Q. If $P(A \cup B) = 0.8\ and\ P(A \cap B) = 0.3,\ then\ P\left( A^{'} \right) + P(B^{'})$ equals to
A) 0.3
B) 0.5
C) 0.7
D) 0.9
Explanation
(d)
Now, $P\left( A^{'} \cap B^{'} \right) = P(A \cup B)'$
$= 1 - P(A \cup B) = 1 - 0.8 = 0.2$
and $P\left( A^{'} \cup B^{'} \right) = 1 - P(A \cap B) = 1 - 0.3 = 0.7$
But $P\left( A^{'} \cup B^{'} \right) = P\left( A^{'} \right) + P\left( B^{'} \right) - P(A^{'} \cap B^{'})$
$\Rightarrow 0.7 = P\left( A^{'} \right) + P\left( B^{'} \right) - 0.2$
$\Rightarrow \ P\left( A^{'} \right) + P\left( B^{'} \right) = 0.9$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.