How to solve Bayes' probabilistic network problem?

Question

Given the following Bayesian Probabilistic Network, let's say I am trying to find the probability of P(!FO|HB). I understand basic Bayes theorem, but not sure how to use it here.

Answer 1

The relevant factorisation is: $\def\P{\operatorname{\sf P}}\P(fo, bp, do, hb)=\P(fo)\P(bp)\P(do\mid fo,bp)\P(hb\mid do)$

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Now, begin with Bayes' Rule.

$$\begin{align}\P(\neg fo\mid hb)&=\dfrac{\P(\neg fo)\P(hb\mid\neg fo)}{\P(\neg fo)\P(hb\mid\neg fo)+\P(fo)\P(hb\mid fo)}\\[1ex]&=\dfrac{0.85\P(hb\mid\neg fo)}{0.85\P(hb\mid\neg fo)+0.15\P(hb\mid fo)}\end{align}$$

But we cannot look up the rest directly from the table, so we need to factorise .

$$\begin{align}\P(hb\mid fo)&=\P(bp)\P(hb\mid fo,bp)+\P(\neg bp)\P(hb\mid fo,\neg bp)\\[1ex]&=0.1\P(hb\mid fo,bp)+0.9\P(hb\mid fo,\neg bp)\\[2ex]\P(hb\mid\neg fo)&=\P(bp)\P(hb\mid\neg fo,bp)+\P(\neg bp)\P(hb\mid\neg fo,\neg bp)\\[1ex]&=0.1\P(hb\mid\neg fo,bp)+0.9\P(hb\mid\neg fp,\neg bp)\end{align}$$

So we factorise some more, $$\begin{align}\P(hb\mid fo, bp)&=\P(do\mid fo,bp)\P(hb\mid do)+\P(\neg do\mid fo,bp)\P(hb\mid\neg do)\\[1ex]&=0.99\cdot0.7+0.01\cdot 0.1\\[1ex]&=0.694\end{align}$$

And.. so forth. You can chocolate the rest .