like others here I am also learning with Russel's and Norvig's book about artificial intelligence. My question is about the conditional probability tables of a clustered multiply connected network with a meganode.

We have the following Bayesian multiply connected network. Multiply connected Bayesian network.

In order to reduce computation (according to the book). We combine Sprinkler and Rain and get a meganode.

Clustered Bayesian network.

I have my difficulties with understanding the conditional probability table for the new meganode.

I assume that the values

  • $P(S+R|C) = .08$ can be calculated with $.08 = P(S|C) \cdot P(R|C) = .10 \cdot .80$.
  • Same goes for $.02$ with $.02 = P(S|C) \cdot P(R|\neg C) = .10 \cdot .20$.

However I do not understand why

  • $P(S+R|\neg C) = .40$ could be computed with $.40 = P(S|\neg C) \cdot P(R|C) = .40 \cdot .80$: Since according to the $TT$ column Sprinkler should be true and Cloudy should be true, same with Rain. However we have to use Sprinkler true but Cloudy false.
  • or $0.72$ could be computed with $.72 = P(R|C) - ( P(S|C) \cdot P(R|C)) = .80 - .10 \cdot .80 = .80 - .08$

Hence my overall question is how $P(S+R=x)$ can be computed.

Thank you!


I think there's a dyslexic typo here. Swap $0.1$'s with $0.4$'s. So for example,

$$P(S=T,R=T|C=F)=P(S=T|C=F)P(R=T|C=F)=0.5\cdot 0.2 =0.1,$$ $$P(S=F,R=T|C=F)=P(S=F|C=F)P(R=T|C=F)=0.5\cdot 0.2 =0.1,$$ $$P(S=T,R=F|C=F)=P(S=T|C=F)P(R=F|C=F)=0.5\cdot 0.8 =0.4,$$ $$P(S=F,R=F|C=F)=P(S=F|C=F)P(R=F|C=F)=0.5\cdot 0.8 =0.4.$$

  • $\begingroup$ Thank you for your answer. I could understand the bottom row of the table. Maybe we should inform Russel & Norvig about the typo. However I cannot apply your approach to the top row: $$ P(S=T,R=T|C=T)=P(S=T|C=T)P(R=T|C=T)=0.1\cdot 0.8 =0.08 \checkmark \\ P(S=T,R=F|C=T)=P(S=T|C=T)P(R=F|C=T)=0.1\cdot 0.2 =0.02 \checkmark \\ P(S=F,R=T|C=T)=P(S=F|C=T)P(R=T|C=T)=0.5\cdot 0.8 =0.4 \bigstar\\ P(S=F,R=F|C=T)=P(S=F|C=T)P(R=F|C=T)=0.5\cdot 0.2 =0.1 \bigstar\\ $$ What is my calculation error in the last two lines? $\endgroup$ – Peter F.well Apr 3 '15 at 20:10

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