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.

combination

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 75
Joined: Sun Jun 14, 2009 1:41 amFollowed by: 1 member
Location: india
Thanked: 2 times
by armaan700+ » Tue Jan 26, 2010 10:02 am
Tanya prepared 4 different letters to 4 different addresses. For each letter, she prepared one envelope with its correct address. If the 4 letters are to be put into the four envelopes at random, what is the probability that only one letter will be put into the envelope with its correct address?

A) 1/24
B) 1/8
C) 1/4
D) 1/3
E) 3/8
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
Legendary Member
Posts: 1560
Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members
Thanked: 137 times

by thephoenix » Tue Jan 26, 2010 10:35 am
Total # of ways of choosing envelopes=4!=24.
Only one letter in the right envelope: 4(# of envelopes)*2(# of ways possible to arrange 3 letters incorrectly in the envelopes, when one is correct)

ABCD(envelopes)
ACDB(letters)
ADBC(letters)
(When A is in the right envelope other three have only 2 possible incorrect arrangements)
As we have 4 letters, total # of ways 4*2=8

P(C=1)=8/24=1/3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1