Solution:

Assume that the number of balls with a digit marked as zero in the experiment of four balls drawn at the same time is $x$.

The trial is a Bernoulli trial because the balls are drawn with replacement, as can be seen.

Probability of a ball drawn from the bag to be marked as digit $0=1 / 10$

It can be clearly observed that $\mathrm{X}$ has a binomial distribution with $\mathrm{n}=4$ and $\mathrm{p}=1 / 10$

Thus, $q=1-p=1-1 / 10=9 / 10$

Thus, $\mathrm{P}(\mathrm{X}=\mathrm{x})={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{x}} \mathrm{q}^{\mathrm{n}-\mathrm{x}} \mathrm{p}^{\mathrm{x}}$, where $\mathrm{x}=0,1,2, \ldots \mathrm{n}$

$={ }^{4} \mathrm{C}_{\mathrm{x}}\left(\frac{9}{10}\right)^{4-\mathrm{x}}\left(\frac{1}{10}\right)^{\mathrm{x}}$

Probability of no ball marked with zero among the 4 balls $=\mathrm{P}(\mathrm{X}=0)$

$={ }^{4} C_{0}\left(\frac{9}{10}\right)^{4-0}\left(\frac{1}{10}\right)^{0}$

$={ }^{4} \mathrm{C}_{\mathrm{x}}\left(\frac{9}{10}\right)^{4}\left(\frac{1}{10}\right)^{0}$

$=1 \times\left(\frac{9}{10}\right)^{4}$

$=\left(\frac{9}{10}\right)^{4}$