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
$$P(A \cap B) \ge P(A)-1+P(B)$$
and this is equivalent to
$$1 \ge P(A \cup B)$$
of which we know it is true.