BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

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.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 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
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
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
GMATOnDemand Start 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

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

There are 5 pairs of white, 3 pairs of black

Expert replies
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.
Join the discussion
Source: — Problem Solving |

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
Join the discussion

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.
Join the discussion

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.)
Join the discussion