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September 26th, 2018, 11:42 AM   #1
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Probability and Independence

Let A, B, C, D, E be five boolean random variables. Assume that
you are give the independence assumptions:

* A and B are independent absolutely.
* D is conditionally independent of both A and C, given B.
* E is conditionally independent of A, B and D, given C.

Assume you also are give the following probabilities:
P(A=T) = 0.9
P(B=T) = 0.6
P(A=F) = 0.1
P(B=F) = 0.4
P(C=T|A=T,B=T) = 0.9
P(C=T|A=T,B=F) = 0.8
P(C=T|A=F,B=T) = 0.4
P(C=T|A=F,B=F) = 0.1
P(C=F|A=T,B=T) = 0.1
P(C=F|A=T,B=F) = 0.2
P(C=F|A=F,B=T) = 0.6
P(C=F|A=F,B=F) = 0.9
P(D=T|B=T) = 0.8
P(D=T|B=F) = 0.2
P(E=T|C=T) = 0.1
P(E=T|C=F) = 0.8
P(D=F|B=T) = 0.2
P(D=F|B=F) = 0.8
P(E=F|C=T) = 0.9
P(E=F|C=F) = 0.2

Compute the following probabilties:
1. P(A=T and B=T) = P(A) * P(B) = 0.54
2. P(A=T or B=T) = P(A) + P(B) = 1.5
3. P(C=T) = ?????????
4. P(C=T|A=T) = ????????
5. P(C=T and D=T) = ???????

I am totally lost of getting the answers of 3,4 5. I would be much appreciate if anyone able to show me how to solve these. Thanks.
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September 26th, 2018, 01:38 PM   #2
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General approach: P(C=T)=P(C=T|A=T,B=T)P(A=T)P(B=T)+P(C=T|A=T,B=F)P( A=T)P(B=F)+P(C=T|A=F,B=T)P(A=F)P(B=T)+P(C=T|A=F,B= F)P(A=F)P(B=F). This uses the independence of A and B.

P(C=T|A=T)=P(C=T|A=T,B=T)P(B=T)+P(C=T|A=T,B=F)P(B= F)

P(D=T)=1 Exercise for you - why? Therefore answer for 5. P(C=T)
Thanks from zollen
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September 27th, 2018, 10:15 AM   #3
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It took me a while but I finally got it!!!!!!!!!!

P(D=T) = P(D=T, B=T) + P(D=T, B=F)
P(D=T) = P(D=T|B=T)P(B=T) + P(D=T|B=F)P(B=F)
P(D=T) = ( 0.8 )( 0.6 ) + ( 0.2 )( 0.4 ) = 0.56

P(C = T and D = T) = P(C) * P (D)
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