How do I find the P(B | D = T) in this bayesina netowrk?

Question

How to find P(B | D = T) in the following Bayesian network?

Answer 1

By Bayes' Rule and the Law of Total Probability $$\begin{align}\mathsf P(B{\,=\,}\mathrm T\mid D{\,=\,}\mathrm T)&=\dfrac{\mathsf P(B{\,=\,}\mathrm T, D{\,=\,}\mathrm T)}{\mathsf P(D{\,=\,}\mathrm T)}\\[1ex]&=\dfrac{\sum_{\small\langle a,c\rangle\in{\{\mathrm F,\mathrm T\}}^2}\mathsf P(A{\,=\,}a, B{\,=\,}\mathrm T, C{\,=\,}c, D{\,=\,}\mathrm T)}{\sum_{\small\langle a,b,c\rangle\in{\{\mathrm F,\mathrm T\}}^3}\mathsf P(A{\,=\,}a, B{\,=\,}b, C{\,=\,}c, D{\,=\,}\mathrm T)}\end{align}$$

Now use the Directed Acyclic Graph to factorise: $\mathsf P(A{\,=\,}a, B{\,=\,}b, C{\,=\,}c, D{\,=\,}d)$