I'm brushing up on probabilistic reasoning and am trying to figure out a problem based on the following information:
There're 5 binary variables: Smart, Study, Prepared, Fair, Pass
I'm supposed to find the probability that a student who passed is smart.
That would translate to finding P(+smart|+pass). In such a case, it breaks down to
P(+smart) P(+pass|+smart)
P(+smart) is .8
I'm not sure how to isolate P(+pass|+smart) when what we're given is P(pass|smart,prepared,fair)
$\newcommand{\pass}{\operatorname{pass}}\newcommand{\smart}{\operatorname{smart}} \newcommand{\study}{\operatorname{study}}\newcommand{\prepared}{\operatorname{prepared}}\newcommand{\fair}{\operatorname{fair}}\newcommand{\plus}[1]{{^+\!}\color{navy}{#1}}\newcommand{\minus}[1]{{^-\!}\color{purple}{#1}}$
From the diagram, $\fair$ is independent of $\prepared$ and $\smart$, so using the Law of Total Probability.
$$\begin{align}\mathsf P(\plus\pass\mid\plus\smart)~&=~{\mathsf P(\plus\pass\mid\plus\smart,\plus\prepared,\plus\fair)~\mathsf P(\plus\prepared\mid\plus\smart)~\mathsf P(\plus\fair)\\+\mathsf P(\plus\pass\mid\plus\smart,\plus\prepared,\minus\fair)~\mathsf P(\plus\prepared\mid\plus\smart)~\mathsf P(\minus\fair)\\+\mathsf P(\plus\pass\mid\plus\smart,\minus\prepared,\plus\fair)~\mathsf P(\minus\prepared\mid\plus\smart)~\mathsf P(\plus\fair)\\+\mathsf P(\plus\pass\mid\plus\smart,\minus\prepared,\minus\fair)~\mathsf P(\minus\prepared\mid\plus\smart)~\mathsf P(\minus\fair)}\end{align}$$
On the other hand, $\prepared$ is dependent on $\smart$ and $\study$, while they are independent of each other.
$$\begin{align}\mathsf P(\plus\prepared\mid\plus\smart)~&=~{\mathsf P(\plus\prepared\mid\plus\smart,\plus\study)~\mathsf P(\plus\study) \\+ \mathsf P(\plus\prepared\mid\plus\smart,\minus\study)~\mathsf P(\minus\study)} \\[2ex]\mathsf P(\minus\prepared\mid\plus\smart)~&=~{\mathsf P(\minus\prepared\mid\plus\smart,\plus\study)~\mathsf P(\plus\study) \\+ \mathsf P(\minus\prepared\mid\plus\smart,\minus\study)~\mathsf P(\minus\study)}\end{align}$$
PS: Also, as $$\mathsf P(\plus\smart\mid\plus\pass) = \dfrac{\mathsf P(\plus\pass\mid\plus\smart)\mathsf P(\plus\smart)}{\mathsf P(\plus\pass)}$$ ...then, you are not quite done.
$$\begin{align}\mathsf P(\plus\pass)~&=~{\mathsf P(\plus\pass\mid\plus\smart)~\mathsf P(\plus\smart)+\mathsf P(\plus\pass\mid\minus\smart)~\mathsf P(\minus\smart)}\end{align}$$