What is the P(A|C) if we know B depends on A and C depends on B? Given a Bayesian network that looks like the following:
A->B->C

How do we compute P(A|C)?  My initial guess would be:
P(A|C) = P(A|B) * P(B|C) + P(A|not B) * P(not B|C)

 A: 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 $\frac{2}{5} = \frac{1}{2} \times \frac{2}{5} + \frac{1}{3} \times \frac{3}{5}$.
A: 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.  
