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.

Probability: getting a seat at an overcrowded table!!

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
Expert replies
by Cúchulainn » Tue May 15, 2012 2:47 am
Hi - i have just joined this forum, so I hope i'm using it correctly. I've just done a problem on one of the kaplan quizzes, which made me think of my own slightly different version, and was hoping someone could confirm my approach to my own version is correct or not!

A square table seats 1 person on each of the four sides, with every person directly facing another person across the table. If there are eight people looking to sit at this table and choose their seats at random, what is the probability that any two of the 8 getting seats directly facing each other?

My approach:
Choose two people A and B.
Probability of A getting a seat at table is 1/2.
Probability of B getting a seat at the table once A has taken one is 3/7.
Probability of B getting a seat once A has taken one and sits opposite A is (1/3)(3/7) = 1/7.

So Answer = (1/2)(1/7) = 1/14

Is this approach correct?
Many thanks for your help.
Join the discussion
Source: — Quantitative Reasoning |

by hey_thr67 » Tue May 15, 2012 11:14 am
A good question indeed. I think the correct answer should be 1/28.

Following is my solution.

Well for first person the probability to have seat is 1/4 ( as one seat is to be selected out of 4). Now, for the second person the probability of choosing the seat on square table is 1. But to the person selected out of rest 7 is 1/7.

Hence, total probability is (1/4) X (1) X ( 1/7) = 1/28
Join the discussion

by Cúchulainn » Tue May 15, 2012 11:57 am
Thanks thr,

I got 1/2 for A, as there 8 people looking for a seat but only four seats, hence 4/8 probability of A getting a seat.

After A has taken a seat, there are 3 seats remaining with 7 people looking to sit down. So this would make a 3/7 probability of B getting a seat. Now only one of those 3 seats will be opposite A, so this 3/7 would need to be multiplied by 1/3 to get the probability of sitting opposite A.

So my final probability was got by multiplying all these probabilities: (1/2)(3/7)(1/3) = 1/14.

But again, I'm not sure if I'm missing something, so it would be good to get it confirmed whether this approach is right or not. Thanks again thr.
Join the discussion