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.

Teacher's Pets (probability)

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Sun Jan 25, 2009 2:34 pm
Thanked: 1 times
by mlaboda » Mon Sep 07, 2009 3:05 pm
A teacher will pick a group of 4 students from a group of 8 students that includes Bart and Lisa. If one of all of the possible four-student groups is picked at random, what is the probability of picking a group that includes both Bart and Lisa?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 399
Joined: Wed Apr 15, 2009 3:48 am
Location: india
Thanked: 39 times

IMO

by xcusemeplz2009 » Mon Sep 07, 2009 10:05 pm
is it 3/14
It does not matter how many times you get knocked down , but how many times you get up
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 77
Joined: Sun Jun 21, 2009 10:25 am
Location: Germany
Thanked: 7 times

by Nermal » Tue Sep 08, 2009 12:08 am
I also got 3/14.

I calculated:
(2C2*6C2)/(8C4)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 156
Joined: Sat Jul 19, 2008 6:41 am
Thanked: 8 times

IMO

by enniguy » Tue Sep 08, 2009 12:36 am
3/10
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Junior | Next Rank: 30 Posts
Posts: 28
Joined: Wed Feb 14, 2007 12:25 am
Thanked: 1 times
GMAT Score:690

by Andrei » Tue Sep 08, 2009 1:33 am
I also think it's 3/14.

Total number of possible groups of 4 persons selected out of 8 is: 8!/(4!*4!) = 70.

If both both Bart and Lisa must be in the group, than there are left only 2 places for the other 6 students, so the total number of possible groups including both Bart and Lista is equal to the number of combinations of 2 selecting persons out of 6: 6!/(4!*2!) = 15.

So the probability of selecting a group that includes Bart and Lista out of total groups of 4 students is: 15/70 = 3/14.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1