Question

A die is thrown. Let A be the event that the number obtained is greater than 2 and B be the event the number is less than 4, then $P(A\cup B)$ is equal to

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

Answer 1

The correct option is C $1$

Given, $A=\{3,4,5,6\}$
$B=\{1,2,3\}$
$A\cap B=\left\{3\right\}$
$\therefore P(A\cup B)=\frac{4}{6}+\frac{3}{6}-\frac{1}{6}=\frac{7-1}{6}=\frac{6}{6}=1$