Question

There are n different object 1, 2, 3, ... n distributed at random in n boxes $A_1,A_2,A_3,\cdots A_N.$ Find the probability that two objects are placed in the boxes corresponding to their number.

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Let $E_i$ be the event that ith object is placed in ithe box,

$P\left(E_i\right)=1.\frac{(n-1)}{n!}=\frac{1}{n},i=1,2\cdots n.$

Now the probability that two boxes contain right objects.$P(E_i\cap E_j)=1.\frac{(n-2)!}{n!}$

and these two objects can be out of n objects in ${}^n{C}_2$ways.

So the probability that two objects are in right boxes.${=}^n{C}_2\frac{(n-2)!}{n!}=\frac{n!}{2!(n-2)!}.\frac{(n-2)}{n!}=\frac{1}{2}.$