I'm looking at this problem as I review for an exam and I would appreciate it if anyone can give me work/answers to Pr(c), Pr(b), Pr(b,c) and Pr(c,d) so I can check my solutions, and use the work to figure out how I was off if so.
so far i have gotten
Pr(c) = Pr(c|a)*Pr(a) + Pr(c| !a)*Pr(!a) = .48
Pr(b) = Pr(b|a)*Pr(a) + Pr(b| !a)*Pr(!a) = .56
not 100% sure or if these are correct or how to proceed when it comes to the next 2
What you have done so far for the calculation of $p(b)$ and $p(c)$ is correct:
$$p(c) = p(c|a)\,p(a) + p(c|\neg a)\,p(\neg a) = 0.48$$
$$p(b) = p(b|a)\,p(a) + p(b|\neg a)\,p(\neg a) = 0.56$$
For the calculation of $p(b,c)$, notice that $$p(b,c)=p(b,c,a)+p(b,c,\neg a)=p(b,c|a)\,p(a)+p(b,c|\neg a)\,p(\neg a)$$
Since $b$ and $c$ are independent given $a$ we have that $$p(b,c)=p(b|a)\,p(c|a)\,p(a) + p(b|\neg a)\,p(c|\neg a)\,p(\neg a)=0.24$$
Finally, we proceed similarly for the calculation of $p(c,d)$. Notice that $$p(c,d)=p(c,d,b) + p(c,d,\neg b)$$
We know from before that $p(b,c)=0.24$ and thus $$p(c,d,b)=p(d|b,c)\,p(b,c)=0.192$$
We can calculate $p(\neg b, c)$ the same way we did for $p(b,c)$ (giving also 0.24) and thus $$p(c,d, \neg b)=p(d|\neg b,c)\,p(\neg b,c)=0.144$$
Therefore, $p(c,d)=0.336$.