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A small company employs 3 men and 5 women

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A small company employs 3 men and 5 women

by rsarashi » Sun May 07, 2017 4:11 am
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

OAD

Men will be selected = 3c2, it will come 3

Women will be selected = 5c2, it will come 10

So total be 13 right?

Please correct me, if anything I am missing.

Thanks.
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by Jay@ManhattanReview » Sun May 07, 2017 4:45 am
rsarashi 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

OAD

Men will be selected = 3c2, it will come 3

Women will be selected = 5c2, it will come 10

So total be 13 right?

Please correct me, if anything I am missing.

Thanks.
Hi rsarashi,

'13' is incorrect. It should be 3*10 = 30.

To compute the total number of way, you must multiply the individual number of ways. When there is a situation of 'OR,' you must add the number of ways.

Hope this is clear.

-Jay

Relevant book: Manhattan Review GMAT Combinatorics and Probability Guide
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by Brent@GMATPrepNow » Sun May 07, 2017 5:30 am
rsarashi 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

OAD
P(exactly 2 women) = [# of teams with exactly 2 women] / [total # of teams possible]

# of teams with exactly 2 women
Take the task of selecting 2 women and 2 men and break it into stages.
Stage 1: Select 2 women for the team. There are 5 women to choose from, so this can be accomplished in 5C2 ways.
Stage 2: Select 2 men for the team. There are 3 men to choose from, so this can be accomplished in 3C2 ways.
By the Fundamental Counting Principle (FCP), the total number of teams with exactly 2 women = (5C2)(3C2) = (10)(3) = 30

# of teams possible
There are 8 people altogether and we must choose 4 of them.
This can be accomplished in 8C4 ways, which equals 70 ways


P(exactly 2 women) = 30/70
= 3/7 = D

Aside: To learn how to calculate combinations (like 5C2) in your head, you can watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=789

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by GMATGuruNY » Sun May 07, 2017 5:35 am
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by [email protected] » Sun May 07, 2017 9:27 am
Hi rsarashi,

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: D

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by Jeff@TargetTestPrep » Fri May 12, 2017 12:23 pm
rsarashi 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

Jeffrey Miller
Head of GMAT Instruction
[email protected]

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