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 of roads

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 107
Joined: Wed Mar 04, 2009 4:39 am
Location: Vancouver
Thanked: 9 times
GMAT Score:750
by pakaskwa » Fri Mar 06, 2009 2:10 am
Two-thirds of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
A. 1/6
B. 1/4
C. 2/3
D. 3/4
E. 11/12

OA is D. But I don't really understand the calculation. Can somebody help me to explain?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Master | Next Rank: 500 Posts
Posts: 418
Joined: Wed Jun 11, 2008 5:29 am
Thanked: 65 times

by bluementor » Fri Mar 06, 2009 2:18 am
"at least 5 miles long" = X

From A to B:
Prob(X_ab) = 2/3
Prob(not X_ab) = 1/3

From B to C:
Prob(X_bc) = 1/4
Prob(not X_bc) = 3/4

Prob of at least 1 X from A to C =
=1 - Prob(not X_ac)
=1 - Prob(not X_ab)*Prob(not X_bc)
=1 - (1/3)*(3/4)
=1- 1/4
=3/4
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 1462
Joined: Thu Apr 09, 2015 9:34 amFollowed by: 22 members
Location: New York, NY
Thanked: 39 times

by Jeff@TargetTestPrep » Mon Dec 11, 2017 9:31 am
pakaskwa wrote:Two-thirds of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
A. 1/6
B. 1/4
C. 2/3
D. 3/4
E. 11/12
We are given that 2/3 of the roads from A to B are at least 5 miles long (which means that 1/3 of the roads from A to B are less than 5 miles long) and 1/4 of the roads from B to C are at least 5 miles long (which means that 3/4 of the roads from A to B are less than 5 miles long). We need to determine the probability that when picking a road from A to B and B to C, at least one of the roads is at least 5 miles long.

We can use the following formula:

P(selecting at least 1 road that is at least 5 miles long) + P(selecting no roads that are at least 5 miles long) = 1

P(selecting at least 1 road that is 5 miles long) = 1 - P(selecting no roads that are at least 5 miles long)

P(selecting at least 1 road that is 5 miles long) = 1 - P(selecting all roads that are less than 5 miles long)

Thus, if we can determine the probability of selecting all roads that are less than 5 miles long, we'll quickly be able to calculate the probability of selecting at least 1 road that is at least 5 miles long.

The probability of selecting all roads that are less than 5 miles long is: 1/3 x 3/4 = 1/4

Thus, the probability of selecting at least 1 road that is at least 5 miles long is: 1 - 1/4 = 3/4

Answer: D

Jeffrey Miller
Head of GMAT Instruction
[email protected]

Image

See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 100
Joined: Wed Nov 29, 2017 4:38 pm
Thanked: 14 times

by GMATWisdom » Mon Dec 11, 2017 10:20 am
pakaskwa wrote:Two-thirds of the roads from A to B are at least 5 miles long, and 1/4 of the roads from B to C are at least 5 miles long. If you randomly pick a road from A to B and then randomly pick a road from B to C, what is the probability that at least one of the roads you pick is at least 5 miles long?
A. 1/6
B. 1/4
C. 2/3
D. 3/4
E. 11/12

OA is D. But I don't really understand the calculation. Can somebody help me to explain?
2/3rd of the roads are A to B are ≥5miles
Therefore 1/3 of the roads are <5miles
¼ th of the roads from C to D are ≥ 5miles
Therefore 3/4th of the roads are <5miles
so probabilility of both the roads <5miles =(1/3)*(3/4)=1/4
hence probabilility of one of the roads ≥5miles=1-1/4=3/4
so D is the correct option .
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1