CUET Preparation Today
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: |
$(\frac{9}{10})^4$ $(\frac{1}{10})(\frac{9}{10})^3$ $(\frac{1}{10})^4$ $(\frac{1}{10})^2(\frac{9}{10})^2$ |
$(\frac{9}{10})^4$ |
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 $=\left(\frac{9}{10}\right)^{4}$ $=\frac{6561}{10000}$ |
