Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
07 Mar 2004, 14:03
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
What is the probability that two persons selected from a row of n persons were not sitting next to each other?
2/n
2/(n-1)(n-2)
1-(2/n-1)
1-(2/n)
None of these
2/n
2/(n-1)(n-2)
1-(2/n-1)
1-(2/n)
None of these
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
07 Mar 2004, 14:09
Kudos
Bookmarks
No of ways N people are arranged in which 2 desired persons are together = 2! * ( n-1)!
P that two are always together = 2! * (n-1)! / n!
P that they are not tgether = 1 - 2! *(n-1)!/n! = 1-2/n
P that two are always together = 2! * (n-1)! / n!
P that they are not tgether = 1 - 2! *(n-1)!/n! = 1-2/n
07 Mar 2004, 14:37
Kudos
Bookmarks
you were right the first time.
The following is the official explanation:
The total # of ways of selecting 2 persons from a set of n = nC2 = {n(n-1)}/2
The number of outcomes in which the two selected will be together = n-1.
Therefore, the # of outcomes in which they will not be together = {{n(n-1)}/2} - (n-1)
= {(n-1)(n-2)}/2.
And the probability = {(n-1)(n-2)/2}/n(n-1)/2 = 1- 2/n.
The following is the official explanation:
The total # of ways of selecting 2 persons from a set of n = nC2 = {n(n-1)}/2
The number of outcomes in which the two selected will be together = n-1.
Therefore, the # of outcomes in which they will not be together = {{n(n-1)}/2} - (n-1)
= {(n-1)(n-2)}/2.
And the probability = {(n-1)(n-2)/2}/n(n-1)/2 = 1- 2/n.
