Bayes Theorum with Multiple conditions with independent ancenstors I have the following Bayes' Net:
{D}->{C}<-{A}->{B}
I need to find P(A|B,C) with B and C both being true.  I have calculated the probability of A given B using the formula
$$
P(A|B)=\frac{P(B|A)P(A)}{P(B)}
$$
but I am unsure how to add C into this equation since it links to D.
Would I use the same method as described in Extended Bayes' theorem: p(A | B, C, D) - Constructing a Bayesian Network ?
The nodes in my diagram have the following values:
P(A)=0.01
P(D)=0.02

+---+---+----------+
| A | D | P(C|A,D) |
+---+---+----------+
| T | T | 0.9      |
| F | T | 0.5      |
| T | F | 0.8      |
| F | F | 0.05     |
+---+---+----------+


+---+---+--------+
| A | B | P(B|A) |
+---+---+--------+
| T | T | 0.6    |
| F | T | 0.02   |
+---+---+--------+

 A: You have $P(C)=0.02$ but I assume you mean $P(D)=0.02$. We can proceed as follows:
\begin{align}
P(A\mid B,C) &= \dfrac{P(A,B,C)}{P(B,C)} = \dfrac{P(A,B,C)}{P(A,B,C) + P(A^c,B,C)}. \\
\end{align}
\begin{align}
& \\
P(A,B,C) &= P(B,C\mid A)\cdot P(A) \\
&= P(B\mid A)\cdot P(C\mid A)\cdot P(A) \qquad\text{since $B,C$ are conditionally independent given $A$} \\
&= P(B\mid A)\cdot P(A)\cdot \left[P(C\mid D,A)\cdot P(D\mid A) + P(C\mid D^c,A)\cdot P(D^c\mid A)\right] \\
&= P(B\mid A)\cdot P(A)\cdot \left[P(C\mid D,A)\cdot P(D) + P(C\mid D^c,A)\cdot P(D^c)\right] \\
& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $A,D$ are independent} \\
&= 0.6\times 0.01\times [0.9\times 0.02 + 0.8\times 0.98] \\
&= 0.004812. \\
\end{align}
Similarly,
\begin{align}
P(A^c,B,C) &= P(B,C\mid A^c)\cdot P(A^c) \\
&= P(B\mid A^c)\cdot P(C\mid A^c)\cdot P(A^c) \qquad\text{since $B,C$ are conditionally independent given $A^c$} \\
&= P(B\mid A^c)\cdot P(A^c)\cdot \left[P(C\mid D,A^c)\cdot P(D\mid A^c) + P(C\mid D^c,A^c)\cdot P(D^c\mid A^c)\right] \\
&= P(B\mid A^c)\cdot P(A^c)\cdot \left[P(C\mid D,A^c)\cdot P(D) + P(C\mid D^c,A^c)\cdot P(D^c)\right] \\
& \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\text{since $A^c,D$ are independent} \\
&= 0.02\times 0.99\times [0.5\times 0.02 + 0.05\times 0.98] \\
&= 0.0011682. \\
\end{align}
Therefore,
\begin{align}
P(A\mid B,C) &= \dfrac{0.004812}{0.004812 + 0.0011682} \approx 0.80466. \\
\end{align}
