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 →

There are 5 pairs of white, 3 pairs of black

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Master | Next Rank: 500 Posts
Posts: 266
Joined: Fri Sep 19, 2014 4:00 amFollowed by: 1 member
Thanked: 4 times
by conquistador » Mon Oct 12, 2015 3:31 pm
Source: Gmatclub

There are 5 pairs of white, 3 pairs of black and 2 pairs of grey socks in a drawer. If four individual socks are picked at random what is the probability of getting at least two socks of the same color?

A. 15
B. 25
C. 1
D. 35
E. 45

naturally in all the probability and combinations problems, a condition with or without replacement is mentioned which plays a key role in deciding the answer.

Isn't the same missing here? Explain.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
Elite Legendary Member
Posts: 10392
Joined: Sun Jun 23, 2013 6:38 pmFollowed by: 511 members
Location: Palo Alto, CA
Thanked: 2867 times
GMAT Score:800

by [email protected] » Mon Oct 12, 2015 3:54 pm
Hi Mechmeera,

This looks like someone took a 'worst-case scenario' question and designed it in 'reverse.'

Since each of the three colors has at least 2 socks in that color, the probability of getting at least one matching pair is 100%. Here's why...

IF....your first 3 socks were one of each color:

1 white, 1 black and 1 grey.....

Then the 4th sock would automatically match one of the prior 3, so you'd have a matching pair no matter what. In the realm of probability math, a 100% probability is written as 1/1 = 1.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 266
Joined: Fri Sep 19, 2014 4:00 amFollowed by: 1 member
Thanked: 4 times

by conquistador » Mon Oct 12, 2015 4:40 pm
[email protected] wrote:Hi Mechmeera,

This looks like someone took a 'worst-case scenario' question and designed it in 'reverse.'

Since each of the three colors has at least 2 socks in that color, the probability of getting at least one matching pair is 100%. Here's why...

IF....your first 3 socks were one of each color:

1 white, 1 black and 1 grey.....

Then the 4th sock would automatically match one of the prior 3, so you'd have a matching pair no matter what. In the realm of probability math, a 100% probability is written as 1/1 = 1.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Yes you are right.
a condition with or without replacement does not matter here.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members
Location: East Bay all the way
Thanked: 625 times
GMAT Score:780

by Matt@VeritasPrep » Wed Oct 14, 2015 11:45 pm
This question is a (basic) illustration of something called the Pigeonhole Principle, one of my favorite ideas in math.

It's incredible how often this comes up in higher math, and it's equally incredible that you can use it to prove that there exist two people in New York City who have the exact same number of hairs on their heads! (Check that link if you're interested.)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1