I am trying to calculate $P(A|B\cap C).$
From my data set I have calculated:
$P(A|B) = 0.58$
$P(A|C) = 0.44$
However, there is not enough data in the data set to calculate $P(A|B\cap C).$
Is there a way of combining $P(A|B) = 0.58$ and $P(A|C) = 0.44$ to calculate $P(A|B\cap C)?$
$P(A|B)$ and $P(A|C)$ are not mutually exclusive and are independant
read that and solve the puzzle by:
list all possible combination of variavles such as P(A), P(B), P(C), P(A,B), P(B,C), P(A,C), P(A,B,C), P(B|A), P(C|A), P(B,C|A), P(A|B), P(C|B), P(A,C|B), P(A|C), P(B|C), P(A,B|C), P(A|B,C), P(B|A,C), P(C|A,B)
and list the equaltions based on baye's thereom
P(B|A) P(A) = P(A|B) P(B)
P(C|A) P(A) = P(A|C) P(C)
P(C|B) P(B) = P(B|C) P(C)
P(B,C|A) P(A)= P(A|B,C) P(B,C)
P(A,C|B) P(B) = P(B|A,C) P(A,C)
P(A,B|C) P(C) = P(C|A,B) P(A,B)
A silly method, fill what you know and get the new iteratively and add more conditions , It still needs 19-6 = 13 equaltions(including known variables)