Bayesian network network problem

I want to find:

$$P(C=0,A=1,D=1)$$.

I know the following:

$$P(C=0,A=1,D=1) = P(C=0|A=1, D=1)*P(A=1|D=1)*P(D=1)$$

From the image, we can see $$P(D=1)$$ is $$P(D=1) = P(D=1|A=0)*P(A=0) + P(D=1|A=1)*P(A=1) = 0 + 0.7 * 0.9 = 0.63$$

I also know $$P(A=1|D=1)=\dfrac{P(D=1|A=1)P(A=1)}{P(D=1)}=\dfrac{0.7*0.9}{0.63}=1$$

$$P(C=0|A=1,D=1)*0.63$$

I don't know how to calculate the first part.

I think it's $$\dfrac{P(C=0,A=1,D=1)}{P(A=1,D=1)}$$, but the numerator is just the entire question again and denominator we are not given.

My textbook says answer should be $$0.3024$$

You have $$\lower{1ex}D\raise{1ex}{\swarrow \raise{2ex}A\searrow}\lower{1ex}{ B\lower{2ex}{\searrow\lower{2ex} C}}$$ and wish to find $$\def\P{\operatorname{\mathsf P}}\P(C^0, A^1, D^1)$$, a shorthand for $$\P(C{=}0,A{=}1,D{=}1)$$.

The goal is to expand this out into terms whose values may be read off the tables.

Well, $$A$$ is root, and $$C,D$$ are leaves of separate branches, so condition over $$A=1$$, then expand out the branch for $$C=0$$.

\begin{align}\P(C^0, A^1, D^1)&=\P(A^1,D^1,C^0)\\[1ex]&=\P(A^1)\P(D^1\mid A^1)\P(C^0\mid A^1)\\[1ex]&=\P(A^1)\P(D^1\mid A^1)\big(\P(C^0\mid B^0)\P(B^0\mid A^1)+\P(C^0\mid B^1)\P(B^1\mid A^1)\big)\end{align}

$$\small\P(C{=}0,A{=}1,D{=}1)=\P(A{=}1)\P(D{=}1\mid A{=}1)\big(\P(C{=}0\mid B{=}0)\P(B{=}0\mid A{=}1)+\P(C{=}0\mid B{=}1)\P(B{=}1\mid A{=}1)\big)$$

• However it seems $C=0$ is not given on the graph, so it has probability $0$. Thus the right most equation all has probability $0$, and this would overall give a value of $0$ Commented Mar 26, 2019 at 21:51
• We can assume the Random Variables are either $1$ or $0$ Commented Mar 26, 2019 at 21:52
• Yes, complements are found the usual way: $\mathsf P(C{=}0\mid B{=}1)=1-\mathsf P(C{=}1\mid B{=}1)=0.3$ Commented Mar 26, 2019 at 21:57
• Do the same rules apply if we are dealing with 4 variables? For example, would $P(A^1,B^0,C^1,D^0)=P(A^1)P(B^0|A^1)P(D^0|A^1)P^(C^1|A^1)$, where $P(C^1|A^1)=P(C^1|B^0)P(B^0|A^1)+P(C^1|B^1)P(B^1|A^1)$? Commented Mar 26, 2019 at 22:15
• Yes but... If you already have a bridge between A and C just use it. $P(A^1)P(B^0\mid A^1)P(C^0\mid B^0)P(D^0\mid A^1)$ Commented Mar 26, 2019 at 22:51