- Event: Not getting a $0$ marked ball
- The probability of not getting a $0$ marked ball remains same because of replacement.
- Probability of not getting a zero marked ball = $\frac{9}{10}$ = $q$
- Bernauli probability distribution can be applied
$$\begin{align*} &\text{No of trials} = n = 4 \\ &\text{Probability of 0 marked} = p = \frac{1}{10} = 0.1 \\ &\text{Probability of not 0 marked} = q = \frac{9}{10} = 0.9 \\ &\text{Required probability } = \binom{4}{4}.p^{0}.q^{4} = 0.6561 \end{align*}$$