Given a Bayes Network, calculate $p(a | b, \neg c)$

Question

I am assigned a homework but I cannot figure out how to solve it.

Given this Bayes' Network:

calculate $p(a | b, \neg c)$.

This is what I have done up to now:

\begin{align} p(a | b, \neg c) & \propto p(a,b,\neg c) \\ & = \sum_{d} p(a,b,\neg c, d) \\ & = \sum_d p(a)p(b|a)p(\neg c | a) p(d|a) \\ & = p(a)p(b|a)p(\neg c | a)\sum_d p(d|a) \end{align}

Now, I can see from the picture that $p(a) = \frac{1}{2}$, but what about the other terms such as $p(b|a)$, $p(\neg c | a)$ and $p(d|a)$?

Answer 1

The solution is:

\begin{align} p(a | b, \neg c) & \propto p(a,b,\neg c) \\ & = \sum_{d} p(a,b,\neg c, d) \\ & = \sum_d p(a)p(b|a)p(\neg c | a) p(d|a) \\ & = p(a)p(b|a)p(\neg c | a)\sum_d p(d|a) \\ & = \frac{1}{2} \times \frac{4}{5} \times (1 - p(c | a)) \\ & = \frac{4}{10} \times (1 - \frac{1}{3}) \\ & = \frac{4}{10} \times \frac{2}{3} \\ & = \frac{4}{15} \end{align}

Since $p(a)$, $p(b|a)$ and $p(\neg c | a) = 1 - p(c | a)$ can be read in the Bayesian Network and $\sum_d p(d|a) = 1$ of course.