Question

In a single throw of two dice, the probability of obtaining a total of $7$ or $9$ is

• A. $\frac{5}{18}$
• B. $\frac{1}{6}$
• C. $\frac{1}{9}$
• D. None of these

Answer 1

The correct option is A

$\frac{5}{18}$

Explanation for the correct option.

Find the probability.

When two dies are thrown, the sample space $S$ is

$$\begin{gathered} S=\left\{\left(1,1\right),\left(1,2\right),\left(1,3\right),\left(1,4\right),\left(1,5\right),\left(1,6\right),\left(2,1\right),\left(2,2\right),\left(2,3\right),\left(2,4\right),\left(2,5\right),\left(2,6\right), \\ \left(3,1\right),\left(3,2\right),\left(3,3\right),\left(3,4\right),\left(3,5\right),\left(3,6\right),\left(4,1\right),\left(4,2\right),\left(4,3\right),\left(4,4\right),\left(4,5\right),\left(4,6\right), \\ \left(5,1\right),\left(5,2\right),\left(5,3\right),\left(5,4\right),\left(5,5\right),\left(5,6\right),\left(6,1\right),\left(6,2\right),\left(6,3\right),\left(6,4\right),\left(6,5\right),\left(6,6\right)\right\} \end{gathered}$$

So the total outcomes is $n\left(S\right)=36$.

Let $A$ be the event of getting a total of $7$ or $9$.

$A=\left\{\left(1,6\right),\left(2,5\right),\left(3,4\right),\left(3,6\right),\left(4,3\right),\left(4,5\right),\left(5,2\right),\left(5,4\right),\left(6,1\right),\left(6,3\right)\right\}$

Thus favorable outcomes are: $n\left(A\right)=10$.

And so the required probability is given as:

$\begin{array}{rcl}P\left(A\right)& =& \frac{n\left(A\right)}{n\left(S\right)}\\ & =& \frac{10}{36}\\ & =& \frac{5}{18}\end{array}$

So the probability of obtaining a total of $7$ or $9$ is $\frac{5}{18}$.

Hence, the correct option is A.