Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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). – Did Mar 30 '12 at 11:06
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 ? – Kaushik Acharya Apr 1 '12 at 11:53
up vote 2 down vote accepted

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?

share|cite|improve this answer
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) ) ? – jaor Jul 30 '14 at 13:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.