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(B – 3 points) Using Network #2 and the probabilities given below, calculate the probability
of the following:
⎧
A = OFF, B = OFF
⎪ 0.1
P(A = ON ) = 0.6
⎪ 0.9
A = ON, B = OFF
P(D = ON | A, B) = ⎨
A = OFF, B = ON
⎪ 0.3
⎧⎪ 0.1, A = OFF
⎪
0.95
P(B = ON | A) = ⎨
A = ON, B = ON
⎩
⎪⎩ 0.95, A = ON
⎧⎪ 0.8, A = OFF
⎧⎪ 0.8, D = OFF
P(E = ON | D) = ⎨
P(C = ON | A) = ⎨
⎩⎪ 0.5, A = ON
⎩⎪ 0.1, D = ON
(i) P(A=ON, B=ON, C=ON, D=ON, E=ON)
P(A=ON, B=ON, C=ON, D=ON, E=ON) = P(A=ON)P(B=ON|A=ON)P(C=ON|A=ON)P(D=ON|A=ON,B=ON)P(E=ON|D=ON) = (0.6)(0.95)(0.5)(0.95)(0.1) = 0.0271 (ii) P(E = ON | A = ON)
B, D, and E are conditionally independent of C given A, so C drops out. Therefore, we
sum over the 4 {B, D} possibilities:
P(E = ON | A = ON) =
∑ P(E = ON | D)P(D | A = ON,B)P(B | A = ON)
B,D={ON,OFF}
B
ON
ON
OFF
OFF
D
ON
OFF
ON
OFF
P(B|A=ON)
0.95
0.95
0.05
0.05
P(D|A=ON,B)
0.95
0.05
0.9
0.1
P(E=ON|D)
0.1
0.8
0.1
0.8
P(E=ON, B, D|A=ON)
0.09025
0.038
0.0045
0.004
Summing over the last column, we obtain P(E=ON | A = ON) = 0.13675. 3