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I haven't been able to find an explanation online; maybe I don't know which search words to use. Let's say I have the following Bayes net:

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How would I compute $P(D=F|A=T)$?

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The trick is to take into account both cases of B being T or F. so you want P(D=F, B, A=T)/P(A=T) if i remember correctly, I'm a bit rusty with those. But you can follow wikipedia example.

So we have: $P(D=F|A=T)=\frac{P_{B\in[T,F]}(D=F, A=T,B)}{P(A=T)}$

Now the hard bit is the probability in the numerator.

$P_{B\in[T,F]}(D=F, A=T,B)=P(D=F,B=T,A=T)+P(D=F,B=F,A=T)$.

Now let's use the joint probability formula: $P(D=F,B=T,A=T) = P(D=F | B=T, A=T) \cdot P(B=T | A=T) \cdot P(A=T) $

Now you only need to substitute in the values and calculate the second part of the sum. You also have to take into account independence of A,B, so $P(B=T| A=T)=0.6 \cdot 0.3$ and other values are just from the tables.

Does that make sense? :) btw. I did the numerics and got 0.078, how about you?

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    $\begingroup$ Yes, I believe your explanation and calculations are correct. This was actually part of a larger computation seen here: cs.nyu.edu/faculty/davise/ai/bayesnet.html (part of the calculations for Prob(A=T|D=F)). In that calculation, $P(A=T)$ cancels, and the numerator is $$\sum_{b\in B} P(D=F|A=T,B=b)P(A=T,B=b)$$ which evaluates to 0.078. The key step for me was this: $$P(D|A)=\frac{P(D,A)}{P(A)}=\frac{\sum_{b\in B} P(D,A,B=b)}{P(A)}$$ $\endgroup$
    – jds
    Jan 14, 2017 at 17:47
  • $\begingroup$ Excellent! :) glad it hepled $\endgroup$
    – Jan Sila
    Jan 14, 2017 at 18:02

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