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What is the probability that two persons selected from a row

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Senior Manager
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What is the probability that two persons selected from a row [#permalink] New post 07 Mar 2004, 13:03
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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
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 [#permalink] New post 07 Mar 2004, 13:09
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
Senior Manager
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 [#permalink] New post 07 Mar 2004, 13:37
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.
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 [#permalink] New post 07 Mar 2004, 13:54
I have another explanation for this.

Assume out of n persons the person at position 1 is selected. P of doign this = 1/n Then we can select anyone other than the person sitting next to him. P of doing this = (n-2)/(n-1). The same is true for the person who is last in the row.
Net P = 2 * 1/n * (n-2)/(n-1)

Now we pick a person from the middle of the row. P of doing this is (n-2)/n
Then se can select any one who is not on either side of this person. P of doing this = (n-3)/(n-1)
Net P = (n-2)/n * (n-3)/(n-1)

Total P = 2 * 1/n * (n-2)/(n-1) + (n-2)/n * (n-3)/(n-1)

= (n-2)/( n * (n-1) ) [ 2 + n-3 ]
= (n-2)/n = 1-2/n
  [#permalink] 07 Mar 2004, 13:54
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