Let $A$ and $B$ are two events and $P\left( A^{'} \right) = 0.3,\ P(B) = 0.4,\ P\left( A \cap B^{'} \right) = 0.5,$ then$P(A \cup B^{'})$ is

Question

Dr. Abubacker · Accepted Answer

(b)
Here, $P\left( B^{'} ight) = 1 - 0.4 = 0.6$
and $P(A) = 1 - 0.3 = 0.7$
$	herefore\ P\left( A \cup B^{'} ight) = P(A) + P\left( B^{'} ight) - P(A \cap B^{'})$
$= 0.7 + 0.6 - 0.5 = 0.8$

Q. Let $A$ and $B$ are two events and $P\left( A^{'} \right) = 0.3,\ P(B) = 0.4,\ P\left( A \cap B^{'} \right) = 0.5,$ then$P(A \cup B^{'})$ is

Explanation

Reviewed by Dr. Abubacker

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