 = (5 x 4)/2! = 10 Number of ways to select 2 men: 3C2 = (3 x 2)/2! = 3 Thus, 2 men and 2 women can be selected in 10 x 3 = 30 ways. Finally, we can determine the number of ways to select 4 people from 8. 8C4 = 8! / [4! x (84)!] = (8x 7 x 6 x 5)/4! = (8 x 7 x 6 x 5)/(4 x 3 x 2) = 7 x 2 x 5 = 70 Thus, the probability is 30/70 = 3/7. Answer: D
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Re: A small company employs 3 men and 5 women. If a team of 4
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06 Jan 2018, 09:35
x2suresh wrote: \(P=\frac{3C2*5C2}{8C4}\) 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



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Re: A small company employs 3 men and 5 women. If a team of 4
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06 Jan 2018, 10:17
Gregsterh wrote: x2suresh wrote: \(P=\frac{3C2*5C2}{8C4}\) 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 C stands for combinations: \(xCy=\frac{x!}{y!(xy)!}=21\). 21. Combinatorics/Counting Methods For more: ALL YOU NEED FOR QUANT ! ! !Ultimate GMAT Quantitative MegathreadHope it helps.
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Re: A small company employs 3 men and 5 women. If a team of 4
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12 Jan 2018, 14:01
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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Re: A small company employs 3 men and 5 women. If a team of 4
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