Question

If $10$ objects are distributed at random among $10$ persons, then the probability that at least one of them will not get anything is

• A. $\frac{10!}{{10}^{10}}$
• B. $\frac{{10}^{10}-10!}{{10}^{10}}$
• C. $\frac{1}{{10}^{10}}$
• D. $\frac{1}{10}$

Answer 1

The correct option is B $\frac{{10}^{10}-10!}{{10}^{10}}$

Since each object can be given to any one of $10$ persons. So, $10$ objects can be distributed among $10$ persons in ${10}^{10}$ ways. Thus, the total number of ways is ${10}^{10}$.
The number of ways of distribution in which each one gets only one thing is $10!.$
So, the number of ways of distribution in which at least one of them does not get anything is ${10}^{10}-10!.$
Hence, required probability is $\frac{{10}^{10}-10!}{{10}^{10}}$