In a box containing $100$ eggs, $10$ eggs are rotten. The probability that out of a sample of $5$ eggs none is rotten if the sampling is with replacement is

The correct option is D

${\left(\frac{9}{10}\right)}^5$

Explanation for the correct option.

Find the probability.

There are $100$ eggs in a box and out of which $10$ are rotten.

Probability of getting a rotten egg is:

$\begin{array}{rcl}p& =& \frac{10}{100}\\ & =& \frac{1}{10}\end{array}$

So, the probability of getting a egg which is not rotten is:

$\begin{array}{rcl}q& =& 1-p\\ & =& 1-\frac{1}{10}\\ & =& \frac{9}{10}\end{array}$

Now, let $X$ be the number of rotten eggs taken out of $5$ trials. Then the binomial distribution is given as:

$\begin{array}{rcl}\mathit{P}\left(X=x\right)& \mathbf{=}& {}^{\mathbf{n}}\mathit{C}_{\mathbf{x}}{\mathit{q}}^{\mathbf{n}\mathbf{-}\mathbf{x}}{\mathit{p}}^{\mathbf{x}}\\ & =& {}^{5}C_x{\left(\frac{9}{10}\right)}^{5-x}{\left(\frac{1}{10}\right)}^x\left[n=5,q=\frac{9}{10},p=\frac{1}{10}\right]\end{array}$

So to find the probability that none of the eggs are rotten substitute $0$ for $x$.

$\begin{array}{rcl}P\left(X=0\right)& =& {}^{5}C_0{\left(\frac{9}{10}\right)}^{5-0}{\left(\frac{1}{10}\right)}^0\\ & =& 1\times {\left(\frac{9}{10}\right)}^5\times 1\left[{}^{n}C_0=1,a^0=1\right]\\ & =& {\left(\frac{9}{10}\right)}^5\end{array}$

So the probability that none of the eggs are rotten is $\begin{array}{rcl}& & {\left(\frac{9}{10}\right)}^5\end{array}$.

Hence, the correct option is D.