# Finding an unconditional joint probability in a Bayesian Belief Network

I have a Bayesian network as drawn in the picture:

We can see that $$B$$ and $$C$$ are conditionally independent given $$A$$. My goal is to find $$P(B\cap C)$$. My first thought was to use the Law of Total Probability: \begin{align} P(B\cap C) &= P (B\cap C\cap A) + P(B\cap C\cap \neg A)\\ &= P(B\cap C \mid A)P(A) + P(B\cap C \mid \neg A)P(\neg A). \end{align} Using conditional independence, we should obtain \begin{align} P(B\cap C \mid A) = P(B \mid A)P(C\mid A)=(0.4)(0.6) = 0.24. \end{align} However, when trying to calculate $$P(B\cap C \mid \neg A)$$, I know that I can't necessarily perform a similar product, as conditional independence given $$A$$ does not imply conditional independence given $$\neg A$$. This is where I am stuck. I thought about utilizing the Markov blanket of $$A$$ or even involving $$D$$ and $$E$$, as in a Bayesian network it is very easy to calculate \begin{align} P(A\cap B\cap C\cap D\cap E) &= P(A)P(B\mid A)P(C\mid A)P(D\mid B\cap C)P(E\mid C)\\ &=(0.8)(0.4)(0.6)(0.8)(0.3)=0.04608. \end{align} However, I am not sure if this is a useful route to take. Any help or comments are greatly appreciated!

It is true that $$P\left(B \cap C \mid \neg A \right) = P\left( B \mid \neg A \right)\times P\left( C \mid \neg A \right).$$
The link that you attached (about conditional independence given $$A$$ not implying conditional independence given $$\neg A$$) is not relevant here. In that link, $$A$$, $$B$$ and $$C$$ are events. Whereas the Bayesian network tells you about dependencies between random variables.
To spell it out, let $$X_A$$ be the random variable that takes the value $$1$$ if $$A$$ is true, and takes the value $$0$$ if $$A$$ is false. Let $$X_B$$ and $$X_C$$ be defined in the same way for $$B$$ and $$C$$. The Bayesian network diagram is a statement about these random variables: it says that the random variables $$X_B$$ and $$X_C$$ are conditionally independent given $$X_A$$, i.e. $$P\left( X_B, X_C \mid X_A \right) = P\left( X_B \mid X_A \right) \times P\left( X_C \mid X_A \right).$$
\begin{align} P\left(B \cap C \mid \neg A \right) & = P\left( X_B = 1, X_C = 1 \mid X_A = 0\right) \\ & = P\left( X_B = 1 \mid X_A = 0 \right) \times P\left( X_C = 1\mid X_A = 0\right)\\ & = P\left( B \mid \neg A \right)\times P\left( C \mid \neg A \right) . \end{align}