Q. A die is thrown. Let $A$ be the event that the number obtained is greater than 3. Let $B$ be the event that the number obtained is less than 5. Then $P(A \cup B)$is
A) $\frac{2}{5}$
B) $\frac{3}{5}$
C) 0
D) 1
Explanation
(d)
Since, $A = \left\{ 4,5,6 \right\} and\ B = \{ 1,2,3,4\}$
$\therefore A \cap B = \{ 4\}$
$\therefore P(A \cup B) = P(A) + P(B) - P(A \cap B)$
$\Rightarrow P(A \cup B) = \frac{3}{6} + \frac{4}{6} - \frac{1}{6} = 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.