How do these conditional probability tables work (resources)? I've taken some probability classes before, but it's been a while. I'm looking for (1) some resource -- website, paper, software, book -- etc on conditional probabilities. Something which accessible to a junior/senior level undergrad. Specifically, I'm interested in the relationships described by the following example: 
Suppose that the nodes $A$ and $B$ are independent, and the node $C$ depends on both  (I understand this may be an example of what's called a directed acyclic graph, if that's important).
The type of dependencies associated with the nodes I described above rely on truth tables. For example, $A$ and $B$'s probability distributions:
$$ D(A) = 
\begin{array}{c|c}
  \text{True} & \text{False} \\ \hline
   0.3 & 0.7 
\end{array}, \quad 
D(B) = 
\begin{array}{c|c}
  \text{True} & \text{False} \\ \hline
   0.01 & 0.99 
\end{array} $$
And $C$'s (in percentages)

Then, without observable evidence, the probability that $C$ is FALSE* is $ \approx 0.9688$.
 How does one arrive at that value?
Furthermore, how would I generalize this behavior to a larger graph, with possibly more complicated stuff like $A$ and $B$ not being independent.
 A: To calculate the probability that $C$ is false we calculate all combinations of $A$ und $B$ which lead to $\overline C$. 
$\overline C$ is the complementary event of $C$.
$$P(\overline C)=P(A)\cdot  P(B|A)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot  P(B|\overline A)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot  P(\overline B| A)\cdot P(\overline C|A\cap \overline   B)+P(\overline A)\cdot  P(\overline B|\overline A)\cdot P(\overline C|\overline A\cap \overline B)$$
The events are: $A:=$Event A is true, $B:=$Event B is true,$\overline A:=$Event A is false, $\overline B:=$Event B is false,$C:=$Event C is true, $\overline C:=$Event C is false
$A$ and $B$ are independent.  Therefore $P(B|A)=P(B)$. This is the same for all combinations with the complementary events
$$P(\overline C)=P(A)\cdot  P(B)\cdot P(\overline C|A\cap B)+P(\overline A)\cdot  P(B)\cdot P(\overline C|\overline A\cap B)+P( A)\cdot  P(\overline B)\cdot P(\overline C|A\cap \overline   B)+P(\overline A)\cdot  P(\overline B)\cdot P(\overline C|\overline A\cap \overline B)$$
And the conditional probability $P(\overline C|A\cap B)=100 \%=1$ is given in the table. 
This is the last row and third column. Similar for the combinations of the events A, B. Therefore 
$$P(\overline C)=0.7\cdot 0.99\cdot 0.999+0.99\cdot 0.3\cdot 0.9+0.01\cdot 0.7\cdot 0.88+0.3\cdot 0.01\cdot 1 \approx 0.9688$$
And $P(C)=1-P(\overline C)$. This result can be also obtained by using the equivalent formula which is used to calculate $P(\overline C)$.  
