Karteikarten
Ohne Titel
Karten in diesem Stapel (2)
- 1—Prob(A=T) = 0.3
Prob(B=T) = 0.6
Prob(C=T|A=T) = 0.8
Prob(C=T|A=F) = 0.4
Prob(D=T|A=T,B=T) = 0.7
Prob(D=T|A=T,B=F) = 0.8
Prob(D=T|A=F,B=T) = 0.1
Prob(D=T|A=F,B=F) = 0.2
Prob(E=T|C=T) = 0.7
Prob(E=T|C=F) = 0.2 - 2-What is Prob(D=F,C=T)? Prob(A=T) = 0.3 Prob(B=T) = 0.6 Prob(C=T|A=T) = 0.8 Prob(C=T|A=F) = 0.4 Prob(D=T|A=T,B=T) = 0.7 Prob(D=T|A=T,B=F) = 0.8 Prob(D=T|A=F,B=T) = 0.1 Prob(D=T|A=F,B=F) = 0.2 Prob(E=T|C=T) = 0.7 Prob(E=T|C=F) = 0.2Prob(D=F,C=T)
P(D=F,C=T) =
P(D=F,C=T,A=T,B=T) + P(D=F,C=T,A=T,B=F) +
P(D=F,C=T,A=F,B=T) + P(D=F,C=T,A=F,B=F) =
P(D=F,C=T|A=T,B=T) P(A=T,B=T) + P(D=F,C=T|A=T,B=F) P(A=T,B=F) +
P(D=F,C=T|A=F,B=T) P(A=F,B=T) + P(D=F,C=T|A=F,B=F) P(A=F,B=F) =
(since C and D are independent given A and B)
P(D=F|A=T,B=T) P(C=T|A=T,B=T) P(A=T,B=T) + P(D=F|A=T,B=F) P(C=T|A=T,B=F) P(A=T,B=F) +
P(D=F|A=F,B=T) P(C=T|A=F,B=T) P(A=F,B=T) + P(D=F|A=F,B=F) P(C=T|A=F,B=F) P(A=F,B=F) =
(since C is independent of B given A and A and B are independent absolutely)
P(D=F|A=T,B=T) P(C=T|A=T) P(A=T) P(B=T) + P(D=F|A=T,B=F) P(C=T|A=T) P(A=T) P(B=F) + P(D=F|A=F,B=T) P(C=T|A=F) P(A=F) P(B=T) + P(D=F|A=F,B=F) P(C=T|A=F) P(A=F) P(B=F) =
0.3*0.8*0.3*0.6 + 0.2*0.8*0.3*0.4 + 0.9*0.4*0.7*0.6 + 0.8*0.4*0.7*0.4 = 0.3032
Prob(A=T|C=T)
P(A=T|C=T) = P(C=T|A=T)P(A=T) / P(C=T).
Now P(C=T) = P(C=T,A=T) + P(C=T,A=F) =
P(C=T|A=T)P(A=T) + P(C=T|A=F)P(A=F) =
0.8*0.3+ 0.4*0.7 = 0.52
So P(C=T|A=T)P(A=T) / P(C=T) = 0.8*0.3/0.52= 0.46.
Informationen
- Beschreibung
- Intro to AI Karteikarten am Ohne Titel, erstellt von tifani.obrien am 09/05/2014.
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- tifani.obrien
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