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

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

Another Permutation

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Wed Apr 22, 2009 10:00 am
Thanked: 1 times
by sportcntr3 » Wed Apr 22, 2009 11:00 am
A pencil box contains 2 red, 3 white and 3 black balls. In how many ways can 2 balls be drawn from the box, if at least one white ball is to be included in the draw?

a. 18
b. 15
c. 16
d. 34
e. 20

OA: 18
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 322
Joined: Fri Mar 27, 2009 3:56 pm
Thanked: 24 times
GMAT Score:710

by mike22629 » Wed Apr 22, 2009 11:13 am
IMO A.

I used 1 - x probability formula to solve this.

First find total possibilities

8!/(6!*2!) = 28

Then find all possibilities of no white balls out of the two picked

Both Red = 1 possibility (there are only 2)

Both Black = 3 possibilities (3!/2!)

One Black and One Red

First Red = 3 possibilities
Second Red = 3 possibilities

Total Possibilities of no white = 1+3+3+3 = 10

28 - 10 = 18 Total Possibilities of at least one white
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 92
Joined: Wed Mar 25, 2009 12:13 pm
Thanked: 10 times

by moutar » Wed Apr 22, 2009 11:15 am
Total number of ways of choosing 2 balls is 8C2 = 8!/6!2! = 28

Prob of picking a white ball first time = 3/8

Prob of picking a white ball second time (having not picked it first time) = 5/8 * 3/7 = 15/56

Total prob of picking at least one white ball = 3/8 + 15/56 = (21+15)/56 = 36/56 = 18/28

Number of ways of picking a white ball = 28 * 18/28 = 18

Long winded but it gets you the right answer. My prob knowledge isnt that good.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 14
Joined: Wed Apr 22, 2009 10:00 am
Thanked: 1 times

by sportcntr3 » Wed Apr 22, 2009 11:16 am
Thanks, great explanation
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1