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

enter image description here

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
    You said you were asked `P(B=1|C=1)` but you computed `P(B=1,C=1)` hence you are off by the factor `P(C=1)` (which you will need to compute).2012-03-30
  • 1
    Is P(C=1) = ( P(C=1|A=0,B=0)+P(C=1|A=0,B=1)+P(C=1|A=1,B=0)+P(C=1|A=1,B=1) )/4 ?2012-04-01
  • 0
    Maybe because 0.99*0.1*0.5 + 0.01*0.1*0.9 = 0.0504 (and not 0.0495) ?2016-10-09
  • 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?

  • 1
    And how do you calculate P(C=1) ? Is it `( P(C=1|A=0,B=0) + P(C=1|A=0,B=1) ) * ( P(C=1|A=1,B=0) + P(C=1|A=1,B=1) )` ?2014-07-30
  • 0
    No. That would give you P(C) > 12016-10-10