Question

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, then the probability that no ball is marked with the digit $0$ is:

• A. $\frac{1}{{10}^4}$
• B. $\frac{9}{{10}^3}$
• C. ${\left(\frac{9}{10}\right)}^4$
• D. $\frac{27}{{10}^4}$

Answer 1

The correct option is C ${\left(\frac{9}{10}\right)}^4$

Let us assume that number of balls with digit marked as zero among the experiment of $4$ balls drawn simultaneously be $X$
As we can see that the balls are drawn with replacement, thus, the trial is a Bernoulli trial.
Probability of a ball drawn from the bag to be marked as digit $0=\frac{1}{10}$
It can be clearly observed that $X$ has a binomial distribution with $n=4$ and $p=\frac{1}{10}$
Thus, $q=1-p=1-\frac{1}{10}=\frac{9}{10}$
Thus, $P(X=x){=}^n{C}_x{q}^{n-x}{p}^x,$ where $x=0,1,2,...n$
${=}^4{C}_x{\left(\frac{9}{10}\right)}^{4-x}{\left(\frac{1}{10}\right)}^x$
Probability of no ball marked with zero among the $4$ balls$=P(X=0){=}^4{C}_0{\left(\frac{9}{10}\right)}^{4-0}{\left(\frac{1}{10}\right)}^0$
$=1\times {\left(\frac{9}{10}\right)}^4={\left(\frac{9}{10}\right)}^4$