A bag contains 10 balls, each marked with one of the digits 0 to 9. If 4 balls are drawn successively with replacement from the bag, then what is the probability that none is marked with the digit 0?

$(\frac{9}{10})^4$

The correct answer is Option (2) → $(\frac{9}{10})^4$

Let E be the event 'ball marked with digit 0', then 

$p = P(E) =\frac{1}{10}$, so $q = 1-\frac{1}{10}=\frac{9}{10}$

As 4 balls are drawn successively with replacement, so there are 4 trials i.e. $n = 4$.

Thus, we have a binomial distribution with $p =\frac{1}{10},q=\frac{9}{10}$ and $n = 4$.

Required probability = P(none marked with digit 0) = P(0)

$={^4C}_0q^4=1.(\frac{9}{10})^4=(\frac{9}{10})^4$