I am working on an exercise that find out certain probability based on a Bayesian Network. The Bayesian Networking showed on the above The following are my work out:
P(C ∩ S ∩ E ∩ ¬P ∩ ¬F) = P(¬F | C ∩ S ∩ E ∩ ¬P) ⋅ P(C ∩ S ∩ E ∩ ¬P)
= P(¬F | C ∩ S ∩ E ∩ ¬P) ⋅ P(¬P | C ∩ S ∩ E) ⋅ P(C ∩ S ∩ E)
= P(¬F | C ∩ S ∩ E ∩ ¬P) ⋅ P(¬P | C ∩ S ∩ E) ⋅ P(E | C ∩ S) ⋅ P(C ∩ S)
= P(¬F | C ∩ S ∩ E ∩ ¬P) ⋅ P(¬P | C ∩ S ∩ E) ⋅ P(E | C ∩ S) ⋅ P(S | C) ⋅
P(C)
= P(¬F | ¬P) ⋅ P(¬P | C) ⋅ P(E | S) ⋅ P(S | C) ⋅ P(C) (the deduction based on the above Bayesian Network)
Based on the Bayesian Network, E is depend on S and P, I am confusing on how to find out P(E | S). Does its probability can be treated as 0?
Exactly, $E$ is conditioned on $S$ and $P$, so don't do that. You use $\mathsf P(E\mid S,\neg P)$.
Just read the map. $C$ has no parents. The parent of $S$ is $C$. The parent of $\neg P$ is $C$. The parents of $E$ are $S$ and $\neg P$. The parent of $F$ is $\neg P$.
Thus your network tells you that the factorisation is:
$$\def\P{\operatorname{\mathsf P}}\P(C\cap S\cap E\cap \neg P\cap \neg F)=\P(C)\P(S\mid C)\P(\neg P\mid C)\P(E\mid S,\neg P)\P(\neg F\mid\neg P)$$
$~$
Now if you really want to find $\P(E\mid S)$ then by the law of total probability and Bayes' Rule:$$\P(E\mid S)=\dfrac{\displaystyle\sum\limits_{p\in\{\neg P,P\}}\sum\limits_{c\in\{\neg C, C\}}\P(E\mid S,p)\P(p\mid c)\P(S\mid c)\P(c)}{\displaystyle\sum\limits_{c\in\{\neg C,C\}}\mathsf P(S\mid c)\mathsf P(c)}$$