Given the following about the three events $A$,$B$, and $C$:
- $P(A \cap B) = P(A)P(B)$,
- $P(A \cap C) = P(A \mid C)P(C)$, and
- $P(B \cap C) = P(B \mid C)P(C)$
In other words, events $A$ and $B$ are independent, but $A$ and $C$ are dependent, as well as $B$ and $C$.
Assume P(A), P(B), and P(C) are known. Also, assume the right hand sides (RHS) of above given equations are all known.
Requirement:
Find $P(A \cap B \mid C)$, noting that $P(C \mid A \cap B)$ is unknown?
My thoughts:
Applying Bayes' theorem:
$P(A \cap B \mid C) = P(A \cap B \cap C)/P(C)$
However, the numerator in RHS is unknown. So, in order to find it, we need to express it in terms of the three given equations above.
I tried to apply below formula
$P(A \cap B \cap C)=P(A \mid B \cap C) P(B \mid C) P(C)$
But the RHS still has the unknown component $P(A \mid B \cap C)$
Any idea please?
The problem does not have a solution: Even if we have $P(A)$,$P(B)$,$P(C)$,$P(A\cap B)=P(A)P(C)$, $P(A\cap C)$, and $P(B\cap C)$ fixed, $P(A\cap B \cap C)$ may still vary.
Consider $[0,1] \times [0,1]$ (endowed with the uniform distribution) and $A,B,C$ such that $C$ has two disjoint parts as shown below.
Apparently $A$ and $B$ are independent. If one shifts $C$ parallel to itself horizontally (within a certain interval) then neither $P(A)$ nor $P(B)$ nor $P(C)$ nor $P(B\cap C)$ nor $P(A\cap B)$ change. However, $P(A\cap B \cap C)$ does.