Given a Bayes Net, compute $p(A|B,C)$

Question

I am given the following Bayes Net:

and I am asked to compute $p(A|B,C)$.

This is what I've done:

\begin{align} p(A|B,C) & := \frac{p(A,B,C)}{p(B,C)} \\ & = \frac{p(A,B,C)}{p(B)\ p(C|A)} \\ & = \frac{p(A)\ p(B)\ p(C|A)}{p(B)\ p(C|A)} \\ & = p(A) \\ & = \frac{1}{3} \end{align}

is that correct?

Answer 1

Not quite. The numerator is correct, but the denominator is found through its total probability.

$$\begin{align}\mathsf P(A\mid B, C)&=\dfrac{\mathsf P(A,B,C)}{\mathsf P(B,C)}\\[1ex]&=\dfrac{\mathsf P(A,B,C)}{\mathsf P(A,B,C)+\mathsf P(\lnot A, B, C)}\\[1ex]&~~\vdots\end{align}$$