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# Probability

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23 May 2004, 23:02
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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

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24 May 2004, 01:43
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
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24 May 2004, 01:56
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
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Mayur

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24 May 2004, 02:30
[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
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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
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24 May 2004, 03:00
Nitin,
ur right. I agree with u. Thanks for correcting me.
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Mayur

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24 May 2004, 03:16
Always Welcome ..
regards
nitin
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24 May 2004, 06:29
Mayur wrote:

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.
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
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25 May 2004, 18:48
Dharmin wrote:
Mayur wrote:

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.
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!?
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26 May 2004, 01:32
The two particular people can be arranged among themselves in 2! ways.
Tghat's the logic.
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Mayur

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26 May 2004, 10:59
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
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26 May 2004, 15:07
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
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Paul

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26 May 2004, 18:48
appreciate the explanation guys
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26 May 2004, 19:59
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.
hi   [#permalink] 26 May 2004, 19:59
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