Conditional independence in bayesian network I am referring a Bayesian Net Example from here.

Prob(A=T) = 0.3
Prob(B=T) = 0.6
Prob(C=T|A=T) = 0.8
Prob(C=T|A=F) = 0.4
Prob(D=T|A=T,B=T) = 0.7
Prob(D=T|A=T,B=F) = 0.8
Prob(D=T|A=F,B=T) = 0.1
Prob(D=T|A=F,B=F) = 0.2
Prob(E=T|C=T) = 0.7
Prob(E=T|C=F) = 0.2 

In this case how do you solve for P(E=T|A=T) ? 
Because C is not known E and A are not independent events. 
 A: A few things.


*

*The Chain Rule:
$$
P(E, C | A) = P(E|C,A) \cdot P(C | A)
$$

*Conditional Independence:  The notation of a Bayes Network implies
$$
P(E|C,A) = P(E|C)
$$
The idea is, if you know that $C$ did or did not happen, it doesn't matter whether or not $A$ happened.  So at this point we have:
$$
P(E,C|A) = P(E|C) \cdot P(C|A)
$$

*Marginalization: Removing a variable from a joint distribution (only variables on the left side of the "|") by summing over the possible values they can take.
$$
P(E|A) = P(E,C=T|A) + P(E,C=F|A).
$$


So now you should be able to answer this question.  Please let me know if you don't understand any of the steps, or don't see how this applies to your question.
A: It seems to me that you would need to compute the probability that E = T for each possible value of C, multiply each by the probability of C being that value given A = T, and add everything up.  That is
$$\sum_{X}P(E=T|C=X)P(C=X)$$
If every node can be either T or F, that would be
$$P(E=T|C=T)P(C=T|A=T) + P(E=T|C=F)P(C=F|A=T) $$
If I am reading jlimahaverford's answer correctly, I think they end up being the same thing.
Let me know if I am incorrect.
