Question

If four dice are thrown together, the probability that the sum of the number appearing on them is 13 is

• A. $\frac{35}{324}$
• B. $\frac{5}{216}$
• C. $\frac{11}{216}$
• D. $\frac{11}{432}$

Answer 1

The correct option is A

$\frac{35}{324}$

Explanation for the correct answer:

Step 1: Calculate total outcomes

The total number of outcomes in throwing $4$ dice is,

$$\begin{gathered} =6^4 \\ =1296 \end{gathered}$$

Step 2: Using Permutation to find the sum of numbers appearing is $13,$

$\mathit{n}{\mathit{P}}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}}$ where, ${\mathrm{nP}}_{\mathrm{r}}=$permutation, $\mathrm{n}=$the total number of objects, $\mathrm{r}=$number of objects selected

Permutation of $(1,1,5,6)=\frac{4!}{2!}=12$

Permutation of $(1,2,4,6)=4!=24$

Permutation of $(1,3,3,6)=\frac{4!}{2!}=12$

Permutation of $(1,2,5,5)=12$

Permutation of $(1,3,5,4)=24$

Permutation of $(2,2,6,3)=12$

Permutation of $(2,2,5,4)=12$

Permutation of $(3,3,2,5)=12$

Permutation of $(3,3,3,4)=4$

Permutation of $(4,4,4,1)=4$

Permutation of $(4,4,3,2)=12$

Therefore, the required probability

$$\begin{gathered} =\frac{12+24+12+12+24+12+12+12+4+4+12}{1296} \\ =\frac{140}{1296} \\ =\frac{35}{324} \end{gathered}$$

Hence, the correct answer is an option (A).