Question

If a dice is thrown $5$ times, then the probability of getting $6$ exact three times is

• A. $\frac{125}{388}$
• B. $\frac{125}{3888}$
• C. $\frac{625}{23328}$
• D. $\frac{250}{2332}$

Answer 1

The correct option is B

$\frac{125}{3888}$

Explanation for the correct option:

Step 1. Find the required probability:

Given, $n=5$

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

probability of not getting a $6$ on a die, $q=1-p$

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

Let $r=3$

Step 2. Probability of getting $6$ exactly three times:

$P(r=3)={}^{5}C_3\times {\left(\frac{1}{6}\right)}^3\times {\left(\frac{5}{6}\right)}^2$ $\left[\mathbf{\because }\mathbf{P}\mathbf{\left(}\mathbf{r}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}{}^{\mathbf{n}}\mathbf{C}_{\mathbf{r}}\mathbf{\times }{\mathbf{p}}^{\mathbf{r}}\mathbf{\times }{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{r}}\right]$

$=10\times \frac{1}{216}\times \frac{25}{36}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{}\mathbf{-}\mathbf{}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$=\frac{\mathbf{125}}{\mathbf{3888}}$

Hence, Option ‘B’ is Correct.