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 Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
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
GMATOnDemand Start 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 559 reviewsTarget Test Prep

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

a rare on GMAT

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
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 » Tue Mar 02, 2010 4:22 am
10 different books and 2 different pens are given to 3 boys so that each boy gets equal number of things. What is the probability that same boy doesn't receive both pens?
(A) 3/11
(B) ¼
(C) 4/11
(D) 8/11
(E) 9/11
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
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 189
Joined: Thu Apr 03, 2008 2:03 pm
Location: USA
Thanked: 21 times

by rohan_vus » Tue Mar 02, 2010 4:38 am
Probability that one boy receives both pens is 10c2*2c2/12c4 = 1/11
So total probability that among 3 boys, anyone will have boths pens, is 3*1/11 = 3/11

So probability that none will have both pens = 1 - 3/11 = 8/11
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 15
Joined: Wed Feb 03, 2010 12:09 am
GMAT Score:540

by stufigol » Wed Mar 03, 2010 12:13 am
u explain the 10c2 and 12c4 plz
thanks
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1132
Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members
Location: India
Thanked: 64 times
GMAT Score:760

by harsh.champ » Wed Mar 03, 2010 4:27 am
stufigol wrote:u explain the 10c2 and 12c4 plz
thanks
Ok.
Lets look at rohan's explanation.
Probability that one boy receives both pens is 10c2*2c2/12c4 = 1/11
So total probability that among 3 boys, anyone will have boths pens, is 3*1/11 = 3/11

So probability that none will have both pens = 1 - 3/11 = 8/11
Suppose the same boy gets both the pens.
Now,10C2 = No. of ways in which the 2 books can be selected.
12C4 = Selection of any 4 items out of the (10 + 2) items

Also,the boy can be selected in 3C1 ways.
So, we have our case (10C2 x 2C2 x 3C1)/12c4 = 3/11
Now, P(favorable case) = 1 - P(all the unfavorable cases)

I hope it is clear now. :)
It takes time and effort to explain, so if my comment helped you please press Thanks button :)



Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker :P
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1