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I have this following question and i would like to know if anyone can answer it. For a given Bayesian network where $P(a) =.6, P(b|a) =.8, P(b|-a)=.4, P(c|a)=.4$ and $P(c|-a) = .3$, compute $P(c|b)$. Note that $a, -a, b$, etc. are propositions: e.g.) $a \leftrightarrow A = true , -a \leftrightarrow A = false$.

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I know that $P(a\land b\land c) = P(b|a)P(c|a)P(a)$ but i don't know how to solve the $P(c|b)$?

Thank you for help in advance.

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Begin with the definition of conditional probability.

$\hspace{25ex}\mathsf P(c\mid b) = \dfrac{\mathsf P(b\cap c)}{\mathsf P(b)}$

Now apply the law of total probability,

$\hspace{25ex}\mathsf P(c\mid b) = \dfrac{\mathsf P(a\cap b\cap c)+\mathsf P(\neg a\cap b\cap c)}{\mathsf P(a\cap b)+\mathsf P(\neg a\cap b)}$

Then use the relations from the DAG (the directed acyclic graph).

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  • $\begingroup$ Thanks for your comment. I have tried but unfortunately i couldn't solve it. I am totally new in Bayesian and i am not familiar with law of total probability and DAG. Would you please give some more help. Thanks $\endgroup$
    – Bilgin
    Dec 12, 2017 at 12:06
  • $\begingroup$ You claimed to know $\mathsf P(a\cap b\cap c)=\mathsf P(a)\mathsf P(b\mid a)\mathsf P(c\mid a)$ in the OP. $\endgroup$
    – Graham Kemp
    Dec 12, 2017 at 12:13
  • $\begingroup$ Thank you for the help. $\endgroup$
    – Bilgin
    Dec 12, 2017 at 12:38

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