Question

A die is rolled $5$ times. What is the probability that $1$ occurs twice, and $2,3$ and $4$ occur once each, and $5$ and $6$ do not occur (round off to fourth decimal place).

Answer 1

Step 1: Find the probability that $\mathbf{1}$ occurs twice, and $\mathbf{2}\mathbf{,}\mathbf{}\mathbf{3}$ and $\mathbf{4}$ occur once each, and $\mathbf{5}$ and $\mathbf{6}$ do not occur.

Given that, the dice are rolled $5$ times.

The probability that $1$ occurs twice, and $2,3$ and $4$ occur once each, and $5$ and $6$ do not occur.

Use the multinomial distribution formula:

$\Rightarrow \mathrm{P}=\frac{\mathrm{n}!}{\left({\mathrm{n}}_1!\right)\left({\mathrm{n}}_2!\right)\dots \left({\mathrm{n}}_{\mathrm{x}}!\right)}{\left({\mathrm{P}}_1\right)}^{{\mathrm{n}}_1}\times {\left({\mathrm{P}}_2\right)}^{{\mathrm{n}}_2}\dots {\left({\mathrm{P}}_{\mathrm{x}}\right)}^{{\mathrm{n}}_{\mathrm{x}}}$.

Here, $\mathrm{P}$ is the probability and ${\mathrm{n}}_1,{\mathrm{n}}_2..{\mathrm{n}}_{\mathrm{x}}$ are the number of occurrences and $\mathrm{n}!$ denoted by the total number of outcomes.

So, substitute the total number of times, the dice rolled $\mathrm{n}!=5$ in the above formula:

$\Rightarrow \mathrm{P}=\frac{5!}{\left({\mathrm{n}}_1!\right)\left({\mathrm{n}}_2!\right)\dots \left({\mathrm{n}}_{\mathrm{x}}!\right)}{\left({\mathrm{P}}_1\right)}^{{\mathrm{n}}_1}\times {\left({\mathrm{P}}_2\right)}^{{\mathrm{n}}_2}\dots {\left({\mathrm{P}}_{\mathrm{x}}\right)}^{{\mathrm{n}}_{\mathrm{x}}}$

Now,

The probability that $1$ occurs twice in $6$ sided dice:

$\Rightarrow {\mathrm{P}}_1={\left(\frac{1}{6}\right)}^2$. Here ${\mathrm{n}}_1=2$

The probability that $2$ occurs once in $6$ sided dice:

$\Rightarrow {\mathrm{P}}_2={\left(\frac{1}{6}\right)}^1$. Here ${\mathrm{n}}_2=1$

The probability that $3$ occurs once in $6$ sided dice:

$\Rightarrow {\mathrm{P}}_3={\left(\frac{1}{6}\right)}^1$. Here ${\mathrm{n}}_3=1$

The probability that $4$ occurs once in $6$ sided dice:

$\Rightarrow {\mathrm{P}}_4={\left(\frac{1}{6}\right)}^1$. Here ${\mathrm{n}}_4=1$

Step 2: Substitute the values and find the final probability.

Substitute all the known values of ${\mathrm{P}}_1,{\mathrm{P}}_2,{\mathrm{P}}_3,{\mathrm{P}}_4$ and ${\mathrm{n}}_1,{\mathrm{n}}_2,{\mathrm{n}}_3,{\mathrm{n}}_4$ in the multinomial distribution formula:

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}=\frac{5!}{2!\times 1!\times 1!\times 1!}{\left(\frac{1}{6}\right)}^2\times {\left(\frac{1}{6}\right)}^1\times {\left(\frac{1}{6}\right)}^1\times {\left(\frac{1}{6}\right)}^1\\ & \Rightarrow & \mathrm{P}=\frac{5\times 4\times 3\times 2\times 1!}{2\times 1!\times 1!\times 1!\times 1!}{\left(\frac{1}{6}\right)}^5\\ & \Rightarrow & \mathrm{P}=\frac{120}{2}\times {\left(\frac{1}{6}\right)}^5\\ & \Rightarrow & \mathrm{P}=\frac{5}{648}\\ & \Rightarrow & \mathrm{P}=0.0077160493\end{array}$

By rounding off to the fourth decimal place:

$\begin{array}{rcl}& \Rightarrow & \mathrm{P}=0.0077\end{array}$

Hence, the probability that $\mathbf{1}$ occurs twice, and $\mathbf{2}\mathbf{,}\mathbf{}\mathbf{3}$ and $\mathbf{4}$ occur once each, and $\mathbf{5}$ and $\mathbf{6}$ do not occur is $\mathbf{0}\mathbf{.}\mathbf{0077}.$