Question

There are $n$ persons sitting in a row. Two of them are selected at random. The probability that the two selected persons are not together, is

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

Answer 1

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

Total persons= $n$

Total number of wage to select two at a time = ${}^n{C}_2$

No. of ways to select two together person= $n-1$

Probability of two together person,

= $\frac{n-1}{{}^n{C}_2}$

= $\frac{n-1}{n!/(n-2)!2!}$

= $\frac{(n-1)(n-2)!2}{n(n-1)(n-2)!}$

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