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I have to solve the following problem:

Suppose we have a bayesian net in which we have the following variables: R, PA and PR

Let:

P(R) = 0.1, P(PA) = 0.5, P(PR|R, PA) = 0.6, P(PR|¬R, PA) = 0.4, P(PR|¬R, ¬PA) = 0.1 and P(PR|R, ¬PA) = 0.2

What is the probability of P(¬R, PR, ¬PA)?

I started with P(¬R) and P(¬PA), because I can compute them as follows:

P(¬R) = 1 - P(R) = 0.9 P(¬PA) = 1 - P(PA) = 0.5

Then I think I can compute P(PR|¬R) and use bayes rule, however:

P(PR|¬R, PA) = 0.4 $\Rightarrow$ P(PR|¬R) * P(PA) = 0.4 \Rightarrow P(PR|¬R) = 0.8

I also have P(PR|¬R, ¬PA) = 0.1 \Rightarrow P(PR|¬R) = 0.2

The same for P(PR|R)... I get different results, so I can't apply bayes rule. This means I am obviously doing something wrong, where is my mistake? How can I solve it?

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  • $\begingroup$ Bayes' Rule says: $\mathsf P(PR\mid \neg R, \neg PA) = \dfrac{ \mathsf P(PR, \neg PA\mid \neg R)}{\mathsf P(\neg PA\mid \neg R)} = \dfrac{\mathsf P(PR,\neg PA, \neg R)}{\mathsf P(\neg PA\mid \neg R)~\mathsf P(\neg R)}$ $\endgroup$
    – Graham Kemp
    Jun 6, 2016 at 3:18

1 Answer 1

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$$\begin{align}\mathsf P(R) =&~ 0.1\\ \mathsf P(PA) =&~ 0.5\\ \mathsf P(PR\mid R, PA) =&~ 0.6\\ \mathsf P(PR\mid ¬R, PA) =&~ 0.4\\\mathsf P(PR\mid ¬R, ¬PA) =&~ 0.1\\ \mathsf P(PR\mid R, ¬PA) =&~ 0.2\end{align}$$

What is the probability of $\mathsf P(¬R, PR, ¬PA)$?

$$\begin{align}\mathsf P(\neg R, PR, \neg PA)=&~\mathsf P(PR\mid \neg PA,\neg R)~\mathsf P(\neg PA, \neg R) \\[1ex] =&~ 0.1~\mathsf P(\neg PA,\neg R) \end{align}$$

We appear to be missing some information.   I suspect that it is that $R$ and $PA$ are independent.   Does your net indicate this?

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  • $\begingroup$ The other information is that:R incides in PR and PA incides in PR (I don't know what that means.. I suppose that R and PR are parent nodes of PR $\endgroup$
    – dpalma
    Jun 6, 2016 at 4:31
  • $\begingroup$ @dpalma Yes, that is it. In that case $\mathsf P(\neg PA,\neg R)=\mathsf P(\neg PA)~\mathsf P(\neg R)$ $\endgroup$
    – Graham Kemp
    Jun 6, 2016 at 5:14
  • $\begingroup$ One last question, how did you get that expression $P(PR|¬PA, ¬R)P(¬PA,¬R)? $\endgroup$
    – dpalma
    Jun 6, 2016 at 21:51
  • $\begingroup$ @dpalma The definition of conditional probability. $$\mathsf P(X\cap Y\cap Z)~=~\mathsf P(X\mid Y\cap Z)~\mathsf P(Y\cap Z)$$ $\endgroup$
    – Graham Kemp
    Jun 6, 2016 at 22:24

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