What is the intuition behind $P(A|B)=P(A|CB)P(C|B)+P(A|C^{c}B)P(C^{c}|B)$?

Question

I'm told that $$P(A|B)=P(A|CB)P(C|B)+P(A|C^{c}B)P(C^{c}|B)$$

Follows from:

$$P(A)=P(A|C)P(C)+P(A|C^{c})P(C^{c})$$

But I do not see how. What is the intuition behind this result?

Answer 1

The intuition is that we are applying the simpler law of total probability formula in a "new world" where everything, absolutely everything, is conditioned on the event $B$ occuring. Notice that if we condition on $B$, we get a conditional distribution, but a conditional distribution is just another probability distribution. So the law of total probability also holds when everything is conditioned on $B$. You can make a mistake if you forget to condition everything on $B$ occuring.

A formal proof is: \begin{align} P[A|B] &= \frac{P[A, B]}{P[B]} \\ &\overset{(a)}{=} \frac{P[A, B |C] P[C] + P[A,B|C^c]P[C^c]}{P[B]}\\ &= \frac{P[A,B,C]}{P[B]} + \frac{P[A,B,C^c]}{P[B]}\\ &= \frac{P[A|B,C]P[B,C]}{P[B]} + \frac{P[A|B, C^c]P[B, C^c]}{P[B]} \\ &= P[A|B,C]\underbrace{\frac{P[B,C]}{P[B]}}_{P[C|B]} + P[A|B,C^c]\underbrace{\frac{P[B,C^c]}{P[B]}}_{P[C^c|B]} \end{align} where step (a) applies the usual law of total probability. $\Box$