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If $P(A)=0.59, P(B)=0.30$ and $P(A \cap B)=0.21$ then $P\left(A^{\prime} \cap B^{\prime}\right)$ is equal to
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The correct answer is:
$0.32$
Given, $P(A)=0.59, P(B)=0.30$
and $P(A \cap B)=0.21$
$\begin{aligned}
P\left(A^{\prime} \cap B^{\prime}\right) &=1-P(A \cup B) \quad[\therefore \text { De Morgan's law }] \\
=& 1-[P(A)+P(B)-P(A \cap B)] \\
=& 1-[0.59+0.30-0.21] \\
=& 1-[0.89-0.21] \\
=& 1-0.68=0.32
\end{aligned}$
and $P(A \cap B)=0.21$
$\begin{aligned}
P\left(A^{\prime} \cap B^{\prime}\right) &=1-P(A \cup B) \quad[\therefore \text { De Morgan's law }] \\
=& 1-[P(A)+P(B)-P(A \cap B)] \\
=& 1-[0.59+0.30-0.21] \\
=& 1-[0.89-0.21] \\
=& 1-0.68=0.32
\end{aligned}$
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