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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]

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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

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Re: Probability Questn  [#permalink]

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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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Re: Probability  [#permalink]

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New post 26 Aug 2008, 21:02
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\(P=\frac{3C2*5C2}{8C4}\)
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A small company employs 3 men and 5 women. If a team of 4  [#permalink]

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New post 23 May 2010, 11:11
1
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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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Re: Proba and combinations (MGMAT)  [#permalink]

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New post 23 May 2010, 11:32
7
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 :wink:
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Re: Proba and combinations (MGMAT)  [#permalink]

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New post 23 May 2010, 11:37
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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  [#permalink]

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New post 07 Jul 2010, 17:24
4
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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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]

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New post 24 Mar 2011, 21:15
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5
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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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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  [#permalink]

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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]

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

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New post 28 Nov 2014, 01:09
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  [#permalink]

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New post 28 Nov 2014, 04:11
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  [#permalink]

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New post 30 Jan 2016, 16:15
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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New post 30 Jan 2016, 17:54
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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

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

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New post 14 Dec 2017, 09:13
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 (8-4)!] = (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  [#permalink]

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New post 06 Jan 2018, 08:35
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  [#permalink]

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New post 06 Jan 2018, 09:17
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: \(xCy=\frac{x!}{y!(x-y)!}=21\).

21. Combinatorics/Counting Methods



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

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New post 12 Jan 2018, 13:01
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

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