Conditional and non-conditional probability in Bayesian network

Question

Question came up on an exam and I'm not sure if what I did was correct. I had to calculate $\mathbb{P}(S)$. Formula for calculating probability of that event would be: $$\sum_{W,U,\lnot W,\lnot U}\mathbb{P}(S,W,U)$$ $\mathbb{P}(S,W,U)$ simplifies to $\mathbb{P}(S|W,U)*\mathbb{P}(W)*\mathbb{P}(U)$

$$0.6 * 0.001*0.5 = 0.0003$$ $$0.007*0.999*0.5 = 0.34965$$ $$0.55*0.001*0.5 = 0.000275$$ $$0.005*0.999*0.5 = 0.0024975$$ $$\mathbb{P}(S) = 0.006569$$

I was also asked to calculate the probabilities of $ \mathbb{P}(W|Z)$ and $\mathbb{P}(Z|U)$ but wasn't sure how to approach these. Any advice regarding the correct way to analyze these probabilities would help me out.

Probabilities of events:
P(W) = 0.001,
P(U) = 0.5,
P(T) = 0.05,
P(S|W,U) = 0.6,
P(S|~W,U) = 0.007,
P(S|W,~U) = 0.55,
P(S|~W,~U) = 0.005,
P(A|T) = 0.1,
P(A|~T) = 0.01,
P(Z|S,A) = 0.95,
P(Z|~S,A) = 0.90,
P(Z|S,~A) = 0.40,
P(Z|~S,~A) = 0.001;

Answer 1

As you know, the definition of conditional probability / Bayes' Rule states:$$\begin{align}\mathsf P(W\mid Z) =&~ \dfrac{\mathsf P(W,Z)}{\mathsf P(Z)} \\[1ex]=&~ \dfrac{\mathsf P(Z\mid W)\,\mathsf P(W)}{\mathsf P(Z)} \end{align}$$

You can find these terms thusly:

$$\begin{align} \mathsf P(Z) =&~ \sum_{s, a} \mathsf P(Z\mid s, a)\,\mathsf P(s)\,\mathsf P(a) \\[1ex] =&~ \sum_{s,a,t,u,w} \mathsf P(Z\mid s,a)\,\mathsf P(s\mid u,w)\,\mathsf P(u)\,\mathsf P(w)\,\mathsf P(a\mid t)\,\mathsf P(t) \\[2ex] \mathsf P(Z\mid W) =&~ \sum_{s,a,t,u} \mathsf P(Z\mid s,a)\,\mathsf P(s\mid u,W)\,\mathsf P(u)\,\mathsf P(a\mid t)\,\mathsf P(t) \\[4ex] \therefore \quad \mathsf P(W\mid Z) =&~ \dfrac{\mathsf P(W)\sum_{s,a,t,u} \mathsf P(Z\mid s,a)\,\mathsf P(s\mid u,W)\,\mathsf P(u)\,\mathsf P(a\mid t)\,\mathsf P(t)}{\sum_{s,a,t,u,w} \mathsf P(Z\mid s,a)\,\mathsf P(s\mid u,w)\,\mathsf P(u)\,\mathsf P(w)\,\mathsf P(a\mid t)\,\mathsf P(t)} \end{align}$$

$\mathsf P(U\mid Z)$ is, of course, very similar.

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NB: Using the shortened notation of $\sum_{x}$ to mean $\sum_{x\in\{X,\neg X\}}$ for any matching pair of lower and upper case letters; and that $\sum_{x,y}$ is the double sum $\sum_{x\in\{X,\neg X\}}\sum_{y\in\{Y,\neg Y\}}$ et cetera.