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If the occurrence of an event $A$ implies that occurrence of an event $B$, the $P\left(A^{C} \cap B^{C}\right)$ is equal to
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Verified Answer
The correct answer is:
$P\left(B^{C}\right)$
$P\left(A^{C} \cap B^{C}\right)=P\left((A \cup B)^{C}\right)=1-P(A \cup B)$
Since, probability of occurrence of an event A implies the occurrence of event $B$,
$$
\begin{aligned}
&\therefore A \subset B \Rightarrow A \cup B=B \\
&\Rightarrow P\left(A^{C} \cap B^{C}\right)=1-P(B)=P\left(B^{C}\right)
\end{aligned}
$$
Since, probability of occurrence of an event A implies the occurrence of event $B$,
$$
\begin{aligned}
&\therefore A \subset B \Rightarrow A \cup B=B \\
&\Rightarrow P\left(A^{C} \cap B^{C}\right)=1-P(B)=P\left(B^{C}\right)
\end{aligned}
$$
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