Q. Let $E_{1},E_{2}$ be two mutually exclusive events of an experiment with $P($not$E_{2}) = 0.6 = P\left( E_{1} \cup E_{2} \right)$.Then, $P(E_{1})$is equal to
A) 0.1
B) 0.3
C) 0.4
D) 0.2
Explanation
(d)
Since, $E_{1},\ E_{2}$are mutually exclusive events, then
$P\left( E_{1} \cap E_{2} \right) = 0$
Now, $P\left( E_{1} \cup E_{2} \right) = P\left( E_{1} \right) + P\left( E_{2} \right) - P{(E}_{1\ } \cap E_{2})$
$\Rightarrow 0.6 = P\left( E_{1} \right) + 1 - P\left( \overline{E_{2}} \right) - 0$
$\Rightarrow 0.6 = P\left( E_{1} \right) + 0.4$
$\Rightarrow P\left( E_{1} \right) = 0.2$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.