Probability of P(A|B) given P(A), P(B)

Question

Given $P(A)=0.9$ and $P(B)=0.8$.

How to prove that $P(A|B)\ge 0.875$ ?

Answer 1

$$P(A|B)=\frac{P(A\cap B)}{ P(B)}$$ $$=\frac{P(A)+P(B)-P(A\cup B)}{ P(B)}$$ $$=\frac{0.9+0.8-P(A\cup B)}{ 0.8}$$ $$\ge \frac{0.9+0.8-1}{ 0.8}$$ $$\ge 0.875$$

Answer 2

Because $\Pr(A\cup B)=\Pr(A)+\Pr(B)-\Pr(A\cap B)$, and $\Pr(A\cup B)\le 1$, we have $\Pr(A\cap B)\ge 0.7$.

Now use the fact that $\Pr(A\cap B)=\Pr(A\mid B)\Pr(B)$. We get $\Pr(A\mid B)\ge \frac{0.7}{0.8}$.