Is my Bayes belief network theory correct?

Question

I am currently trying to learn Bayes' theorem, and in turn, Bayesian belief networks. I haven't done any 'real' maths in nearly 20 years, so I am rusty to say the least. I am trying to determine the probability of A given B and C. Therefore if:

B(y/n)

By = 0.2
Bn = 0.8

C(h/s)

Ch = 0.2
Cs = 0.8

A(h/l)

By, Ch = 0.2, 0.8
By, Cs = 0.5, 0.5
Bn, Ch = 0.6, 0.4
Bn, Cs = 0.8, 0.2

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So if: $$ P(A∣BC)=\frac{P(A∣B)P(A∣C)}{P(A)} $$

Surely:

$$ P(Ah∣BnCs)=\frac{P(Ah∣Bn)P(Ah∣Cs)}{P(Ah)} $$

Must equal 0.512?

I have assigned the values based on prior knowledge/belief of what they should be from my own experience.

Answer 1

Your table says $\mathsf P(Ah\mid Bn,Cs)=0.8$ .$$\boxed{\begin{array}{|l|l|l|l|}\hline & & \color{red}{Ah} & Al\\\hline By & Ch & 0.2 & 0.8\\ By & Cs & 0.5 & 0.5\\ Bn & Ch & 0.6 & 0.4\\ \color{red}{Bn} &\color{red}{Cs} &\color{red}{ 0.8} & 0.2\\\hline\end{array}}$$

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Nothing indicates: $\mathsf P(A\mid B,C)=\tfrac{\mathsf P(A\mid B)\mathsf P(A\mid C)}{\mathsf P(A)}$ . What makes you think this would be true? That's not Bayes' Theorem. You cannot even read those terms directly from the given tables, so it would not even be useful.

Answer 2

Bayes' theorem states:

$$P(A | B, C) = \frac{ P(B | A, C) P(A | C)}{P(B | C)}$$

which is not at all what you have.