Question

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

• A. $2/m$
• B. $1/m$
• C. $\frac{m(m-1)}{(m+1)(m+2)}$
• D. $1-\frac{2}{m}$

Answer 1

The correct option is D $1-\frac{2}{m}$

The total number of ways of selecting two persons out of $m$ is ${}^m{C}_2=\frac{m(m-1)}{2}$
The number of ways in which the two selected person are together is ($m-1$).
Therefore, the number of ways in which the two selected persons are not together is ${}^m{C}_2(m-1)=\frac{(m-1)(m-2)}{2}$
Thus, the probability of the required event is $\frac{(m-1)(m-2)/2}{m(m-1)/2}=\frac{m-2}{m}=1-\frac{2}{m}$