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If $A$ and $B$ are events having probabilities, $P(A)=0.6, P(B)=0.4$ and $P(A \cap B)=0$, then probability that neither $A$ nor $B$ occurs is
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Given, $P(A)=0.6, P(B)=0.4$ and $P(A \cap B)=0$, then $P$ (neither $A$ nor $B)=P(\bar{A} \cap \bar{B})$
$$
\begin{aligned}
& =P(\overline{A \cup B})=1-P(A \cup B) \\
& =1-P(A)-P(B)+P(A \cap B) \\
& \quad \quad[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)] \\
& =1-0.6-0.4+0=1-1=0
\end{aligned}
$$
$$
\begin{aligned}
& =P(\overline{A \cup B})=1-P(A \cup B) \\
& =1-P(A)-P(B)+P(A \cap B) \\
& \quad \quad[\because P(A \cup B)=P(A)+P(B)-P(A \cap B)] \\
& =1-0.6-0.4+0=1-1=0
\end{aligned}
$$
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