# Belief network problem

1. Nov 16, 2007

### sensitive

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

2. Nov 16, 2007

### EnumaElish

How do you get P(F = N|S = N) = P(S = N|F = N)/P(F) ?

P(F = N|S = N) = P(S = N & F = N)/P(S = N) and P(S = N|F = N) = P(S = N & F = N)/P(F = N) so P(S = N & F = N) = P(F = N|S = N)P(S = N) = P(S = N|F = N)P(F = N).

3. Nov 16, 2007

### sensitive

So the Bayesian approach was right. I taught I was wrong at the first place because using Bayesian ended up with the following

P(S = N|F = N) P(F = N)/ P(S = N)

but from the diagram I have and as you can see from the probabilities, S depend on both F and T and in the expression above we want to know the probability of S = N given that F = N (in other words the probability that the engine will not start given that the fuel tank was empty).

4. Nov 17, 2007

### sensitive

I am still having trouble solving P(S = N|F = N).

plz help....

5. Nov 19, 2007

### EnumaElish

Below, I assume that the notation (A|B,C) means (A|B)|C = A|(B|C), and neither A|(B & C) nor (A|B) & C. (If anyone disagrees, please post your opinion.)

P(S = N|F = N) = P(S = N|T = Y, F = N) P(T = Y) + P(S = N|T = N, F = N) P(T = N) so you should first derive P(T = Y) and P(T = N).

You can derive P(T=N) from:
P(B = N) = 0.02
P(T = N|B = Y) = 0.03
P(T = N|B = N) = 0.98
using a formula similar to the one in the previous paragraph of this post.

Then, P(T=Y) = 1 - P(T=N).

Last edited: Nov 19, 2007