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Given $A,B,C$ such that: $$ P(A\mid B),P(A\mid B^c),P(B\mid C),P(B\mid C^c) \text{ are known } $$ and that $A,C$ are conditionally independent given $B$, so that: $$ P(A\mid B\cap C)=P(A\mid B),P(A\mid B^c\cap C)=P(A\mid B^c) $$ (the equations above are not the definition of conditionally independent, but follow from it)

I want to obtain $P(A\mid C),P(A\mid C^c)$.

I've managed to show that: $P(A\mid C)=P(A\mid B)P(B\mid C)+P(A\mid B^c)P(B^c\mid C)$. But I'm stuck on what to do with $P(B^c\mid C)$. Any ideas?

Thanks for helping! :DDDD

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  • $\begingroup$ P(Bc|C)=1-P(B|C). $\endgroup$ – Did Aug 27 '15 at 23:49
  • $\begingroup$ Indeed! Thanks! $\endgroup$ – Guilherme Salomé Aug 28 '15 at 0:55
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In order to calculate $P(B^c\mid C)$ observe that: $$ P(B^c\mid C)+P(B|C)=\frac{P(B^c\cap C)}{P(C)}+\frac{P(B\cap C)}{P(C)}=\frac{P((B^c\cap C)\cup(B\cap C)}{P(C)}=\frac{P(C)}{P(C)}=1 $$ So that $P(B^c\mid C)=1-P(B\mid C)$ and $P(B\mid C)$ is known. The same applies to $P(B^c\mid C^c)=1-P(B\mid C^c)$, which is used to obtain $P(A|C^c)$.

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