I have been asked to work through this question on a homework, and am quite stuck.
Given three events A, B, and C such that P(A ∩ B ∩ C) =/= 0 and P(C|A ∩ B) = P(C|B), show that P(A|B ∩ C) = P(A|B).
So far, I have played around with P(C|A ∩ B) = P(C|B)but this leads me to P(C (∩ A ∩ B)) = P(C ∩ B) and the same with P(A|B ∩ C) = P(A|B).
Does this mean that P(A ∩ B) = P(B)? Do you have a different perspective on this problem that could lead me to an answer?
Thank you for any help.
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$P(C \mid A \cap B) = P(C \mid B)$ is equivalent to $\frac{P(A \cap B \cap C)}{P(A \cap B)} = \frac{P(B \cap C)}{P(B)}$, so if you prove this last equality is true, you have solved the problem.
$P(A \mid B \cap C) = P(A \mid B)$ is equivalent to $\frac{P(A \cap B \cap C)}{P(B \cap C)} = \frac{P(A \cap B)}{P(B)}$. This gives you an expression for $P(A \cap B \cap C)$; plug it into the earlier equation above and check if it holds.
$$P(A|B\cap C)=\frac{P(B\cap C|A)P(A)}{P(B\cap C)}$$ by Bayes' rule $$=\frac{P(C|B\cap A)P(B|A)P(A)}{P(C|B)P(B)}$$ by conditional probability definition $$=\frac{P(B|A)P(A)}{P(B)}$$ by the given condition $$=P(A|B)$$ by Bayes' rule