Find all School-related info fast with the new School-Specific MBA Forum

It is currently 21 Oct 2014, 11:58

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

Probability

  Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:
Manager
Manager
avatar
Joined: 24 Jan 2004
Posts: 87
Location: Valsad
Followers: 1

Kudos [?]: 0 [0], given: 0

Probability [#permalink] New post 23 May 2004, 22:02
Please solve this:-

13 persons take their places at a round table. What is the probability that 2 particular persons do not sit together?
_________________

Mayur

Intern
Intern
User avatar
Joined: 24 May 2004
Posts: 3
Location: Kharagpur , India
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 24 May 2004, 00:43
lets suppose the 2 particular persons to be one things called 'A'
now for the Sample ......we have to calculate the no. of ways ..the 13 ppl can sit around the round table ..
it goes like this
fix one person...now we have to arrange 12 persons around the round table
so Sample Space : ! 12 . ( ! n === Factorial n );
now
the reqd permutation when two particular persons sit together...is
: ! 11
( i suppose u dont have any problem in understanding this )

hecne the reqd probability :
=== 1 - ( ! 11 / ! 12 );
==== 11/12 ;


Hope i m correct........
correct me if i m not ;
regards
nitin
Manager
Manager
avatar
Joined: 24 Jan 2004
Posts: 87
Location: Valsad
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 24 May 2004, 00:56
Continuing from where you've left: -

"lets suppose the 2 particular persons to be one things called 'A'
now for the Sample ......we have to calculate the no. of ways ..the 13 ppl can sit around the round table ..
it goes like this
fix one person...now we have to arrange 12 persons around the round table
so Sample Space : ! 12 . ( ! n === Factorial n );
now
the reqd permutation when two particular persons sit together...is
: ! 11 "

The remaining 10 people can be arranged among themselves in 10 ways.

So the prob. of 2 people together is (11! X 10)/12! = 10/12 = 5/6

The prob. of 2 people not together is 1-5/6 = 1/6
_________________

Mayur

Intern
Intern
User avatar
Joined: 24 May 2004
Posts: 3
Location: Kharagpur , India
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 24 May 2004, 01:30
[quote]
the reqd permutation when two particular persons sit together...is
: ! 11 "

The remaining 10 people can be arranged among themselves in 10 ways.

So the prob. of 2 people together is (11! X 10)/12! = 10/12 = 5/6

The prob. of 2 people not together is 1-5/6 = 1/6
_________________
Mayur
[\quote]
oops i think u went wrong here..
it shud be
after taking 2 ppl together we have total 12 ppl in hand..
so no. of ways to arrange them is...... ! 11
and the 2 ppl can be arranged too ( HERE I MISSED )
hence it will be......
(! 11 x 2 )/ ! 12



and Prob reqd. === 1 - (( ! 11 x 2 ) / ! 12 ) ;
regards
nitin
Manager
Manager
avatar
Joined: 24 Jan 2004
Posts: 87
Location: Valsad
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 24 May 2004, 02:00
Nitin,
ur right. I agree with u. Thanks for correcting me.
_________________

Mayur

Intern
Intern
User avatar
Joined: 24 May 2004
Posts: 3
Location: Kharagpur , India
Followers: 0

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 24 May 2004, 02:16
Always Welcome .. :P
regards
nitin
Senior Manager
Senior Manager
User avatar
Joined: 06 Dec 2003
Posts: 366
Location: India
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: Probability [#permalink] New post 24 May 2004, 05:29
Mayur wrote:
Please solve this:-

13 persons take their places at a round table. What is the probability that 2 particular persons do not sit together?


Mayur, The type of question in which it asks you to count no of ways in which particular things do not go together. It would surely help you to follow this procedure.
Count the total no results - 12! in this case
If we keep two things together - 2! * 11!

So the no of ways in which 2 particular persons do not sit together
= 12! - ( 2!*11! )

So the probability = [12! - (2 * 11!)] / 12! = 1- (1/6) = 5/6

Hope this should help,

Dharmin
Senior Manager
Senior Manager
User avatar
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

Kudos [?]: 6 [0], given: 0

Re: Probability [#permalink] New post 25 May 2004, 17:48
Dharmin wrote:
Mayur wrote:
Please solve this:-

13 persons take their places at a round table. What is the probability that 2 particular persons do not sit together?


Mayur, The type of question in which it asks you to count no of ways in which particular things do not go together. It would surely help you to follow this procedure.
Count the total no results - 12! in this case
If we keep two things together - 2! * 11!

So the no of ways in which 2 particular persons do not sit together
= 12! - ( 2!*11! )

So the probability = [12! - (2 * 11!)] / 12! = 1- (1/6) = 5/6

Hope this should help,

Dharmin


Hey Dharmin,
Trying to understand the logic here,
Why are you multiplying 2! by 11!?
Manager
Manager
avatar
Joined: 24 Jan 2004
Posts: 87
Location: Valsad
Followers: 1

Kudos [?]: 0 [0], given: 0

 [#permalink] New post 26 May 2004, 00:32
The two particular people can be arranged among themselves in 2! ways.
Tghat's the logic.
_________________

Mayur

Senior Manager
Senior Manager
User avatar
Joined: 06 Dec 2003
Posts: 366
Location: India
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: Probability [#permalink] New post 26 May 2004, 09:59
lastochka wrote:

Hey Dharmin,
Trying to understand the logic here,
Why are you multiplying 2! by 11!?


2! is there to arrange the 2 people within themselves, who are going to adhere to each other. They can be placed at either side of each other.
more over, 12 people are being seated at circular table which leads to 11! there

Hope, this would help
Cheers, Dharmin
GMAT Club Legend
GMAT Club Legend
avatar
Joined: 15 Dec 2003
Posts: 4318
Followers: 23

Kudos [?]: 172 [0], given: 0

 [#permalink] New post 26 May 2004, 14:07
Say the particular individuals are X and Y and every other person is M

When you have a round table, you have to fix one person down: Let's say we fix the first M down. We are then left with (X - Y) and 10 other M's to sit down. The scheme can be represented as follows:

(X - Y) - M - M - M - M - M - M - M - M - M - M

(X - Y) can be interchanged 2! ways. It could be (X - Y) or (Y - X).

(X - Y) or (Y - X) can be seen as a single unit in and of itself just as any other M. Hence, (X -Y) along with other 10 M's can be seated 11! ways

Round table arrangement total possible outcomes: (n-1)! = (13-1)!
Unfavorable outcomes when 2 particular individuals are arranged next to each other: 2!*11!
Favorable outcomes: (12! - 2!*11!)
When we talk about probability, we have to further divide above favorable outcomes by the total outcomes: (12! - 2!*11!) / 12! = 1 - 1/6 = 5/6
_________________

Best Regards,

Paul

Senior Manager
Senior Manager
User avatar
Joined: 07 Oct 2003
Posts: 358
Location: Manhattan
Followers: 2

Kudos [?]: 6 [0], given: 0

 [#permalink] New post 26 May 2004, 17:48
appreciate the explanation guys
Intern
Intern
User avatar
Joined: 13 May 2004
Posts: 21
Location: USA
Followers: 0

Kudos [?]: 0 [0], given: 0

hi [#permalink] New post 26 May 2004, 18:59
hi,

i read this formula on another site and it might help some people.

the number of ways to arrange n unlike object when clockwise and counterclockwise arrangements are different is (n-1)!. when clockwise and counterclockwise are the same then the number of ways is 1/2(n-1)!.

i'm not sure if this will work for this problem. let me know if anyone gets 15 million. :-D
hi   [#permalink] 26 May 2004, 18:59
    Similar topics Author Replies Last post
Similar
Topics:
probability zainab 4 08 Dec 2005, 05:14
Probability Priti 8 19 Aug 2005, 11:12
Probability mirhaque 9 29 Jun 2005, 18:06
Probability peterpan 4 18 Dec 2004, 19:21
Probability oxon 5 07 Nov 2004, 12:13
Display posts from previous: Sort by

Probability

  Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Privacy Policy| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.