Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.

See classes
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.

Newsweek Readers’ Choice Awards 2026 winner: Target Test Prep, #1 Test Prep

Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.

★★★★★5.0563 reviews
GMATLiveTeachStarts Nov 22
Josh Braslow, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Josh Braslow, 20+ years in test prep

Nov 22 · Josh Braslow
Sun · 9:30 AM to 12:30 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
EALiveTeachOnDemandStart anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

EXECUTIVE ASSESSMENT

with

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
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
GMATOnDemandStart 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 563 reviewsTarget Test Prep

Trivial Probability?

Expert replies
Post new topic Post Reply
Subscribe topic
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Tue Jul 05, 2011 1:13 am
by StoneBlack » Thu Dec 29, 2011 9:13 pm
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
A. 1/4
B. 1/2
C. 5/8
D. 2/3
E. 3/4

This question is from 300+ GMAT questions document available on BeatTheGMAT forums.
The solution is provided there but I am not really convinced of it.
Top
Join the discussion
Source: — Problem Solving |
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members
Thanked: 230 times

by shankar.ashwin » Thu Dec 29, 2011 9:31 pm
Each ball picked has a 3/8th probability of being yellow and 5/8th probability of being black. The number of selection doesn't matter.

Ofcourse theres a mathematical proof which is lengthy, but think about it, each ball is likely to be either yellow or black, now how does the number of pick matter..

If the probability was any greater or lesser you would have to say, the first 3 picks were biased towards picking a particular color, which is not the case. Hence probability at any pick shouldn't change.

P (nth pick) = P (1st pick) = [spoiler]5/8[/spoiler]
Top
Join the discussion
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Thu Dec 29, 2011 10:21 pm
Check my posts here:

https://www.beatthegmat.com/probablity-ques-t60161.html
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Senior | Next Rank: 100 Posts
Posts: 39
Joined: Tue Jul 05, 2011 1:13 am

by StoneBlack » Thu Dec 29, 2011 11:11 pm
Thanks for the pointer.
Apparently i tried the expanded way by considering all possibilities:
3Y,1B
2Y2B
1Y3B
4B

but failed to take into consideration the different arrangement possible in 2Y2B and 1Y3B.
Thanks again!
Top
Join the discussion