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 question

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Legendary Member
Posts: 727
Joined: Sun Jun 08, 2008 9:32 pmFollowed by: 1 member
Thanked: 8 times
by umaa » Tue Aug 05, 2008 1:00 am
In a particular lottery game, 6 numbers are chosen out of 4 even numbers and 4 odd numbers. What is the probability that the four numbers chosen will have an equal number of even and odd numbers?

A) 1/35
B) 2/7
C) 1/2
D) 4/7
E) 3/4
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 661
Joined: Tue Jul 08, 2008 12:58 pm
Location: France
Thanked: 48 times

by pepeprepa » Tue Aug 05, 2008 1:49 am
Number of total possibilities: 8C6

Number of possibilities you want: (4C3)x(4C3)

Probability is the number of possibilities you want divided by the total number of possibilities:
( (4C3)x(4C3) ) / ( 8C6 ) = 4*4 / 28 = 4/7

That's D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1153
Joined: Wed Jun 20, 2007 6:21 amFollowed by: 2 members
Thanked: 146 times

by parallel_chase » Tue Aug 05, 2008 2:02 am
I agree with you pepeprepa but we need to find out the probability of 4 numbers having equal odd and even integers.

Therefore, after selecting 6, these 4 numbers could be starting 4, middle 4 or ending 4.

Let me know what you think.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 661
Joined: Tue Jul 08, 2008 12:58 pm
Location: France
Thanked: 48 times

by pepeprepa » Tue Aug 05, 2008 2:08 am
The question does not take into account "order" and factorials neither. For me that is a non-order exercice... I have to work back combinatory to deeply understand but I think it is ok for this question. If you have clues, tell me more.
For me, this one is in the same spirit as forming a group of 4 with 3 boys and 3 girls and having 2 boys and 2 girls. When you make a group like that, order does not matter, you just choose people.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1