Q. Let $0 < P(A) < 1,\ 0 < P(B) < 1$ and $P(A \cap B) = P(A) + P(B) - P(A)P(B),$ then
A) $P\left( B \middle| A \right) = P(B) - P(A)$
B) $P\left( A^{c} \cup B^{c} \right) = P\left( A^{c} \right) + P(B^{c})$
C) $P(A \cup B)^{c} = P\left( A^{c} \right)P(B^{c})$
D) $P\left( A \middle| B \right) = P(A) + P(B^{c})$
Explanation
(c)
Since, $P(A \cap B) = P(A)P(B)$
$\Rightarrow A$ and $B$ are independent events
$\Rightarrow A^{c}$and $B^{c}$will also independent events
Hence, $P(A \cup B)^{c} = P(A^{c} \cap B^{c})$
$= P\left( A^{c} \right)P(B^{c})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.