# A bag consists of 10 balls each marked with one of the digits 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?

$\begin{array}{1 1} (\large \frac{9}{10})^4 \\ (\large \frac{3}{10})^4 \\ (\large \frac{1}{10})^4 \\ (\large \frac{7}{10})^4 \end{array}$

Toolbox:
• For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
• Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.
The experiment is to draw $n=4$ balls successively w replacement from a bag of 10 balls. It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.
The balls are marked w a number from 0 to 9. Therefore P (drawing a ball marked w a 0) $= p = \large\frac{1}{10}$
P (drawing a ball marked w number other than 0) $= q = 1- p = 1- \large\frac{1}{10} = \frac{9}{10}$
Let X be the number of balls marked with 0, X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
P (drawing a ball not marked 0) = P (X = 0) = $\large^{4}C_0\large\frac{1}{10}^0$$\times$$\large\frac{9}{10}^4 = (\frac{9}{10})^4$