Question

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

• A. $1-\frac{2}{n}$
• B. $\frac{2}{n-1}$
• C. $1-\frac{1}{n}$
• D. None of these

Answer 1

The correct option is A $1-\frac{2}{n}$

The total number of ways of choosing two persons out of $n$ is ${}^n{C}_2=\frac{n(n-1)}{2}$
$\therefore$ The number of ways in which two chosen persons are together is $(n-1)$
$\therefore$ The number of favorable ways $={}^n{C}_2-(n-1)$
$=\frac{n(n-1)}{2}-(n-1)=\frac{(n-1)(n-2)}{2}$

$\therefore$ The required probability $=\frac{\frac{(n-1)(n-2)}{2}}{\frac{n(n-1)}{2}}=\frac{n-2}{n}=1-\frac{2}{n}$