If ten objects are distributed at random among ten persons, the probability that at least one of them will not get any thing

$\frac{10^{10}-10!}{10^{10}}$

Since each object can be given to any one of ten persons. So, ten objects can be distributed among 10 persons in $10^{10}$ ways.

Thus, the total number of elementary events = $10^{10}$

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 any thing is $10^{10}-10!$

Hence, required probability $=\frac{10^{10}-10!}{10^{10}}$