How to combine Bayes network, total probability and probability of parent event given child event?

Question

Given the following Bayes network:

how can I calculate $\Pr(C|\lnot A,E)$?

I think first we need to use Bayes theorem, then we can use chain rule: $$ \Pr(C|\lnot A, E)=\frac{\Pr(\lnot A, E|C)\cdot\Pr(C)}{\Pr(\lnot A,E)}=\\\frac{\Pr(\lnot A, E,C)\cdot\Pr(C)}{\Pr(\lnot A,E)}=\\ \frac{\Pr(\lnot A)\Pr(C|A)\Pr(E|C)\Pr(C)}{\sum_{C\in \{T,F\}}\Pr(\lnot A, C, E)} $$

Am I on the right track?

Answer 1

$P(\neg A,C,E)$ is the marginalization of the joint distribution $$\sum_{B,D} P(\neg A,B,C,D,E) $$ and the joint is given via the network (which you've already used).

Joint probability:

$P(\neg A,B,C,D,E) = P(\neg A)P(C\mid\neg A) P(E\mid C) P(B) P(D\mid \neg A,B)$.

Answer 2

$$\begin{align}\mathsf P(C\mid \neg A, E)&=\dfrac{\mathsf P(\neg A,C, E)}{\sum_{C\in\{\top,\bot\}}\mathsf P(\neg A,C,E)}\\[1ex]&=\dfrac{\mathsf P(\neg A)\mathsf P(C\mid\neg A)\mathsf P(E\mid C)}{\mathsf P(\neg A)\mathsf P(C\mid\neg A)\mathsf P(E\mid C)+\mathsf P(\neg A)\mathsf P(\neg C\mid\neg A)\mathsf P(E\mid\neg C)}\\[1ex]&=\dfrac{\mathsf P(C\mid\neg A)\mathsf P(E\mid C)}{\mathsf P(C\mid\neg A)\mathsf P(E\mid C)+\mathsf P(\neg C\mid\neg A)\mathsf P(E\mid\neg C)}\end{align}$$