# Why P(A|B)≠P(A|B,C)+P(A|B,¬C)P(A|B) \neq P(A | B,C) + P(A | B, \neg C)?

I suppose that

is correct, whereas

is incorrect.

However, I have got an “intuition” about the later one, that is, you consider the probability P(A | B) by splitting two cases (C or Not C). Why this intuition is wrong?

Suppose, as an easy counter example, that the probability $P(A)$ of $A$ is $1$, regardless of the value of $C$. Then, if we take the incorrect equation, we get:
$P(A | B) = P(A | B, C) + P(A | B, \neg C) = 1 + 1 = 2$
That obviously can’t be correct, a probably cannot be greater than $1$. This helps to build the intuition that you should assign a weight to each of the two cases proportional to how likely that case is, which results in the first (correct) equation..