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# A small company employs 3 men and 5 women. If a team of 4

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VP
Joined: 30 Jun 2008
Posts: 1022
A small company employs 3 men and 5 women. If a team of 4 [#permalink]

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17 Sep 2008, 07:30
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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 at least 2 women?

NO OA guys

apnew

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Current Student
Joined: 31 Aug 2007
Posts: 365

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17 Sep 2008, 07:58
maybe 13/14?

total number of ways being 70, then adding up all the different ways you can get at least 2 women on the team (65).
Manager
Joined: 14 Jun 2007
Posts: 158
Location: Vienna, Austria

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17 Sep 2008, 08:09
guys- is this with the scenarios: 2w 2m , 3w 1m , 4w 0m ?
cheers
VP
Joined: 30 Jun 2008
Posts: 1022

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17 Sep 2008, 08:16
young_gun wrote:
maybe 13/14?

total number of ways being 70, then adding up all the different ways you can get at least 2 women on the team (65).

How did you determine that total number of ways are 70 and different ways of getting at least 2 women is 65 ?
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SVP
Joined: 07 Nov 2007
Posts: 1765
Location: New York

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17 Sep 2008, 08:16
= (5C2*3C2+5C3*3C1+5C4*3C0 )/(8C4)
= 65/70= 13/14
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VP
Joined: 30 Jun 2008
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17 Sep 2008, 08:18
x2suresh wrote:
= (5C2*3C2+5C3*3C1+5C4*3C0 )/(8C4)
= 65/70= 13/14

why do we have + in the numerator suresh ?

Thanks
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"You have to find it. No one else can find it for you." - Bjorn Borg

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Joined: 07 Nov 2007
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17 Sep 2008, 08:30
amitdgr wrote:
x2suresh wrote:
= (5C2*3C2+5C3*3C1+5C4*3C0 )/(8C4)
= 65/70= 13/14

why do we have + in the numerator suresh ?

Thanks

Ateleast 2 women = 2 women 2 men + 3 women 1 man + 4 woman and 0 man
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Manager
Joined: 11 Jan 2008
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17 Sep 2008, 12:39
Total ways = 70

Condition that has all 3 men and 1 woman = 3c3 and 5c1 = 5.

probablity with all 3 men and 1 women is 5/70 = 1/14.

therefore with atleast 2 women is 1-1/14 = 13/14.

is this correct and faster way of doing ?

-Jack
VP
Joined: 30 Jun 2008
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17 Sep 2008, 22:34
What is the best method of approaching combinatorics or probability problems ? Please guide
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Intern
Joined: 02 Sep 2008
Posts: 45

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19 Sep 2008, 10:43
yep..

I also end up with 13/14.
Senior Manager
Joined: 21 Apr 2008
Posts: 265
Location: Motortown

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30 Oct 2008, 18:07
I got a similar question in MGMAT CAT

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?

Here are the choices :
A 1/14
B 1/7
C 2/7
D 3/7
E 1/2
Intern
Joined: 30 Oct 2008
Posts: 30

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30 Oct 2008, 22:25
LiveStronger wrote:
I got a similar question in MGMAT CAT

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?

Here are the choices :
A 1/14
B 1/7
C 2/7
D 3/7
E 1/2

Select 2 out of 5 women = 5C2
Select 2 out of 3 men = 3C2
Favorable events = 5C2 * 3C2
Total events = 8C4

Probab = Fav/Total = 3/7
SVP
Joined: 17 Jun 2008
Posts: 1507

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30 Oct 2008, 22:53
amitdgr wrote:
What is the best method of approaching combinatorics or probability problems ? Please guide

To me, most of the time, it is common sense. For example, in this case, there are four possibilities. Either, 2 women, 2 men or, 3 women, 1 man or 4 women, 0 man.

And, since we are talking of union here and each of the possibilities is independent (2women and 2 women cannot happen with 3 women and 1 man), we can simply add the values from these possibilities.

Hope, this clarifies.
Re: Probability   [#permalink] 30 Oct 2008, 22:53
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# A small company employs 3 men and 5 women. If a team of 4

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