Out of 10 persons sitting at a round table, two persons are selected at random then the probability that they are not adjacent to each other is

The number of ways of arranging 10 persons in a round table is $n\left(S\right)=9!$

Let $E$ be the event  in whcih two particular persons are not adjacent each other

Suppose that two perticular persons are taken as one bunch, they can arrange among themselves in 2 ways.

The remiaining 8 plus this bunch, total 9 things can be arranged in arround table in $2\left(8!\right)$ ways

Hence, $n\left(\overline{E}\right)=2\left(8!\right)$

Therefore, 

   $\begin{array}{rcl}P\left(E\right)& =& 1-P\left(\stackrel{⏜}{E}\right)\\ & =& 1-\frac{2\left(8!\right)}{9!}\\ & =& 1-\frac{2}{9}\\ & =& \frac{7}{9}\end{array}$