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:
1.$\dfrac{{17}}{{36}}$
2. $\dfrac{4}{9}$
3. $\dfrac{5}{{12}}$
4. $\dfrac{1}{2}$

Answer 1

Hint: In this question, we are given the condition that two dice are rolled and the numbers on the six faces of dice are $1,2,3,5,7$ and $11$. We have to calculate the probability of those pairs whose sum of the numbers on the top faces is equal to less than $8$. Count them and apply probability formula \[Probability = \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]. Solve further.

Formula Used:
Probability formula –
\[Probability = \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]

Complete step by step Solution:
Given that,
Two dice are rolled, and the dices have six faces numbered $1,2,3,5,7$ and $11$.
Sample space of two rolled dices will be

${I}/ {II}$	$1$	$2$	$3$	$5$	$7$	$11$
$1$	$\left( {1,1} \right)$	$\left( {1,2} \right)$	$\left( {1,3} \right)$	$\left( {1,5} \right)$	$\left( {1,7} \right)$	$\left( {1,11} \right)$
$2$	$\left( {2,1} \right)$	$\left( {2,2} \right)$	$\left( {2,3} \right)$	$\left( {2,5} \right)$	$\left( {2,7} \right)$	$\left( {2,11} \right)$
$3$	$\left( {3,1} \right)$	$\left( {3,2} \right)$	$\left( {3,3} \right)$	$\left( {3,5} \right)$	$\left( {3,7} \right)$	$\left( {3,11} \right)$
$5$	$\left( {5,1} \right)$	$\left( {5,2} \right)$	$\left( {5,3} \right)$	$\left( {5,5} \right)$	$\left( {5,7} \right)$	$\left( {5,11} \right)$
$7$	$\left( {7,1} \right)$	$\left( {7,2} \right)$	$\left( {7,3} \right)$	$\left( {7,5} \right)$	$\left( {7,7} \right)$	$\left( {7,11} \right)$
$11$	$\left( {11,1} \right)$	$\left( {11,2} \right)$	$\left( {11,3} \right)$	$\left( {11,5} \right)$	$\left( {11,7} \right)$	$\left( {11,11} \right)$

Number of possible outcomes$ = 36$
Elements whose sum of the numbers on the top faces is less than or equal to $8$ are $\left( {1,1} \right),\left( {1,2} \right),\left( {1,3} \right),\left( {1,5} \right),\left( {1,7} \right),\left( {2,1} \right),\left( {2,2} \right),\left( {2,3} \right),\left( {2,5} \right),\left( {3,1} \right),\left( {3,2} \right),\left( {3,3} \right),\left( {3,5} \right),\left( {5,1} \right),\left( {5,2} \right),\left( {5,3} \right),\left( {7,1} \right)$
Therefore, Number of favourable outcomes $ = 17$
Using Probability formula,
\[Probability = \dfrac{{Number{\text{ }}of{\text{ }}favourable{\text{ }}outcomes}}{{Number{\text{ }}of{\text{ }}possible{\text{ }}outcomes}}\]
\[Probability = \dfrac{{17}}{{36}}\]

Hence, the correct option is 1.

Note: The key concept involved in solving this problem is a good knowledge of Probability. Students must remember that the probability of an event occurring is calculated by dividing the number of favorable outcomes by the total number of possible outcomes. A coin flip is the most basic example. When a coin is flipped, there are only two possible outcomes: heads or tails. In this question, the numbers on the faces are changed which means the sample space will be different. Don’t assume the favorable outcomes from the sample space of dice whose faces are $1,2,3,4,5,6$. Make a separate table for a better understanding.