Q. Let $A$ and $B$ two events such that $P(\overline{A \cup B}) = \frac{1}{6},P(A \cap B) = \frac{1}{4}$ and $P\left( \overline{A} \right) = \frac{1}{4,},$where $\overline{A}$ stands for complement of event $A$. Then, events $A$ and $B$ are
A) Mutually exclusive and independent
B) Independent but not equally likely
C) Equally likely but not independent
D) Equally likely and mutually exclusive
Explanation
(b)
Given, $P(\overline{A \cup B}) = \frac{1}{6}$
$\Rightarrow 1 - P(A \cup B) = \frac{1}{6}$
$\Rightarrow 1 - P(A) - P(B) + P(A \cap B = \frac{1}{6}$
$\Rightarrow P\left( \overline{A} \right) - P(B) + P(A \cap B) = \frac{1}{6}$
$\Rightarrow \frac{1}{4} - P(B) + \frac{1}{4} = \frac{1}{6}$
$\Rightarrow P(B) = \frac{1}{3}$
and $P(A) = 1 - P\left( \overline{A} \right) = 1 - \frac{1}{4} = \frac{3}{4}$
Since, $P(A \cap B) = P(A).P(B)$ so the events $A$ and $B$ are independent events but not equally likely
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.