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Given $P(A|B \cap C)$, where $B$ is independent of $A$, and $C$ is dependent of $A$. In this case, how should we calculate the probability?

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    @SeyhmusGüngören The sample space must have _at least_ $4$ elements because if we assume that independent events $A$ and $B$ have probabilities such that 0 < P(A), P(B) < 1 to avoid trivial special cases, then all $4$ disjoint events $AB, AB^c, A^cB, A^cB^c$ _must_ be non-empty since all four events have nonzero probability: $P(AB) = P(A)P(B), P(AB^c) = P(A)P(B^c),P(A^B)=P(A^C)P(B), P(A^cB^c) = P(A^c)P(B^c).$2012-09-16

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Regardless of whether $B$ is independent or not from $C$, one way to calculate this probability should be: $P(A|B\cap C)=P(A\cap B \cap C)/P(B\cap C)$and, $P(A\cap B \cap C)=P(A \cap B)+P(C)-P((A\cap B) \cup C)\\=P(A)P(B)+P(C)-P((A\cap B) \cup C)$ It appears to me that no further simplification can be done given the information in the question.

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P(A|B∩C)=P(A∩B∩C)/P(B∩C). If B is independent of C this can be further broken down to

P(A∩B∩C)/[P(B)P(C)]=P[A∩C|B]/P(C) Also it is NOT equal to P(A). There does not appear to be any further simplification without further assumptions.

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    @SeyhmusGüngören Thanks for the kind words (but you might relax about your abilities, to me you seem to have a brain and to be willing to use it--and this is all one needs, ain't it...).2012-09-15
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When A and C are dependent on each other but A is independent of B, it simply turns out:

P(A|B∩C)=P(A|C).