A box contains 10 balls, each marked with one of the digits 0 to 9. If four balls are drawn successively with replacement from the bag, then the probability that none is marked with the digit 0 is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → $(\frac{9}{10})^4$
Total balls = $10$ (digits $0$ to $9$).
Probability of drawing a ball not marked $0 =$ $\frac{9}{10}$.
Since the draws are with replacement and four draws are made:
Required probability $=\left(\frac{9}{10}\right)^{4}$
$=\frac{6561}{10000}$