Marginalising out $B$ in $P(A \mid B,C)$ Let's say that I have $P(A \mid B,C)$ - is it accurate to say that $P(A \mid C)$ can be found like this:
$P(A \mid C) = \sum_B P(A \mid B,C)$
I know the values of all $P(A \mid B,C)$ as well as $P(B)$ and $P(C)$

You probably know (marginalization, or total probability) 
$$P(A) = \sum_B P(A,B)$$
Because this is true for any $A,B$, it's also true for the variables conditioned on $C$ (which amounts to restricting the universe).
$$P(A \mid C) = \sum_B P(A,B\mid C)$$
Further $P(A,B\mid C)=P(A \mid B,C)P(B\mid C)$. 
(Why? Because $P(A,B)=P(A \mid B) P(B)$; again, that must be also true if conditioning everything over $C$). 
Hence
$$P(A \mid C) =\sum_B P(A \mid B,C)P(B\mid C)$$
