$A$ and $B$ are two events such that $P(A)=0.58$, $P(B)=0.32$ and $P(A \cap B)=0.28$. Then the probability that neither $A$ nor $B$ occurs is

Question

Aicharya · Accepted Answer

$\begin{aligned} & P(\bar{A} \cap \bar{B})=P(A \cup B)=1-P(A \cup B) \ & =1-\{P(A)+P(B)-P(A \cap B)\} \ & =1-\{0.58+0.32-0.28\}=1-0.62=0.38\end{aligned}$

$A$ and $B$ are two events such that $P(A)=0.58$, $P(B)=0.32$ and $P(A \cap B)=0.28$. Then the probability…

Solution