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

 It is currently 29 Mar 2017, 02:23

### 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

Author Message
TAGS:

### Hide Tags

Director
Joined: 10 Feb 2006
Posts: 657
Followers: 3

Kudos [?]: 492 [5] , given: 0

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

### Show Tags

22 Nov 2007, 19:46
5
KUDOS
32
This post was
BOOKMARKED
00:00

Difficulty:

45% (medium)

Question Stats:

64% (02:18) correct 36% (01:27) wrong based on 1182 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

GMAT the final frontie!!!.

SVP
Joined: 07 Nov 2007
Posts: 1821
Location: New York
Followers: 36

Kudos [?]: 900 [10] , given: 5

### Show Tags

26 Aug 2008, 22:02
10
KUDOS
2
This post was
BOOKMARKED
$$P=\frac{3C2*5C2}{8C4}$$
_________________

Smiling wins more friends than frowning

Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 2

Kudos [?]: 147 [0], given: 3

### Show Tags

27 Sep 2009, 23:24
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?

1/14
1/7
2/7
3/7
1/2

Soln:
= 3C2 * 5C2/8C4
= 3/7
Intern
Joined: 06 Jul 2010
Posts: 18
Followers: 1

Kudos [?]: 2 [1] , given: 1

### Show Tags

07 Jul 2010, 18:24
1
KUDOS
For this question answer is: 3/7

Solution:
There are 5 women and 3 men. Total number of employees are 8. So, out of 8 employees we have to select 4 employees = 8C4

Out of 8 employees, we must select exactly 2 women, this implies that we must select exactly 2 men.

(5C2 X 3C2) / 8C4

Ans: 3/7

Cheers
/Milap
Joined: 31 Dec 1969
Location: Russian Federation
GMAT 1: 710 Q49 V0
GMAT 2: 700 Q V
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 7: Q42 V44
GMAT 8: Q42 V44
GMAT 9: 740 Q49 V42
GMAT 10: 740 Q V
GMAT 11: 500 Q47 V33
GMAT 12: 670 Q V
GMAT 13: 680 Q V
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)
Followers: 0

Kudos [?]: 214 [1] , given: 103828

### Show Tags

11 Jul 2010, 03:05
1
KUDOS
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
Manager
Joined: 24 Apr 2010
Posts: 62
Followers: 1

Kudos [?]: 9 [0], given: 0

### Show Tags

10 Aug 2010, 21:22
is probablity the same if
rather than "2 women" it says "jena and monica" must be in (ie 2 in particular among 4)?
bit confused in that
Manager
Joined: 20 Mar 2010
Posts: 84
Followers: 2

Kudos [?]: 92 [0], given: 1

### Show Tags

11 Aug 2010, 00:17
frank1 wrote:
is probablity the same if
rather than "2 women" it says "jena and monica" must be in (ie 2 in particular among 4)?
bit confused in that

Then there is only $$C^2_2$$ or 1 way to pick women instead of $$C^4_2$$.So probability will change.
_________________

___________________________________
Please give me kudos if you like my post

Manager
Joined: 09 Aug 2010
Posts: 107
Followers: 1

Kudos [?]: 44 [9] , given: 7

### Show Tags

24 Mar 2011, 22:15
9
KUDOS
1
This post was
BOOKMARKED
I used to have trouble with this but for those having trouble with probablility and combinatorics. I think you can have some help from ANAGRAM.

Example: Ways to rearrange LEVEL

5! = because we have 5 letters
2! = because we have 2 Ls
2! = because we have 2 Es

Formula is 5!/2!2! = 30

YOU CAN USE THIS WITH THE PROBLEM ABOVE.

SOLUTION:

How many ways to select 4 from 8 people?
(imagine this as rearranging YYYYNNNN) 8!/4!4! = 70

How many ways to select 2 women from 5 women?
(imagine this as rearranging YYNNN) 5!/2!3! = 10

How many ways to select 2 men from 3 men?
(imagine this as rearranging YYN) 3!/2! = 3

Probability = 3 x 10 / 70 = 3/7
Current Student
Joined: 07 Aug 2009
Posts: 49
GMAT 1: 680 Q V
Followers: 0

Kudos [?]: 6 [0], given: 1

### Show Tags

24 Apr 2011, 15:52
Hi all,

I was working on a practice test when i saw this question. Now, the problem is; i have solved it in two ways, one way(using combinations) leads to the correct answer, the other gives me something different. i would appreciate if you could point out what im doing wrong here.

Questn
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

first method:
5C2 * 3C2 -> it gives combination of exactly 2 women and 2 men.
8C4 -> gives total possibilities of 4 people from 5 women and 3 men.

Probability = 5C2*3C2 / 8C4 = 3/7

SECOND METHOD:
Probability of two women -> 5/8 * 4/7.

probability of two men -> 3/6 * 2/5.

Probability: (5/8 * 4/7) * (3/6 * 2/5) = 1/14.

I know something is wrong with the second method but i can't really figure out why its flawed.

Any pointers ?? thanks
Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7255
Location: Pune, India
Followers: 2205

Kudos [?]: 14369 [4] , given: 222

### Show Tags

24 Apr 2011, 16:17
4
KUDOS
Expert's post
8
This post was
BOOKMARKED
coldteleporter wrote:
Hi all,

SECOND METHOD:
Probability of two women -> 5/8 * 4/7.

probability of two men -> 3/6 * 2/5.

Probability: (5/8 * 4/7) * (3/6 * 2/5) = 1/14.

I know something is wrong with the second method but i can't really figure out why its flawed.

Any pointers ?? thanks

The problem here is that you are arranging the people. When you select a woman out of 8 as 5/8, you are saying that you are picking a woman first. You are arranging them in this way:
WWMM
Now, if you want to un-arrange them, multiply it by the total number of arrangements i.e. 4!/(2!*2!) (because there are 2 men and 2 women so you divide by 2!s to get the total number of arrangements)

When you do that, you get 1/14 * 4!/(2!*2!) = 3/7
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Current Student Joined: 07 Aug 2009 Posts: 49 GMAT 1: 680 Q V Followers: 0 Kudos [?]: 6 [0], given: 1 ### Show Tags 25 Apr 2011, 11:22 VeritasPrepKarishma wrote: coldteleporter wrote: Hi all, SECOND METHOD: Probability of two women -> 5/8 * 4/7. Quote: The problem here is that you are arranging the people. When you select a woman out of 8 as 5/8, you are saying that you are picking a woman first. You are arranging them in this way: WWMM Now, if you want to un-arrange them, multiply it by the total number of arrangements i.e. 4!/(2!*2!) (because there are 2 men and 2 women so you divide by 2!s to get the total number of arrangements) When you do that, you get 1/14 * 4!/(2!*2!) = 3/7 Hello karishma, From what i understand, you have added all probabilities like so: 1) WWMM 2) WMMW 3) WMWM 4) MWMW 5) MMWW 6) MWWM ...such that you do not repeat arrangements that are the same, such as W1W2 and W2W1. ... right ? Thanks Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7255 Location: Pune, India Followers: 2205 Kudos [?]: 14369 [1] , given: 222 Re: [#permalink] ### Show Tags 25 Apr 2011, 11:53 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED Yes. What you have to do is select a group. Hence you do not have to arrange them. In your second method, you got 1/14 which is the probability of getting wwmm. Since there are other such arrangements too which are all acceptable to us e.g. Wmwm since we just need a group of 2 men and 2 women irrespective of their arrangement, we multiply 1/14 by 6 (since 6 such arrangements are possible as shown by you) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Current Student
Joined: 07 Aug 2009
Posts: 49
GMAT 1: 680 Q V
Followers: 0

Kudos [?]: 6 [0], given: 1

### Show Tags

25 Apr 2011, 13:55
VeritasPrepKarishma wrote:
Yes. What you have to do is select a group. Hence you do not have to arrange them. In your second method, you got 1/14 which is the probability of getting wwmm. Since there are other such arrangements too which are all acceptable to us e.g. Wmwm since we just need a group of 2 men and 2 women irrespective of their arrangement, we multiply 1/14 by 6 (since 6 such arrangements are possible as shown by you)

Thank you Karishma. You were very helpful.
Intern
Joined: 18 Jun 2012
Posts: 10
Location: United States
Concentration: Social Entrepreneurship, General Management
GMAT 1: 700 Q38 V47
GPA: 3.37
WE: Marketing (Non-Profit and Government)
Followers: 0

Kudos [?]: 7 [0], given: 0

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

### Show Tags

25 May 2013, 10:58
Can someone please explain to me why you can't solve this by multiplying:

5/8 * 5/8 * 3/8 * 3/8?

Is it because this method doesn't take into account different order possibilities?

In general how do you know whether to solve a probability by multiplying plain fractions like this or by using factorals (!s) ?
Math Expert
Joined: 02 Sep 2009
Posts: 37661
Followers: 7411

Kudos [?]: 99876 [0], given: 11061

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

### Show Tags

25 May 2013, 11:16
dawip wrote:
Can someone please explain to me why you can't solve this by multiplying:

5/8 * 5/8 * 3/8 * 3/8?

Is it because this method doesn't take into account different order possibilities?

In general how do you know whether to solve a probability by multiplying plain fractions like this or by using factorals (!s) ?

Check here: a-small-company-employs-3-men-and-5-women-if-a-team-of-56037.html#p913073

Hope it helps.
_________________
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 59 [0], given: 23

### Show Tags

10 May 2014, 15:24
VeritasPrepKarishma wrote:
Yes. What you have to do is select a group. Hence you do not have to arrange them. In your second method, you got 1/14 which is the probability of getting wwmm. Since there are other such arrangements too which are all acceptable to us e.g. Wmwm since we just need a group of 2 men and 2 women irrespective of their arrangement, we multiply 1/14 by 6 (since 6 such arrangements are possible as shown by you)

Hi Karishma,

Very helpful answer. Maybe you could clarify something for me: When we use the combinatorics method in this problem - (2c5*2c3)/4c8 - why don't we add in the permutations part like we do for the probability method?

What I mean by that is -- If i use the probability approach (favorable outcomes/total outcomes) -- I get (5/8)(4/7)(3/6)(2/5) and then I multiply that by the permutations, which means that I multiply it 4!/2!2! -- why don't we do this last part when it comes to probability?

Thanks!
Manager
Joined: 07 Apr 2014
Posts: 145
Followers: 1

Kudos [?]: 23 [0], given: 81

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

### Show Tags

15 Aug 2014, 12:12
Hi,

4W- 5C4
3W, 1M - 5C3*3C1
1W,3M - 5C1*3C3

Total - 8C4

1-( 4W+3W,1M +1W,3M) / TOTAL....

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 14496
Followers: 609

Kudos [?]: 174 [0], given: 0

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

### Show Tags

26 Sep 2015, 10:31
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
Manager
Joined: 18 Aug 2014
Posts: 175
Location: United States
GMAT 1: 740 Q47 V45
GMAT 2: 730 Q48 V42
GPA: 3.13
Followers: 0

Kudos [?]: 31 [0], given: 77

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

### Show Tags

30 Jan 2016, 17:15
Can someone explain to me why the answer is not 1/7?

I did Desired Outcomes/Total Outcomes. Obviously total is 70 we all agree on that. We also all agree that 5 women choose 2 = 10.

However, I don't see why we then do 3 men choose 2. Why does it matter which men are chosen?

Can someone explain logically rather than mathematically?
_________________

VP
Joined: 12 Sep 2015
Posts: 1187
Followers: 121

Kudos [?]: 1212 [1] , given: 219

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

### Show Tags

30 Jan 2016, 18:54
1
KUDOS
redfield wrote:
Why does it matter which men are chosen?
Can someone explain logically rather than mathematically?

We need to consider which men are chosen since a committee with Ann, Bea, Joe and Ed is DIFFERENT FROM a committee with Ann, Bea, Joe and Kevin

Cheers,
Brent
_________________

Brent Hanneson – Founder of gmatprepnow.com

Re: A small company employs 3 men and 5 women. If a team of 4   [#permalink] 30 Jan 2016, 18:54

Go to page    1   2    Next  [ 23 posts ]

Similar topics Replies Last post
Similar
Topics:
7 There are 5 women and 4 men including 2 couples, 7 31 Dec 2015, 03:33
7 A team of 6 cooks is chosen from 8 men and 5 women. The team 19 16 Nov 2012, 07:38
1 At a conference, one team is made up of 4 men and 5 women. Four presen 2 04 Oct 2010, 22:12
14 A small company employs 3 men and 5 women. If a team of 4 7 23 May 2010, 12:11
3 A small company employs 3 men and 5 women. If a team of 4 employees is 6 27 Mar 2010, 21:53
Display posts from previous: Sort by