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 →

probability

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 104
Joined: Sun Dec 11, 2011 1:25 pmFollowed by: 3 members
Thanked: 11 times

by MakeUrTimeCount » Sat Jan 28, 2012 12:49 pm
Total number of balls = 2 + 3 + 4 = 9
Probability to get 2 balls = 9C2

Probability to get 2 Red balls = 2C2
Probability to get 2 White balls = 3C2
Probability to get 2 Black balls = 4C2

Probability to get 2 balls of same color = (2C2 + 3C2 + 4C2)/ 9C2 = 5/18
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 406
Joined: Mon Jan 25, 2010 11:36 amFollowed by: 4 members
Location: Syracuse, NY
Thanked: 23 times
GMAT Score:740

by tomada » Sat Jan 28, 2012 12:52 pm
I would think that you want the sum of the following probabilities:

a) P (2 red balls)
b) P (2 white balls)
c) P (2 black balls)

If I'm correct about that, then...

a) P (2 red balls) = (2/9) * (1/8) = 2/72
b) P (2 white balls) = (3/9) * (2/8) = 6/72
c) P (2 black balls) = (4/9) * (3/8) = 12/72

Summing, (2/72) + (6/72) + (12/72) = 20/72 = 5/18

Of course, if my initial assumption is incorrect, then ignore everything. :-)
I'm really old, but I'll never be too old to become more educated.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1084
Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members
Thanked: 158 times

by pemdas » Sat Jan 28, 2012 1:50 pm
do you count probability or the number of subsets?
below 9C2 should return the total number of subsets, it's not probability
if the problem asked not *without replacement* the method of Tomada would be the best one. However, if we approach with the method of MakeYourTimeCount here we must consider this only for *the same color* balls selected *without replacement*.

@nikhilgupta, to avoid confusion in test use method of Tomada
use the method of MakeYourTimeCount to improve your strategy with combinatorics question types
MakeUrTimeCount wrote:Total number of balls = 2 + 3 + 4 = 9
Probability to get 2 balls = 9C2

Probability to get 2 Red balls = 2C2
Probability to get 2 White balls = 3C2
Probability to get 2 Black balls = 4C2

Probability to get 2 balls of same color = (2C2 + 3C2 + 4C2)/ 9C2 = 5/18
Success doesn't come overnight!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1