Question

A die is thrown. If $\mathrm{A}$ is the event that the number obtained is greater than $3$ and $\mathrm{B}$ is the event that the number obtained is less than$5$. Then, $\mathrm{P}(\mathrm{A}\cup \mathrm{B})$ is

• A. $\frac{2}{5}$
• B. $\frac{3}{5}$
• C. $0$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for correct option:

Step1. Define possible events

Sample space $=\left\{1,2,3,4,5,6\right\}$

Number of sample space $=6$

Let the events be defined as

The number obtained is greater than $3=\mathrm{A}$

The number obtained is less than $5=\mathrm{B}$

$\therefore \mathrm{P}\left(\mathrm{A}\right)=\frac{3}{6}$

$\mathrm{P}\left(\mathrm{B}\right)\hspace{0.17em}=\frac{4}{6}$

Step2. Calculate $\mathrm{P}(\mathrm{A}\cap \mathrm{B})$

$\mathrm{A}\cap \mathrm{B}=\left\{4\right\}$

$\therefore \mathrm{P}(\mathrm{A}\cap \mathrm{B})=\frac{1}{6}$

Step3. Calculate $\mathrm{P}(\mathrm{A}\cup \mathrm{B})$

$\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\mathrm{P}\left(\mathrm{A}\right)+\mathrm{P}\left(\mathrm{B}\right)-\mathrm{P}(\mathrm{A}\cap \mathrm{B})$

Substitute the value of $\mathrm{P}\left(\mathrm{A}\right),\mathrm{P}\left(\mathrm{B}\right)\mathrm{and}\mathrm{P}(\mathrm{A}\cap \mathrm{B})$ in the formula

$\mathrm{P}(\mathrm{A}\cup \mathrm{B})=\frac{3}{6}+\frac{4}{6}-\frac{1}{6}$

$\therefore \mathrm{P}(\mathrm{A}\cup \mathrm{B})=1$

Hence, Option(D) is the correct answer.