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Hi,

for total outcome, I only got 65:

3m 1w = 5 arrangements
2m 2w = 3C2 * 5C2 = (3*10) = 30 arrangements
1m 3w = 3C1 * 5C3 = (3*10) = 30 arrangements

where are the missing 5? I go crazy
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Chinaski
Hi,

for total outcome, I only got 65:

3m 1w = 5 arrangements
2m 2w = 3C2 * 5C2 = (3*10) = 30 arrangements
1m 3w = 3C1 * 5C3 = (3*10) = 30 arrangements

where are the missing 5? I go crazy

Hi Chinaski,

Your list doesn't include the groups that are 0m 4w. Since there are 5 women in total, there are 5 potential groups of 4 women that can be formed.

As an aside, you don't have to break down this part of the calculation into 'pieces.' You can just use the Combination Formula. With 8 total people, and groups of 4, there are:

8c4 = 8!/4!4! = 70 possible groups of 4 that can be formed.

GMAT assassins aren't born, they're made,
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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.

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.

==============================================================

Once we do 5C2 for women, why do we need to do 3C2 for men as well. 5C2 means 2 women out of 4 people in the group so remaining two should naturally be men.

Am I missing something here?
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Total employees: 8
Team: 4 employees

Total outcome: \(^8{C_4}= 70\)

Men: 3
Women: 5
Condition: Exactly 2 women

4 members : 2 women and 2 men = \(^5{C_2}= 10\) * \(^3{C_2}= 3 =\) 10 * 3 = 30

Probability: \(\frac{30}{70} = \frac{3}{7}\)

Answer D
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Bunuel
nifoui
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.

==============================================================

Once we do 5C2 for women, why do we need to do 3C2 for men as well. 5C2 means 2 women out of 4 people in the group so remaining two should naturally be men.

Am I missing something here?

We have 3 men to choose from (say A, B, C) and 2 men out of 3 can be chosen in different ways: AB, AC or BC. So, we should account for these different cases also.
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Deconstructing the Question

Topic: Combinations & Probability

We need to select a team of 4 employees from a group of 3 men and 5 women. What is the probability that the team contains exactly 2 women?

1. Calculate Total Outcomes (Denominator)
The number of ways to choose 4 employees out of a total of 8:
\(8C4 = (8 * 7 * 6 * 5) / (4 * 3 * 2 * 1) = \) 70

2. Calculate Favorable Outcomes (Numerator)
For the team to have exactly 2 women, the remaining 2 members must be men.
  • Ways to choose 2 women from 5: \(5C2 = 10\)
  • Ways to choose 2 men from 3: \(3C2 = 3\)
\(Total Favorable = 10 * 3 = \) 30

3. Final Probability
\(P = 30 / 70 = \) 3/7

The correct answer is D.
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Are these question still in the test? I feel like I haven't seen this
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Are these question still in the test? I feel like I haven't seen this

Yes. It’s a probability question, and probability is still part of the GMAT Focus syllabus.
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