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I am having problem solving this exercise. The problem actually comes with a diagram but I do not know and I do not think i can draw it in the forum. The exercise is based on car starting(Heckerman 1995)
Since I can't draw the network diagram here but values of probability are given but first let me define all the variables
B - Battery
G - Gauge
F - Fuel
T - Turnover
S - Start
N - No
Y - Yes
P(B = N) = 0.02
p(F = N) = 0.05
P(G = N|B = Y, F = Y) = 0.04
P(G = N|B = Y, F = N) = 0.97
P(G = N|B = N, F = Y) = 0.10
P(G = N|B = N, F = N) = 0.99
P(T = N|B = Y) = 0.03
P(T = N|B = N) = 0.98
P(S = N|T = Y, F = Y) = 0.01
P(S = N|T = Y, F = N) = 0.92
P(S = N|T = N, F = Y) = 1.0
P(S = N|T = N, F = N) = 1.0
It was asked to calculate p(F = N|S = N)
Im thinking of Bayesian but I got stuck somewhere so I think it is the wrong approach since S depend on F and T NOT F alone.
Im thinking of the other approach and came up with an expression
P(F = N|S = N) = P(S = N|F = N)/P(F)
= P(S = N, B, G, T|F = N)/P(F)
But I am not sure how to compute and put the figures together.
Any input/help is appreciated. Thank you
Since I can't draw the network diagram here but values of probability are given but first let me define all the variables
B - Battery
G - Gauge
F - Fuel
T - Turnover
S - Start
N - No
Y - Yes
P(B = N) = 0.02
p(F = N) = 0.05
P(G = N|B = Y, F = Y) = 0.04
P(G = N|B = Y, F = N) = 0.97
P(G = N|B = N, F = Y) = 0.10
P(G = N|B = N, F = N) = 0.99
P(T = N|B = Y) = 0.03
P(T = N|B = N) = 0.98
P(S = N|T = Y, F = Y) = 0.01
P(S = N|T = Y, F = N) = 0.92
P(S = N|T = N, F = Y) = 1.0
P(S = N|T = N, F = N) = 1.0
It was asked to calculate p(F = N|S = N)
Im thinking of Bayesian but I got stuck somewhere so I think it is the wrong approach since S depend on F and T NOT F alone.
Im thinking of the other approach and came up with an expression
P(F = N|S = N) = P(S = N|F = N)/P(F)
= P(S = N, B, G, T|F = N)/P(F)
But I am not sure how to compute and put the figures together.
Any input/help is appreciated. Thank you