TITLE: Probability Bayesian network problem
QUESTION [5 upvotes]: 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!

REPLY [2 votes]: 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.