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Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
02 May 2020, 17:02
02 May 2020, 17:02
Bunuel wrote:
nifoui 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
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.
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.
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21843
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
20 Jul 2020, 11:15
20 Jul 2020, 11:15
Expert Reply
KaranB1 wrote:
Bunuel wrote:
nifoui 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
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.
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.
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!
GMAT assassins aren't born, they're made,
Rich
GMAT Club Legend
Joined: 03 Oct 2013
Affiliations: CrackVerbal
Posts: 4950
Location: India
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
08 Jun 2021, 03:01
08 Jun 2021, 03:01
Top Contributor
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)
Devmitra Sen
GMAT SME
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)
Devmitra Sen
GMAT SME
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
11 Jun 2021, 07:13
11 Jun 2021, 07:13
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
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
GMAT Club Legend
Joined: 19 Dec 2014
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Posts: 21843
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
11 Jun 2021, 14:17
11 Jun 2021, 14:17
Expert Reply
Chinaski wrote:
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
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,
Rich
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
13 Jul 2021, 21:46
13 Jul 2021, 21:46
Bunuel wrote:
nifoui 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
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.
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?
Math Revolution GMAT Instructor
Joined: 16 Aug 2015
Posts: 10372
GMAT 1: 760 Q51 V42
GPA: 3.82
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
13 Jul 2021, 22:42
13 Jul 2021, 22:42
Expert Reply
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
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
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
13 Jul 2021, 23:41
13 Jul 2021, 23:41
Expert Reply
Vivaan7627 wrote:
Bunuel wrote:
nifoui 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
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.
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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Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
20 Nov 2022, 15:54
20 Nov 2022, 15:54
ankur17 wrote:
Ok I get it. The order is not given. its random.
So we have 4*3*2/2*2 = 6 ways to arrange wwmm
so final probability is 1/14 * 6 = 3/7
So we have 4*3*2/2*2 = 6 ways to arrange wwmm
so final probability is 1/14 * 6 = 3/7
Why are their 6 ways to arrange wwmm? I don't understand the formula
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
20 Nov 2022, 16:18
20 Nov 2022, 16:18
[quote="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
where can i find the general formula for 5!/2! 3!?
I don't understand why we divide by 2!*3! (i know that we pick 2 women out of 5 and 3 remain)
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
where can i find the general formula for 5!/2! 3!?
I don't understand why we divide by 2!*3! (i know that we pick 2 women out of 5 and 3 remain)
Re: A small company employs 3 men and 5 women. If a team of 4
[#permalink]
21 Nov 2023, 02:48
21 Nov 2023, 02:48
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Re: A small company employs 3 men and 5 women. If a team of 4 [#permalink]
21 Nov 2023, 02:48
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