<s>[INST] Inter-causal reasoning: How to solve probability with two conditions? Below is the scheme of conditional dependence and the probabilities of events:

P(A=1) = 0.01
P(A=0) = 0.99
P(B=1) = 0.1
P(B=0) = 0.9
P(C=1|A=0,B=0) = 0.1
P(C=1|A=0,B=1) = 0.5
P(C=1|A=1,B=0) = 0.6
P(C=1|A=1,B=1) = 0.9

Given the probabilities above I wanted to calculate P(B=1|C=1) and P(B=1|C=1,A=1) but didn't get the correct result.
I wrote the probabilistic function the following way:
P(A, B, C) = P(A)P(B)P(C|A, B)

and then set the variables
P(B=1, C=1) = P(A=0, B=1, C=1) + P(A=1, B=1, C=1)=
=P(A=0)P(B=1)P(C=1|A=0, B=1) + P(A=1)P(B=1)P(C=1|A=1, B=1)=
=0.99*0.1*0.5 + 0.01*0.1*0.9 = 0.0495

The result however is not correct and don't know where is the error. I would be very thankful if anyone could correct/explain what's wrong.
 [/INST] The typical way I do inter-causal reasoning is to flip the conditional probabilities around --
$$
\begin{align}
P(B = 1 \vert C = 1) & = \frac{P(B = 1, C = 1)}{P(C = 1)} \\
& = \frac{P(C = 1 \vert B = 1)P(B = 1)}{P(C = 1)} \\ \\
P(B = 1 \vert C = 1, A = 1) & = \frac{P(B = 1, C = 1, A = 1)}{P(C = 1, A = 1)} \\
& = \frac{P(C = 1 \vert B = 1, A = 1)P(B = 1)P(A = 1)}{P(C = 1, A = 1)} \\
& = \frac{P(C = 1 \vert B = 1, A = 1) P(B = 1) P(A = 1)}{P(C = 1 \vert A = 1)P(A = 1)} \\
& = \frac{P(C = 1 \vert B = 1, A = 1) P(B = 1)}{P(C = 1 \vert A = 1)}
\end{align}
$$
Does that help?
</s>