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 $P(A=1|D=1)=\dfrac{P(D=1|A=1)P(A=1)}{P(D=1)}=\dfrac{0.7*0.9}{0.63}=1$
So our answer is:
$P(C=0|A=1,D=1)*0.63$
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 $C=0$.
$$\begin{align}\P(C^0, A^1, D^1)&=\P(A^1,D^1,C^0)\\[1ex]&=\P(A^1)\P(D^1\mid A^1)\P(C^0\mid A^1)\\[1ex]&=\P(A^1)\P(D^1\mid A^1)\big(\P(C^0\mid B^0)\P(B^0\mid A^1)+\P(C^0\mid B^1)\P(B^1\mid A^1)\big)\end{align}$$
$$\small\P(C{=}0,A{=}1,D{=}1)=\P(A{=}1)\P(D{=}1\mid A{=}1)\big(\P(C{=}0\mid B{=}0)\P(B{=}0\mid A{=}1)+\P(C{=}0\mid B{=}1)\P(B{=}1\mid A{=}1)\big)$$