Q. An experiment yields 3 mutually exclusive and exhaustive events $A,B,C$.If $P(A) = 2P(B) = 3P(C),$ then $P(A)$ is equal to
A) $\frac{1}{11}$
B) $\frac{2}{11}$
C) $\frac{3}{11}$
D) $\frac{6}{11}$
Explanation
(d)
Clearly, $P(A \cup B \cup C) = 1$
$\Rightarrow \ P(A) + P(B) + P(C) = 1$
$\Rightarrow P(A) + \frac{1}{2}P(A) + \frac{1}{3}P(A) = 1$
$\Rightarrow \ \frac{11}{6}P(A) = 1$
$\Rightarrow P(A) = \frac{6}{11}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.