Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 May 2017, 23:57

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
VP
Joined: 30 Jun 2008
Posts: 1034
Followers: 14

Kudos [?]: 613 [0], given: 1

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

### Show Tags

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

_________________

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

Current Student
Joined: 31 Aug 2007
Posts: 369
Followers: 1

Kudos [?]: 143 [0], given: 1

### Show Tags

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: 169
Location: Vienna, Austria
Followers: 1

Kudos [?]: 8 [0], given: 0

### Show Tags

17 Sep 2008, 08:09
guys- is this with the scenarios: 2w 2m , 3w 1m , 4w 0m ?
cheers
VP
Joined: 30 Jun 2008
Posts: 1034
Followers: 14

Kudos [?]: 613 [0], given: 1

### Show Tags

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 ?
_________________

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

SVP
Joined: 07 Nov 2007
Posts: 1806
Location: New York
Followers: 37

Kudos [?]: 931 [0], given: 5

### Show Tags

17 Sep 2008, 08:16
= (5C2*3C2+5C3*3C1+5C4*3C0 )/(8C4)
= 65/70= 13/14
_________________

Smiling wins more friends than frowning

VP
Joined: 30 Jun 2008
Posts: 1034
Followers: 14

Kudos [?]: 613 [0], given: 1

### Show Tags

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
_________________

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

SVP
Joined: 07 Nov 2007
Posts: 1806
Location: New York
Followers: 37

Kudos [?]: 931 [0], given: 5

### Show Tags

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
_________________

Smiling wins more friends than frowning

Manager
Joined: 11 Jan 2008
Posts: 54
Followers: 0

Kudos [?]: 30 [0], given: 0

### Show Tags

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
Posts: 1034
Followers: 14

Kudos [?]: 613 [0], given: 1

### Show Tags

17 Sep 2008, 22:34
What is the best method of approaching combinatorics or probability problems ? Please guide
_________________

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

Intern
Joined: 02 Sep 2008
Posts: 45
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

19 Sep 2008, 10:43
yep..

I also end up with 13/14.
Senior Manager
Joined: 21 Apr 2008
Posts: 269
Location: Motortown
Followers: 2

Kudos [?]: 150 [0], given: 0

### Show Tags

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
Followers: 0

Kudos [?]: 2 [0], given: 0

### Show Tags

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: 1553
Followers: 11

Kudos [?]: 264 [0], given: 0

### Show Tags

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
Display posts from previous: Sort by

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

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.