Question

There are $\mathrm{n}$ different object $1,2,3,.\dots ,\mathrm{n}$ distributed at random in $\mathrm{n}$ places marked $1,2,3,\dots ,\mathrm{n}$. The probability that at least three of the objects occupy places corresponding to their number is

• A. $\frac{1}{6}$
• B. $\frac{5}{6}$
• C. $\frac{1}{3}$
• D. None of these.

Answer 1

The correct option is A

$\frac{1}{6}$

Find the required probability

Let ${\mathrm{A}}_{\mathrm{i}}$ be the event that the ${\mathrm{i}}^{\mathrm{th}}$ object will occupy the ${\mathrm{n}}^{\mathrm{th}}$ place

For $\mathrm{i}<\mathrm{j}<\mathrm{k}$, then, $\mathrm{P}({\mathrm{A}}_{\mathrm{i}}\cap {\mathrm{A}}_{\mathrm{j}}\cap {\mathrm{A}}_{\mathrm{k}})=\frac{(\mathrm{n}-3)!}{\mathrm{n}!}$

Out of $“\mathrm{n}”$ numbers, we can choose $3$ places in ${}^{\mathrm{n}}\mathrm{C}_3$ ways.

Hence, the required probability $={}^{\mathrm{n}}\mathrm{C}_3.\left[\frac{(\mathrm{n}-3)!}{\mathrm{n}!}\right]$

$=\frac{\mathrm{n}!}{3!(\mathrm{n}-3)!}\left[\frac{(\mathrm{n}-3)!}{\mathrm{n}!}\right]$

$=\frac{1}{6}$

Hence option $\left(\mathrm{A}\right)$, $\frac{1}{6}$ is the correct answer.