# Conditional Probability based on intersection of two sets, independent and dependent respectively

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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It is a fact of life that we cannot. Even the (different) hypothesis that B is independent of A and that C is independent of A would not suffice. – Did Sep 15 '12 at 12:42
I really don't know what did was thinking. Maybe he will eventually tell us. But he has a penchant for just making snide remarks. – Michael Chernick Sep 15 '12 at 16:03
@MichaelChernick Stop your bickering. If you fail to understand what I explain (which happens to you a lot, lately), just ask (politely, but I know this is difficult to you). At the moment, I am rather surprised that somebody claiming to be a biostatistician, may fall in such a simple trap: contrary to what you assert in your answer, nothing guarantees that B and C are independent. – Did Sep 15 '12 at 16:36
@did The reason I don't understand some things you say is because you are often not clear. The OP wanted to know how the conditions A independent of B and A and C dependent affect the computation of the conditional probability. The answer would be that you would need to know P(A∩B∩C) and P(B∩C) since the conditions do not allow further simplification. Saying you can't calculate it does not really answer the question. – Michael Chernick Sep 15 '12 at 17:56
@MichaelChernick Suuure... If only you could stop polluting the discussion. – Did Sep 15 '12 at 19:59

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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If B was dependent on C, and C is dependent on A (as given) then doesn't this imply A is dependent on B, and therefore since B is independent of A, then B is independent of C? – Mew Sep 15 '12 at 13:19
@chris He didn't say that B is dependent on C. – Michael Chernick Sep 15 '12 at 13:51
Agreed. My above comment is a proof by Reductio ad absurdum that B is independent of C. – Mew Sep 15 '12 at 13:59
I see B is independent of C and the above reduction does work... Absolutely not. @Chris Sorry but your comment does not prove that B is independent of C (the implication does not hold). Exercise: Find events A, B and C such that A is independent of B, A is not independent of C and B is not independent of C. Hint: This can be done on a probability space of size 4. – Did Sep 15 '12 at 16:40
@SeyhmusGüngören This is just the smallest size for which I saw a straightforward example. Still smaller sizes might be possible, I did not check (but I think that to determine the simplest possible (counter-)example is often a good way to really understand what is going on). – Did Sep 15 '12 at 20:27

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).

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