Answer
$P(A\cap B)=1$
Work Step by Step
$P(A)=0.2$
$P(B)=0.4$
$P(A|B)+P(B|A)=0.75$
By the definition of conditional probability,
$P(A|B)=\frac{P(A\cap B)}{P(B)}$
And
\[P(B|A)=\frac{P(A\cap B)}{P(A)}\]
Consider,
\[\begin{align}
& 0.75=P(A|B)+P(B|A) \\
& 0.75=\frac{P(A\cap B)}{P(B)}+\frac{P(A\cap B)}{P(A)} \\
& 0.75=P(A\cap B)\left( \frac{1}{P(B)}+\frac{1}{P(A)} \right) \\
& 0.75=P(A\cap B)\left( \frac{1}{0.4}+\frac{1}{0.2} \right) \\
& 0.75=(0.75)P(A\cap B) \\
& P(A\cap B)=1 \\
\end{align}\]
