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

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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 09 Apr 2016, 02:31
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


if there are exactly two women in the team then other to two will be the men.

So we have to choose 2 women out of 5 and 2 men out of 3. and total 4 out of 8.

Probability = 3c2 * 5c2 / 8c4 = 30/(10*7) = 3/7
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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 25 Feb 2017, 09:26
1) The total number of 4 people combinations selected from 3+5=8 people - Combination: 8!/4!(8-4)!=8*7*6*5/4*3*2=7*2*5=70
2) The probability that there will be exactly two women: First woman can be any one of the 5; second woman - one of the remaining 4; first man - one of the 3; second man - one of the 2: 5*4*3*2=120. Since it does not matter the order of women and men we have to divide each category by 2: 5*4/2=10; 3*2/2=3; 10*3=30
3) Probability that there will be exactly 2 women in a group is 30/70=3/7

The correct answer is 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 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 (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 14 Dec 2017, 17:50
redfield wrote:
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?


You agree there are 70 possible groups. It matters what men are chosen because they form more groups. Let's assume the three men are A,B, and C. For each of the 10 female pairs chosen two men have to be chosen (one man has to be left out. So for a given female pair the left out man could be A,B, or C. When you don't multiply 10 by 3 you're basically asserting female pair A B, female pair A C, and female pair B C are the same group.

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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, 09:35
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, 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: \(xCy=\frac{x!}{y!(x-y)!}=21\).

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, 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

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Re: A small company employs 3 men and 5 women. If a team of 4 &nbs [#permalink] 12 Jan 2018, 14:01

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