$$W\lower{2ex}{\searrow\lower{2ex} S\swarrow }U\lower{6ex}{\hspace{-5ex}\searrow\lower{2ex}{Z}\swarrow\raise{2ex}{A\raise{2ex}{\swarrow\raise{2ex}T}}}$$
Here are known probabilities
$P(W) = 0.002,\\
P(U) = 0.5, \\
P(T) = 0.06, \\
P(S|W,U) = 0.3, \\
P(S|-W,U) = 0.2, \\
P(S|W,-U) = 0.07, \\
P(S|-W,-U) = 0.02, \\
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) = 0.11 $
I'm trying to figure out how to calculate probability P(U|Z) and P(U|Z,T).
I have my joint probability like
P(Z,S,A,W,U,T) = P(Z|S,A)P(S|W,U)P(A|T)P(W)P(U)P(T)
but I don't know hot to transform this equation. Thanks for help
Use Bayes' Rule.
$$\small\def\P{\operatorname{\sf P}}\begin{align}\P(W,U,T,A,S,Z) &= \P(W)\P(U)\P(T)\P(A\mid T)\P(S\mid W,U)\P(Z\mid A,S)\\[2ex]\P(U\mid Z)&=\dfrac{\P(U)\P(Z\mid U)}{\P(Z)} &\llap{=\dfrac{\P(U)\P(Z\mid U)}{\sum_{\cal U}\P(\mathcal U)\P(Z\mid\mathcal U)}}\\[1ex]&=\dfrac{\P(U)\sum_{\cal W,T,A,S}\P(\mathcal W)\P(\mathcal T)\P(\mathcal A\mid\mathcal T)\P(\mathcal S\mid\mathcal W,U)\P(Z\mid\mathcal A,\mathcal S)}{\sum_{\cal W,U,T,A,S}\P(\mathcal W)\P(\mathcal U)\P(\mathcal T)\P(\mathcal A\mid\mathcal T)\P(\mathcal S\mid\mathcal W,\mathcal U)\P(Z\mid\mathcal A,\mathcal S)}\\[4ex]\P(U\mid Z,T) &=\dfrac{\P(U)\P(Z,T\mid U)}{\P(Z,T)}&\llap{=\dfrac{\P(U)\P(Z,T\mid U)}{\sum_{\cal U}\P(\mathcal U)\P(Z,T\mid\cal U)}}\\[1ex]&~~\vdots \end{align}$$