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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 ⇒⇒ 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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You don't need Bayes Rule for this. From the information you have given for the Bayes network the joint probability factors like this:

$$ P(R, PR, PA) = P(R)P(PA)P(PR|R,PA) $$

so you can calculate the joint probability directly from the conditional probabilities. Like you said you know $P(\neg r)$ and $P(\neg pa)$ and you also know $P(pr| \neg r, \neg pa)$.

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  • $\begingroup$ Doesn't this require that R and PA to be independent? $\endgroup$ – Cagdas Ozgenc Jun 6 '16 at 13:57
  • $\begingroup$ From the description of the BN they are independent. $\endgroup$ – Dave Kincaid Jun 6 '16 at 14:28
  • $\begingroup$ one last question, why the joint probability factors are these? I know the answer is correct, but do not know how to get that answer $\endgroup$ – dpalma Jun 6 '16 at 14:39
  • $\begingroup$ In a short answer it is because of the independencies that are encoded by the graph. The factorization theorem tells us that the joint probability distribution, P, can be written as the product of the conditional probabilities of the individual random variables conditioned only on its parents in the graph. $\endgroup$ – Dave Kincaid Jun 6 '16 at 19:09

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