Answer
$P((A\cap B)|{{(A\cup B)}^{C}})=0$
Work Step by Step
Suppose events A and B are such that,
$P(A\cap B)=0.1$
$P{{(A\cup B)}^{C}}=0.3$
$P(A)=0.2$
By definition of conditional probability,
$P((A\cap B)|{{(A\cup B)}^{C}})=\frac{P((A\cap B)\cap {{(A\cup B)}^{C}})}{P({{(A\cup B)}^{C}})}$
Therefore, \[(A\cap B)\cap {{(A\cup B)}^{C}}=\varnothing =0\]
$\begin{align}
& P((A\cap B)|{{(A\cup B)}^{C}})=\frac{P((A\cap B)\cap {{(A\cup B)}^{C}})}{P({{(A\cup B)}^{C}})} \\
& P((A\cap B)|{{(A\cup B)}^{C}})=\frac{0}{0.3} \\
& P((A\cap B)|{{(A\cup B)}^{C}})=0 \\
\end{align}$
Hence, $P((A\cap B)|{{(A\cup B)}^{C}})=0$
