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The diagram above is the Bayesian network of my problem. I want to find $$\Pr(B=F \mid E=F, A=T)$$ I have evaluated it into the following steps, then I got a bit stuck:

$$\Pr(B=F \mid E=F, A=T) = \frac{\Pr(B=F,E=F,A=T)}{\Pr(E=F,A=T)}$$ $$=\frac{\Pr(B=F,E=F,A=T)}{\Pr(A=T\mid E=F)\times\Pr(E=F)}$$ $$$$

I was able to get $\Pr(B=F,E=F,A=T)$ from: $\Pr(B=F,E=F,A=T)=\Pr(A=T\mid B=F,E=F) \times \Pr(B=F,E=F)$ $\Pr(B=F,E=F,A=T)=\Pr(A=T\mid B=F,E=F) \times \Pr(B=F)\times \Pr(E=F)$

Am I right so far? So now the part that I got stuck is I am not sure how to get $$\Pr(A=T\mid E=F)$$

Any help would be appreciated. Thanks!


I think you're right so far.

\begin{eqnarray*} P(A=T\mid E=F) &=& P(A=T\mid E=F,B=T)P(B=T\mid E=F) \\ && + P(A=T\mid E=F,B=F)P(B=F\mid E=F) \\ && \qquad\text{by the Law of Total Probability} \\ && \\ &=& P(A=T\mid E=F,B=T)P(B=T) \\ && + P(A=T\mid E=F,B=F)P(B=F) \\ && \qquad\text{by independence of $B$ and $E$.} \\ \end{eqnarray*}

This can now be evaluated from values on the network diagram.


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