# Inequality regarding Conditional Probability

I am working through Rohatgi and Saleh's "An Introduction to Probability and Statistics", and I am stuck on Exercise 1.5.1 :

Let A and B be two events such that $$PA = p_1 >0$$, $$PB = p_2 >0$$, and $$p_1 + p_2 >1$$. Show $$P(B|A) \geq 1 - \frac{1-p_2}{p_1}$$.

Here is my work so far:

$$P(B|A)=\frac{(P A \cap B)}{P(A)}$$, and $$P(A \cap B) = P(A) + P(B) -P(A \cup B)$$. Hence $$P(B|A) =\frac{P(A) + P(B)}{P(A)}-\frac{P(A \cup B)}{P(A)}$$. Since $$P(A)+P(B) >1$$ and $$P(B) \in (0,1]$$, we know $$\frac{P(A)+P(B)}{P(A)}>1$$. So I have $$P(B|A) \geq 1 - \frac{P(A \cup B}{P(A)}$$, so the only thing stopping me from finishing the proof is to show that $$P(A \cup B) \leq 1 -P(B)$$, which is evading me.

Any help is appreciated.

To show $$P(B|A) \ge 1-\frac{1-p_2}{p_1}$$
Multiplying by $$P(A)$$ everywhere, it is equivalent to show that
$$P(A \cap B) \ge P(A)-1+P(B)$$
$$1 \ge P(A \cup B)$$