Conditional probability: Proof $P(C|A) = P(B)P(C|(A \cap B) + P(B^c)P(C|A \cap B^c)$

Question

I'm struggling with an exercise.
I have to proof that $$P(C|A) = P(B)P(C|(A \cap B) + P(B^c)P(C|A \cap B^c)$$ But I don't know where to start.This $(C|(A \cap B)$ and this $P(B^c)P(C|A \cap B^c)$ are confusing me a lot.
Thank you very much for your attention.

Answer 1

\begin{multline} P(C|A)=P(A|C)\times\frac{P(C)}{P(A)}=(P(A\cap B|C)+P(A\cap B^c|C))\times\frac{P(C)}{P(A)}\\=P(C|A\cap B)\times\frac{P(A\cap B)}{P(A)}+P(C|A\cap B^c)\times\frac{P(A\cap B^c)}{P(A)}\\ =P(C|A\cap B)\times P(B|A)+P(C|A\cap B)\times P(B^c|A) \end{multline} So what you were trying to prove is only correct when $B$ is independent of $A$, because then $P(B|A)=P(B)$ and $P(B^c|A)=P(B^c)$.

Answer 2

For visualization, not proof :

$$Red = n(C|(A\cap B)) $$

$$Blue = n(C|(A\cap \bar{B})) $$

$$ Blue + Red = n(C|A)$$

$$ \therefore \boxed{P(C|A) = P(B)P(C|(A\cap B) + P(\bar B)(C|(A\cap \bar{B})} $$