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KaranB1
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A small company employs 3 men and 5 women. If a team of 4 02 May 2020, 18:02
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Bunuel
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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.
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Bunuel

Should we not divide 30 with 2! as order of the selection does not matter. I'm getting 15/70 or 3/14 as an answer.
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A small company employs 3 men and 5 women. If a team of 4 20 Jul 2020, 12:15
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KaranB1
Bunuel
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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.

Bunuel

Should we not divide 30 with 2! as order of the selection does not matter. I'm getting 15/70 or 3/14 as an answer.
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Hi KaranB1,

Since the calculation has already used the Combination Formula to determine all of the groups of '2 men' and all of the groups of '2 women', that aspect of the calculation (re: order does NOT matter) has already been accounted for - and you do NOT need to divide by 2!

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A small company employs 3 men and 5 women. If a team of 4 08 Jun 2021, 04:01
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Solution:

The total number of possibilities of choosing/selecting 4 of 8 people = 8C4

To have exactly 2 women we need 2 men and 2 women in 3C2 and 5C2 ways

=>Number of ways = 3C2 * 5C2 (AND condition)

=30

Probability = 30/70 =3/7 (option d)

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A small company employs 3 men and 5 women. If a team of 4 11 Jun 2021, 08:13
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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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A small company employs 3 men and 5 women. If a team of 4 11 Jun 2021, 15:17
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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
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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.

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A small company employs 3 men and 5 women. If a team of 4 13 Jul 2021, 22:46
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Bunuel
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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.
Show more

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

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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A small company employs 3 men and 5 women. If a team of 4 13 Jul 2021, 23:42
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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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A small company employs 3 men and 5 women. If a team of 4 14 Jul 2021, 00:41
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Vivaan7627
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?
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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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A small company employs 3 men and 5 women. If a team of 4 21 Jul 2025, 05:50
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Hi there, just wondering why is the way to choose total amount of choice not just 8x7x6x5? Because there would be 8 choices for position A, 7 for B etc and this would not repeat
Detran
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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