Suppose I want to set an exam question on conditional probability. There will be two events (not necessarily independent) A and B, so the question will contain the usual expressions: $P(A), P(A|B), P(A \cap B)$ etc.
1) Let's say I fix the values of $P(A)$ and $P(B|A)$. What are the constraints I have to set on $P(B)$ in terms of these probabilities?
2) For fixed $P(B|A)$ and $P(B|\bar{A})$, what will now be the constraints on $P(B)$?
From Bayes' theorem: $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ so for (1) I need to figure out what would be the physical conditions for $$P(A|B)=\frac{P(A \cap B)}{P(B)}$$ If A and B are independent, then $P(A \cap B) = P(A)P(B)$, so $P(B)=P(B|A)$. I'm not sure what would be the other limit and how to proceed with 2).
$P(B)=P(A \cap B)+P(\bar{A} \cap B)$ so
$P(B|A)P(A) \le P(B) \le P(B|A)P(A) + 1- P(A)$
$\min(P(B|A), P(B|\bar{A})) \le P(B) \le \max(P(B|A), P(B|\bar{A})) $