Question

Suppose $n(\ge 3)$ persons are sitting in 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 $2$ persons out of $n$ is ${}^n{C}_2$
After selecting two persons when the remaining $n-2$ persons sit in a row $(n-1)$ places are created between them in which $2$ persons can be arranged in ${}^{n-1}{C}_2.2!$ ways.
So, the required probability $=\frac{{}^{n-1}{C}_2.2!}{{}^n{C}_2}=\frac{n-2}{n}=1-\frac{2}{n}$