Question

Two fair dice are rolled. Then the probability of getting a composite number as the sum of the face values is equal to
A. $\dfrac{7}{12}$
B. $\dfrac{5}{12}$
C. $\dfrac{1}{12}$
D. $\dfrac{3}{4}$
E. $\dfrac{2}{3}$

Answer 1

Hint: To solve this question, we should know about the sample space and the favourable cases for the required event. The sample space when two dice are rolled is given by $S=\left\{ \left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)......\left( 6,4 \right)\left( 6,5 \right)\left( 6,6 \right) \right\}$ where the elements in the ordered pair are the numbers appeared on the two dice. We can write that the total number of elements in the sample space is $n\left( S \right)=6\times 6=36$. We should write the favourable cases out of the sample space whose sum is a composite number and the probability is given by ${{p}_{required}}=\dfrac{n\left( favorable \right)}{n\left( S \right)}$.

Complete step by step answer:
We are given a sample space that has two dice. When two dice are rolled, the sample space is as follows
$S=\left\{ \left( 1,1 \right)\left( 1,2 \right)\left( 1,3 \right)......\left( 6,4 \right)\left( 6,5 \right)\left( 6,6 \right) \right\}$ where $\left( x,y \right)$ denotes that the first dice is rolled the number-x and the second dice is rolled the number-y.
There are 6 possibilities each for the first dice and second dice. So, the total number of elements is given by $n\left( S \right)=6\times 6=36$
The value of the sum ranges from 2 to 12.
Let us consider the sum of the two numbers is S and an event E which has the value of S as a composite number.
The favorable cases for the event E are
$S=4,6,8,9,10,12$ as these are the composite numbers in the range of the sum.
Let us consider $S=4$. The ordered pairs for which we get the sum as 4 are
$\left( 1,3 \right),\left( 3,1 \right),\left( 2,2 \right)$
So, $N\left( S=4 \right)=3$
Let us consider $S=6$. The ordered pairs for which we get the sum as 4 are
$\left( 1,5 \right),\left( 5,1 \right),\left( 2,4 \right),\left( 4,2 \right)\left( 3,3 \right)$
So, $N\left( S=6 \right)=5$
Let us consider $S=8$. The ordered pairs for which we get the sum as 4 are
$\left( 2,6 \right),\left( 6,2 \right),\left( 3,5 \right),\left( 5,3 \right),\left( 4,4 \right)$
So, $N\left( S=8 \right)=5$
Let us consider $S=9$. The ordered pairs for which we get the sum as 4 are
$\left( 3,6 \right),\left( 6,3 \right),\left( 4,5 \right),\left( 5,4 \right)$
So, $N\left( S=9 \right)=4$
Let us consider $S=10$. The ordered pairs for which we get the sum as 4 are
$\left( 4,6 \right),\left( 6,4 \right),\left( 5,5 \right)$
So, $N\left( S=10 \right)=3$
Let us consider $S=12$. The ordered pairs for which we get the sum as 4 are
$\left( 6,6 \right)$
So, $N\left( S=12 \right)=1$
We can write that
$\begin{align}
  & N\left( E \right)=N\left( S=4 \right)+N\left( S=6 \right)+N\left( S=8 \right)+N\left( S=9 \right)+N\left( S=10 \right)+N\left( S=12 \right) \\
 & N\left( E \right)=3+5+5+4+3+1=21 \\
\end{align}$

We can write the required probability as
$P\left( E \right)=\dfrac{N\left( E \right)}{N\left( S \right)}=\dfrac{21}{36}=\dfrac{7}{12}$
$\therefore $The required probability is $\dfrac{7}{12}$. The answer is option-A.


Note:
There is a shortcut to get the number of ordered pairs for the sum when two dice are rolled. The below figure has the sum as the x-axis and the number of ordered pairs on the y-axis. The number starts at 1 for the sum=2 and increases one by one until it reaches the value of 6 at sum=7. Again, it decreases one by one to reach the value of 1 when sum=12. Using this method, we can easily write the number of ordered pairs instead of writing the ordered pairs for every question.