Suppose n ge 3 persons are sitting in a row. | vedclass

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(A) Total number of ways to select $2$ persons out of $n$ is given by $^nC_2 = \frac{n(n-1)}{2}$.To find the number of ways they are together,we treat

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$1 - \frac{2}{n}$

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(A) Total number of ways to select $2$ persons out of $n$ is given by $^nC_2 = \frac{n(n-1)}{2}$.To find the number of ways they are together,we treat the $2$ adjacent persons as a single unit. There are $(n-1)$ such pairs in a row of $n$ persons.So,the number of ways they are together is $(n-1)$.The probability that they are together is $P(	ext{together}) = \frac{n-1}{^nC_2} = \frac{n-1}{\frac{n(n-1)}{2}} = \frac{2}{n}$.The probability that they are not together is $P(	ext{not together}) = 1 - P(	ext{together}) = 1 - \frac{2}{n}$.

Suppose $n \ge 3$ persons are sitting in a row. Two of them are selected at random. The probability that they are not together is

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