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Below is the scheme of conditional dependence and the probabilities of events:

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P(A=1) = 0.01 P(A=0) = 0.99 P(B=1) = 0.1 P(B=0) = 0.9 P(C=1|A=0,B=0) = 0.1 P(C=1|A=0,B=1) = 0.5 P(C=1|A=1,B=0) = 0.6 P(C=1|A=1,B=1) = 0.9 

Given the probabilities above I wanted to calculate P(B=1|C=1) and P(B=1|C=1,A=1) but didn't get the correct result.

I wrote the probabilistic function the following way:

P(A, B, C) = P(A)P(B)P(C|A, B) 

and then set the variables

P(B=1, C=1) = P(A=0, B=1, C=1) + P(A=1, B=1, C=1)= =P(A=0)P(B=1)P(C=1|A=0, B=1) + P(A=1)P(B=1)P(C=1|A=1, B=1)= =0.99*0.1*0.5 + 0.01*0.1*0.9 = 0.0495 

The result however is not correct and don't know where is the error. I would be very thankful if anyone could correct/explain what's wrong.

  • 0
    http://math.stackexchange.com/questions/884728/how-to-solve-probability-with-two-conditions-with-explanation?noredirect=1&lq=12017-04-01

1 Answers 1

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The typical way I do inter-causal reasoning is to flip the conditional probabilities around --

P(B=1|C=1) = P(B=1,C=1) / P(C=1)            = P(C=1|B=1) P(B=1) / P(C=1)  P(B=1|C=1,A=1) = P(B=1,C=1,A=1) / P(C=1,A=1)                = P(C=1|B=1,A=1) P(B=1,A=1) / P(C=1,A=1)                = P(C=1|B=1,A=1) P(B=1) P(A=1) / P(C=1|A=1)P(A=1)                = P(C=1|B=1,A=1) P(B=1) / P(C=1|A=1) 

Does that help?

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    No. That would give you P(C) > 12016-10-10