# Bayesian Prior in Conditional Probabilities

I have $A$, $B$, and $C$, all are inter-dependent. The question is whether

$$\Pr(A | B \cap C) \propto \Pr(A|B)\times \Pr(C|A)$$

Specifically, I want to infer the most likely outcome of $A$ given $B$ and $C$, and I know

• $\Pr(A)$
• $\Pr(C)$
• $\Pr(A | B)$
• $\Pr(A | C)$
• $\Pr(C | A)$
• $\Pr(A \cap C)$

I don't have

• $\Pr(B)$
• $\Pr(B | A)$
• $\Pr(C | B)$
• $\Pr(B | C)$

My math fails me. Is the proportional equality above correct for dependent events? If not, what's the correct way to estimate $\Pr(A | B \cap C)$ with what I have?

By Bayes' Rule, we have $$P(A|B\cap C) = \frac{P(C|A \cap B)P(A|B)}{P(C|B)}.$$ Now, if $C$ is conditionally independent of $B$ given $A$, i.e. $$P(C|A \cap B) = P(C|A)$$ then $$P(A|B\cap C) = \frac{P(C|A \cap B)P(A|B)}{P(C|B)} = \frac{P(C|A)P(A|B)}{P(C|B)} \propto P(C|A)P(A|B)$$ where the proportionality drops $P(C|B)$.