conditional probability question Given bayesian network could not understand solution given

Question

Given Bayesian network
can't understand the two last steps in
why the p(C=c|E=e,~H) can get out of the e sum?
and why sum p(E=e|A,S,~H) and sum p(C=c|E=e,~H) and been neglected?
Thank for the answer.

Answer 1

why the $p(C=c\mid E=e,\neg H)$ can get out of the $e$ sum?

It's not - it's just loose notation in that answer. Adding extra parentheses would be clearer, (though it should be understood by the appearance of $e$ in the inner sum) making the denominator in Line (4) read:

$$P(A)P(S)P(\neg H)\sum_e{[P(E=e\mid A,S,\neg H) \sum_c{P(C=c\mid E=e,\neg H)} ]}.$$

why $\sum p(E=e|A,S,\neg H)$ and $\sum p(C=c|E=e,\neg H)$ have been neglected?

Because $\sum_c p(C=c|E=e,\neg H) = 1$ (because we're summing over all possible values of $C$).

Similarly, $\sum_e p(E=e|A,S,\neg H) = 1$ (because we're summing over all possible values of $E$).