Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemandStart anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 563 reviewsTarget Test Prep

Probability Question

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Wed Oct 06, 2010 11:35 am
Thanked: 1 times
by chrisjim5 » Mon Oct 18, 2010 6:48 am
Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Top
Join the discussion
Source: — Problem Solving |
User avatar
Master | Next Rank: 500 Posts
Posts: 324
Joined: Mon Jul 05, 2010 6:44 amFollowed by: 3 members
Location: London
Thanked: 70 times

by kmittal82 » Mon Oct 18, 2010 7:01 am
Total employeed = 8
Number to be selected = 5

Total ways to select 5 employees out of 8 = 8C5

#ways of selecting all women = #ways of selecting 1 man * #ways of selecting 4 women = 4C1x4C4 = 4x1 = 4

Thus, probability = 4/8C5 = 4/56 = 1/14

OA?
Top
Join the discussion
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Oct 18, 2010 7:15 am
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Newbie | Next Rank: 10 Posts
Posts: 8
Joined: Wed Oct 06, 2010 11:35 am
Thanked: 1 times

by chrisjim5 » Mon Oct 18, 2010 7:20 am
Thats right ...

Thanks for explaining ...
Top
Join the discussion
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Mon Oct 18, 2010 7:23 am
GMATGuruNY wrote:
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Why the following method is getting wrong Please clarify?

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 possibilities with 4 W and 1 M

Total possibilities = 8C5 = 56

Why not 5/56 ?
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
User avatar
GMAT Instructor
Posts: 905
Joined: Sun Sep 12, 2010 1:38 amFollowed by: 123 members
Thanked: 378 times
GMAT Score:760

by Geva@EconomistGMAT » Mon Oct 18, 2010 7:33 am
shovan85 wrote:
GMATGuruNY wrote:
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Why the following method is getting wrong Please clarify?

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 possibilities with 4 W and 1 M

Total possibilities = 8C5 = 56

Why not 5/56 ?
The two are just incompatible: 8C5 is the number of ways of choosing 5 out of 8, order doesn't matter plays around with combinations of all 4 men), while the 5 options actually assumes a single man and order matters. Whatever, you do, be consistent: either order matters in both top and bottom, or it doesn't.
Geva
Senior Instructor
Master GMAT
1-888-780-GMAT
https://www.mastergmat.com
Top
Join the discussion
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Oct 18, 2010 7:35 am
shovan85 wrote:
GMATGuruNY wrote:
chrisjim5 wrote:Here is a question:

The employees of a certain company include exactly 4 men and 4 women. If 5 employees are to be selected at random, what is the probability that all of the women are selected?

Please let me know how to solve this ...

Thanks ...
Another approach:

P(WWWWM) = 4/8 * 3/7 * 2/6 * 1/5 * 4/4 = 1/70.

Since M can appear in any of the 5 positions, we multiply by 5:

5*(1/70) = 1/14.
Why the following method is getting wrong Please clarify?

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 possibilities with 4 W and 1 M

Total possibilities = 8C5 = 56

Why not 5/56 ?
You're mixing up methods.

8C5 counts the number ways to combine 5 out of 8 choices.

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 = the number of ways to arrange the 5 letters WWWWM. (We could also determine this by saying the number of ways to arrange WWWWM = 5!/4! = 5.)

You can't put arrangements/combinations.

Does this help?
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 49
Joined: Tue May 11, 2010 5:46 am
Thanked: 3 times

by abhishekg21 » Mon Oct 18, 2010 7:42 am
you are taking into account the order or arrangement.Where as in commitee it does not matter. wwwwm is same as mwwww.
so remove the arrangment from your answer and you will correct figure of 1/14
Top
Join the discussion
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Mon Oct 18, 2010 7:49 am
Geva Stern wrote: The two are just incompatible: 8C5 is the number of ways of choosing 5 out of 8, order doesn't matter plays around with combinations of all 4 men), while the 5 options actually assumes a single man and order matters. Whatever, you do, be consistent: either order matters in both top and bottom, or it doesn't.
Finally I guess I got it...

5 possibilities if I take as WWWWM, WWWMW, WWMWW, WMWWW, MWWWW then order matters. Using Combinatorics I got 1/14 but the picture was not getting clear in my mind as I had taken W1,W2...

So I was taking the upper part with respect to Arrangement but was considering the formula of Combinatorics.

Using Permutations it will be 5*(4P4 * 4P1)/8P5 = 1/14.

Not a good idea but I have a clear picture now ;)

Thanks Geva Stern I was not able to see this
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion
Community Manager
Posts: 991
Joined: Thu Sep 23, 2010 6:19 amFollowed by: 24 members
Location: Bangalore, India
Thanked: 146 times

by shovan85 » Mon Oct 18, 2010 7:51 am
GMATGuruNY wrote:You're mixing up methods.

8C5 counts the number ways to combine 5 out of 8 choices.

WWWWM, WWWMW, WWMWW, WMWWW, MWWWW = 5 = the number of ways to arrange the 5 letters WWWWM. (We could also determine this by saying the number of ways to arrange WWWWM = 5!/4! = 5.)

You can't put arrangements/combinations.

Does this help?
Thanks it helps a lot Never mind it was a stupid Question after all
and no more Maths for me today ;)
If the problem is Easy Respect it, if the problem is tough Attack it
Top
Join the discussion