Question

Two dice are rolled.

If both dices have six faces numbered $1,2,3,5,7$ and $11$, then the probability that the sum of the numbers on the top faces is less than or equal to $8$ is

• A. $\frac{17}{36}$
• B. $\frac{4}{9}$
• C. $\frac{5}{12}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A

$\frac{17}{36}$

Explanation for the correct answer:

Two dice are rolled.

The total number of outcomes in sample space $n\left(S\right)$ are given as

$\Rightarrow$ $\mathrm{n}\left(\mathrm{S}\right)=6^2=36$

The numbers on the faces of the dice are $1,2,3,5,7,11$

The possible scenarios where the sum of the digits is less than or equal to eight is with the combination of $\left(1,2\right),\left(1,3\right),\left(1,5\right),\left(1,7\right),\left(2,3\right),\left(2,5\right),\left(3,5\right)$

Hence the ordered pairs with the combination of these digits are $\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,5\right),\left(1,7\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,5\right),\left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,5\right),\left(5,1\right)\left(5,2\right)\left(5,3\right),\left(7,1\right)\right\}$

Number of ordered pairs $=17$

Hence the $\mathrm{required}\mathrm{probability}=\frac{\mathrm{Number}\mathrm{of}\mathrm{ordered}\mathrm{pairs}}{\mathrm{n}\left(\mathrm{S}\right)}$

$\mathrm{required}\mathrm{probability}=\frac{17}{36}$

Hence, the probability of the sum of the digits appearing on the faces of the dice less than or equal to eight is $\frac{17}{36}$.

Hence, option $\left(\mathrm{A}\right)$ is the correct answer.