Question

A bag contains 10 balls, each marked with one of the digits from 0 to 9. If four balls are drawn successively with replacement from the bag, what is the probability that none is marked with the digit 0?

Answer 1

Let X be the number of balls marked with the digit 0 when 4 balls are drawn successfully with replacement.

As this is with replacement, X follows a binomial distribution with n = 4;
$p\mathit{}=\text{p}\mathrm{robabilty}\mathrm{that}\mathrm{a}\mathrm{ball}\mathrm{randomly}\mathrm{drawn}\mathrm{bears}\mathrm{digit}0=\frac{1}{10};q=1-p=\frac{9}{10};$

$$\begin{gathered} P(X=\mathrm{r})={}^{4}C_r{\left(\frac{1}{10}\right)}^r{\left(\frac{9}{10}\right)}^{4-r} \\ P(\mathrm{none}\mathrm{bears}\text{the}\mathrm{digit}0)=\mathit{}P(X=0) \\ ={}^{4}C_0{\left(\frac{1}{10}\right)}^0{\left(\frac{9}{10}\right)}^{4-0} \\ ={\left(\frac{9}{10}\right)}^4 \end{gathered}$$