I suppose that

P(A|B)=P(A|B,C)∗P(C)+P(A|B,¬C)∗P(¬C)

is correct, whereas

P(A|B)=P(A|B,C)+P(A|B,¬C)

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?

**Answer**

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,¬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.~~.

That brings you closer to your first equation, but the weights are not completely right. See A. Rex’ comment for the correct weights.

**Attribution***Source : Link , Question Author : zell , Answer Author : Dennis Soemers*