Bayes' Net Conditional Probability I have a Bayes' Net with 4 boolean nodes connected in a diamond shape.  I want to find the probability of one of the middle nodes being true given that the ones above and below are both true.  So given the diagram below, I want to find P(B|A and D).
  A
 / \ 
B   C
 \ /
  D

I'm a bit lost because I'm not sure how to relate the conditional probabilities coming from both higher up and further down the net.  Any tips on how to approach this would be welcome!
 A: Consider the Theorem of Bayes for two competing events $A, B$ (with "$c$" for $complement$):
$P(B|A) = \dfrac{P(A|B) \cdot P(B)}{P(A|B) \cdot P(B) + P(A|B^c) \cdot P(B^c)}$ 
In the case of the Bayesian network problem described above [P(B|A and D)] we have:
$P(B|A \cap D) = \dfrac{P(A \cap D|B) \cdot P(B)}{P(A \cap D|B) \cdot P(B) + P(A \cap D|B^c) \cdot P(B^c)}$
Assume a causal network of the type:
$A \rightarrow B$; $A \rightarrow C$; 
and
$B \rightarrow D$; $C \rightarrow D$ 
Now, and more specifically, what does $P(A \cap D|B)$ equal? Basic probability allows us to assume $0.00 \leq P(A \cap D|B) \le 1.00$.
If $P(A \cap D|B)$ equal $0.00$, then $P(B|A \cap D) = 0$. If $P(A \cap D|B)$ equal $1.00$, then $P(B|A \cap D) = 1.00$. 
Let us now consider the case of $0.00 < P(A \cap D|B) < 1.00$. One question that come to mind: Are Events $B$ and Events $C$ $mutually$ $exclusive$? Are we given any actual probability values. Are we looking for a general solution?
One tool that helps me is a probability tree - i have started one:

A: 
I'm a bit lost because I'm not sure how to relate the conditional probabilities coming from both higher up and further down the net. Any tips on how to approach this would be welcome!

The diamond inheritance of the diagram indicates the following factorisation. $$\mathsf P(A,B,C,D)=\mathsf P(A)\,\mathsf P(B\mid A)\,\mathsf P(C\mid A)\,\mathsf P(D\mid B,C)$$
Then by the Law of Total Probability.$$\begin{align}\mathsf P(A,B,D)&=\mathsf P(A)\,\mathsf P(B\mid A)\sum_{\small c\in\{C,\lnot C\}}\mathsf P(c\mid A)\,\mathsf P(D\mid B,c)\\[1ex]\mathsf P(A,D)&=\mathsf P(A)\sum_{\small b\in\{B,\lnot B\}}\mathsf P(b\mid A)\sum_{\small c\in\{C,\lnot C\}}\mathsf P(c\mid A)\,\mathsf P(D\mid b,c)\end{align}$$
Finally by definition of conditional probability.
$$\mathsf P(B\mid A, D)=\dfrac{\mathsf P(A)\,\mathsf P(B\mid A)\sum\limits_{\small c\in\{C,\lnot C\}}\mathsf P(c\mid A)\,\mathsf P(D\mid B,c)}{\mathsf P(A)\sum\limits_{\small b\in\{B,\lnot B\}}\mathsf P(b\mid A)\sum\limits_{\small c\in\{C,\lnot C\}}\mathsf P(c\mid A)\,\mathsf P(D\mid b,c)}$$
