Bayes Network Homework

Question

I have a homework due in 3 hours and absolutely no idea how to solve this :

To 1: $B$ can be either $0$ OR $1$ .. $P(D=1| A=1) = P(A=1) * ...$ (I have no idea what to put here)

To 2: $B$ and $C$ are independent. Therefore : $P(B=0|C=0) = P(B=0) = 1-\alpha + 1-\beta$

Answer 1

For the first question, note that: \begin{align*} &\Pr(D = 1 \mid A = 1) \\ &= \frac{1}{\Pr(A = 1)} \Pr(A = 1, D = 1) \\ &= \frac{1}{\Pr(A = 1)} \sum_b \sum_c \Pr(A = 1, B = b, C = c, D = 1) \\ &= \frac{1}{\Pr(A = 1)} \sum_b \sum_c \Pr(A = 1)\Pr(B = b \mid A = 1)\Pr(C = c)\Pr(D = 1 \mid B = b, C = c) \\ &= \sum_b \Pr(B = b \mid A = 1)\sum_c \Pr(C = c)\Pr(D = 1 \mid B = b, C = c) \\ \end{align*} Now just plug in the four possible pairs of $(b, c)$ to get your answer.

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For the second question, note that you need to weight the two options according to $\Pr(A)$: \begin{align*} \Pr(B = 0 \mid C = 0) &= \Pr(B = 0) \\ &= \sum_a \Pr(A = a, B = 0) \\ &= \sum_a \Pr(A = a)\Pr(B = 0 \mid A = a) \\ &= \Pr(A = 0)\Pr(B = 0 \mid A = 0) + \Pr(A = 1)\Pr(B = 0 \mid A = 1) \\ &= 0.3(1 - \beta) + 0.7(1 - \alpha) \\ \end{align*}

Answer 2

For part 1: Naively I would say you could "replace" the terms in the table for D, so that B=1 becomes a and B=0 1-a, C=1 to g and C=0 is 1-g. And then it's just a matter of writing all terms for D=1 out without interpunction. P(D=1|A=1) = 0.4ag+0.2a(1-g)+0.7(1-a)g+0.5(1-a)(1-g)... Which you can simplify..