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.

Bayes Network


1 Answer 1


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

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 .

  • $\begingroup$ Thank you so much. May I ask one more thing: P( hb | !fo) = (P(hb)*P(!fb | hb)) / (P(hb)*P(!fb | hb) +P(!hb)*P(!fb | !hb)) ---- P(!fo | !hb) = P(bp)*P(!fo | !hb, bp) + P(!bp)*P(!fo | !hb, !bp) ---- P(fo | hb) = P(bp)*P(fo | hb, bp) + P(!bp)*P(fo | hb, !bp) ----- P(fo | !hb, bp) = P(do | !hb, bp)P(!hb | do) + P(!do | !hb, bp)*P(!hb | !do) ---- P(fo | !hb, bp) = P(do | !hb, bp) * 0.3 + P(!do | !hb, bp) * 0.99 --- But when I come here I am not sure how to calculate P(do | !hb, bp)? What am I missing? $\endgroup$
    – vojtak
    Apr 21, 2020 at 9:13
  • $\begingroup$ Sorry, I cannot read that. $\endgroup$ Apr 21, 2020 at 9:32
  • $\begingroup$ Sorry, I'm not able to edit that comment now. I am trying to calculate the P( hb | !fo) and I am not able to continue after I reach P(do | !hb, bp). Because I don't know where to find P(do | !hb, bp). $\endgroup$
    – vojtak
    Apr 21, 2020 at 9:48
  • $\begingroup$ I see, you appear to be trying $\mathsf P(hb\mid {!}fo)=\dfrac{\mathsf P({!}fo\mid hb)\mathsf P(hb)}{\mathsf P({!}fo)}$. Don't do that. Look instead to the factorisation derived from the diagram.$$\mathsf P(hb\mid{!}fo)=\sum_{\mathrm{do}}\mathsf P(hb\mid\mathrm{do})\sum_{\mathrm {bp}}\mathsf P(\mathrm{bp})\,\mathsf P(\mathrm{do}\mid{!}fo,\mathrm{bp})\,$$ $\endgroup$ Apr 21, 2020 at 10:26
  • $\begingroup$ The key is: You were not given $\mathsf P(hb)$, so using Bayes' Rule to put it in the numerator will not make things easier.$~$ The diagram does, however, give $\mathsf P(hb\mid\pm do)$ , $\mathsf P(\pm do\mid{!}fo,\pm bp)$ , and $\mathsf P(\pm bp)$. @vojtak $\endgroup$ Apr 23, 2020 at 2:53

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