Solved – Why $P(A|B) \neq P(A | B,C) + P(A | B, \neg C)$

bayesianprobability

I suppose that

$$P(A|B) = P(A | B,C) * P(C) + P(A|B,\neg C) * P(\neg C)$$

is correct, whereas

$$P(A|B) = P(A | B,C) + P(A|B,\neg C) $$

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?

Best 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:

$P(A | B) = P(A | B, C) + P(A | B, \neg C) = 1 + 1 = 2$

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..

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

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