BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
Live EA class + 6 months of EA OnDemand
  • Expert-led weekly online sessions
  • EA Masterclass access between classes
View EA class & enroll
Limited cohort · enrollment openTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

130-point score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions
Start free 5-day trial
Read 559 reviewsTarget Test Prep

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Only Jim and not Jill

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 57
Joined: Thu Feb 26, 2009 3:25 pm
Thanked: 5 times
GMAT Score:730

wrong calculation

by 007.r.mason » Sat Feb 28, 2009 7:09 pm
I think 8C3 is 56 not 48...
Aim high!!
Achieve excellence!!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 3650
Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members
Location: India
Thanked: 267 times
GMAT Score:760

by sanju09 » Mon Mar 02, 2009 4:20 am
DanaJ wrote:The wording of your problem makes me wonder what you actually mean. On a first glace (and I'm probably not going to give it a second glance, since I HATE probabilities), you're looking for (the number of 3 person groups containing Jack but not Jill)/(total number of groups of 3 people)
The total number of groups will be 8C3 = 56.
Groups that contain Jack have "two open positions" for the other 6 people. There are 6C2 possible groups of two people from the pool of 6 left or 15.
This makes the solution to this problem (IMHO, of course, since I DESPISE probabilities) 15/56.
:) So, Jack is also in the pool, great info DanaJ!
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 16
Joined: Mon Mar 09, 2009 10:56 am

by keeyu2 » Thu Mar 12, 2009 6:34 am
I think the answer is 5/56.

The proability of Jim is 1/8
The proability of not Jill is 6/7
The proability of not Jill again is 5/6

These are dependent P(Jim & Not Jill & Not Jill).

1/8*6/7*5/6 is 5/56.


What is the OG answer?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 20
Joined: Sat Mar 07, 2009 4:15 pm
Thanked: 2 times

by Reader » Sun Mar 15, 2009 10:34 am
I think it's 15/56,

the chance of jim being select = 3/8, the chance both jim and jill being selected = 3/8 x 2/7 =6/56. only jim not jill = 3/8 - 6/56 = 15/56
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply