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
I tried so many times but don't know why I am not getting an answer.
probability
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The number of ways that we can select 2 women out of 5 is 5C2.
The number of ways that we can select 2 men out of 3 is 3C2.
The number of ways that we can select 2 women out of 5 AND 2 men out of 3 is
(5C2)(3C2)
The total number of ways of selecting 4 people out of 8 is 8C4.
Therefore the probability is:
(5C2)(3C2)/8C4 = 3/7
The number of ways that we can select 2 men out of 3 is 3C2.
The number of ways that we can select 2 women out of 5 AND 2 men out of 3 is
(5C2)(3C2)
The total number of ways of selecting 4 people out of 8 is 8C4.
Therefore the probability is:
(5C2)(3C2)/8C4 = 3/7
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P(exactly n times) = P(one way) * total possible ways.j_shreyans 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
Let W = woman and M = man.
P(one way):
P(1st person is W) = 5/8. (8 people, 5 of them W.)
P(2nd person is W) = 4/7. (7 people left, 4 of them W.)
P(3rd person is M) = 3/6. (6 people left, 3 of them M.)
P(4th person is M) = 2/5. (5 people left, 2 of them M.)
Since we want all of these events to happen together, we multiply the fractions:
P(WWMM) = 5/8 * 4/7 * 3/6 * 2/5 = 1/14.
Total possible ways:
Any arrangement of the letters WWMM represents one way to get exactly 2 W's and 2 M's.
Thus, to account for ALL OF THE WAYS to get exactly 2 W's and 2 M's, the result above must be multiplied by the total number of ways to arrange the letters WWMM.
Number of ways to arrange 4 distinct elements = 4!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 2! to account for the 2 identical W's and by another 2! to account for the 2 identical M's:
4!/(2!2!) = 6.
Thus, P(exactly 2 W) = 1/14 * 6 = 3/7.
The correct answer is D.
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One approach is to apply counting methods:prernamalhotra 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
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
--------------------------
Note: the Fundamental Counting Principle (FCP) can be used to solve the MAJORITY of counting questions (and many probability questions) on the GMAT. For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
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MEDIUM
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DIFFICULT
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Cheers,
Brent