Answer
The proof is below.
Work Step by Step
We know that,
\[\begin{align}
& P(A\cup B)=P(A)+P(B)-P(A\cap B) \\
& P(A\cup B)=a+b-P(A\cap B)\text{ } \\
\end{align}\]
Since $P(A\cup B)\le 1$. Therefore
\[\begin{align}
& P(A\cup B)=a+b-P(A\cap B)\le 1 \\
& P(A\cap B)\ge a+b-1 \\
\end{align}\]
By definition of condition probability,
$\begin{align}
& P(A|B)=\frac{P(A\cap B)}{P(B)} \\
& P(A|B)\ge \frac{a+b-1}{b} \\
\end{align}$
Hence, it is verified that $P(A|B)\ge \frac{a+b-1}{b}$.
