If a dice is thrown twice, the probability of occurrence of $4$ at least once is

The correct option is A

$\frac{11}{36}$

Explanation for the correct option:

Step 1. Find the required probability:

Given, $n=2$

Now, Probability of getting a $4$ on a die , $p=\frac{1}{6}$

probability of not getting a $4$ on a die, $q=1-p$

$$\begin{gathered} =1-\frac{1}{6} \\ =\frac{5}{6} \end{gathered}$$

Step 2. Probability of getting $4$ at least one times:

$\begin{array}{rcl}P(X& =& 1)+P(X=2)={}^{2}C_1\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)+{}^{2}C_2{\left(\frac{1}{6}\right)}^2{\left(\frac{5}{6}\right)}^0\end{array}$ $\left[\because P(X=x)={}^{n}C_x{p}^x{q}^{n-x}\right]$ $=\frac{10}{36}+\frac{1}{36}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$=\frac{\mathbf{11}}{\mathbf{36}}$

Hence, Option ‘A’ is Correct.