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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

mfl
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user2970105
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  • Have a look at this related question http://math.stackexchange.com/questions/817507/pa-given-that-pab-and-pb-are-known/817513#817513 – mfl Jun 29 '14 at 19:50

1 Answers1

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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)

  • What do the commas mean? Set union? – nickalh Feb 03 '24 at 12:38
  • And what is the order of operations between union vs. conditional probability- $P(B,C|A)= P( (B \cup C)|A)$ or $= P(B \cup (C|A))$ Also, is my second option even a valid probability notation? – nickalh Feb 03 '24 at 12:39