What is the P(A|C) if we know B depends on A and C depends on B?

I would prefer $\Pr(A|C) = \Pr(A|C,B) \Pr(B|C) + \Pr(A|C, \text{not } B) \Pr(\text{not } B|C)$ and the following counterexample shows why there is a difference. Prob A B C 0.1 T T T 0.1 F T T 0.1 T F T 0.2 F F T 0.2 T T F 0.1 F T F 0.1 T F F 0.1 F F F Then in your formulation $\Pr(A|C)=\frac{2}{5}$, $\Pr(A|B)=\frac{3}{5}$, $\Pr(B|C)=\frac{2}{5}$, $\Pr(A|\text{not } B)= \frac{2}{5}$, $\Pr(\text{not } B|C)= \frac{3}{5}=6$ but $\frac{2}{5} \not = \frac{3}{5} \times \frac{2}{5} + \frac{2}{5} \times \frac{3}{5}$. In my formulation $\Pr(A|C,B) = \frac{1}{2}$ and $\Pr(A|C, \text{not } B)=\frac{1}{3}$ and we have the equality $\frac{2}{5} = \frac{1}{2} \times \frac{2}{5} + \frac{1}{3} \times \frac{3}{5}$.

While Henry's equation is correct in the general case, the probability assignment in his counterexample is not compatible with the Bayesian network $A \rightarrow B \rightarrow C$. For the probability assignments required by the locally Markov property in this case, @ArKitect's solution is correct. – Dilip Sarwate Nov 28 at 2:42.

You are correct; you are integrating out B, conditional upon C. Good job! To extend this answer in response to Henry's answer and Dilip's comment - yes, the network structure implies that $P(A|B,C) = P(A|B)$, so, taking a problem-specific shortcut, Dilip's original answer is correct.

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