Question

There are $\mathrm{m}$ persons sitting in a row. Two of these are selected at random. The probability that the two selected persons are not together is

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

Answer 1

The correct option is B

$1-\left(\frac{2}{\mathrm{m}}\right)$

Find the required probability.

Given : $\mathrm{m}$ persons are sitting in a row.

The number of ways,

selecting two persons out of m persons $={}^{\mathrm{m}}\mathrm{C}_2$

$=\frac{\mathrm{m}\left(\mathrm{m}-1\right)}{2}$

Two selected persons are not together $=\frac{(\mathrm{m}-1)(\mathrm{m}-2)}{2}$

Thus, the required probability is ,

$$\begin{gathered} =\frac{{\displaystyle \frac{(\mathrm{m}-1)(\mathrm{m}-2)}{2}}}{{\displaystyle \frac{\mathrm{m}(\mathrm{m}-1)}{2}}} \\ =\frac{(\mathrm{m}-2)}{\mathrm{m}} \\ =\left(1-\frac{2}{\mathrm{m}}\right) \end{gathered}$$

Hence option $\left(\mathrm{B}\right)$, $1-\left(\frac{2}{\mathrm{m}}\right)$ is the correct answer.