Bayesian Network calculation questions update
The solution follows obtain the right answer now.
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The Question is here

And my answer is here
$P(a_0)=P(a_0|r_0) + P(a_0|r_1) = 0.4$
$P(d_0)=\sum_{r,c}P(d_0|r,c)\cdot P(r)\cdot P(c)=0.64$
\begin{align*}
P(c_0|s_0,d_0,a_0) &= \frac{P(c_0,s_0,d_0,a_0)}{P(s_0,d_0,a_0)}\\                   &=\frac{P(c_0,s_0,d_0,a_0,r_0)+P(c_0,s_0,d_0,a_0,r_1)}{P(s_0,d_0,a_0)}\\
&=\frac{P(c_0) P(s_0|a_0,d_0)[P(r_0)P(d_0|c_0,r_0) P(a_0|r_0)+ P(r_1) P(d_0|c_0,r_1) P(a_0|r_1)]}{P(s_0|a_0,d_0) \cdot \sum_{r,c}P(c)P(r)P(a_0|r)P(d_0|c,r)}\\
&=\frac{0.6\times 0.6 \times (0.5 \times 1 \times 0.2 + 0.5 \times 0.6 \times 0.6)}{0.6 \times 0.208}\\
&=0.807
\end{align*}
Where
\begin{align*}
\sum_{r,c}P(c)P(r)P(a_0|r)P(d_0|c,r)
&=0.5\sum_{r,c}P(c)P(a_0|r)P(d_0|c,r)\\
&=0.5(0.6\times 1\times 0.2 + 0.6\times 0.6\times 0.6 + 0.4\times 0.7\times 0.2 + 0.4\times 0.1\times 0.6)\\
&=0.208
\end{align*}
I learned Bayesian Network point for the first time |-_-|, what's wrong with my answer?
Any advice will be appreciated!
 A: For the moment I write my brute force calculation as a reference. Maybe there are smarter ways using more properly the properties of Bayesian Networks, but at least I get the correct answer $(d)$. In case I find something smarter will update.
We first define:
$A=p(c_0,s_0,d_0,a_0)$, $B=p(c_1,s_0,d_0,a_0)$
And note that the probability that we want is:
$R=A/(A+B)$
Now:
$A=\sum_{r,f} p(c_0,r,f,d_0,a_0,s_0)=p(c_0)p(s_0|d_0,a_0)\sum_{r,f}p(r)p(d_0|c_0,r)p(a_0|r)p(f|c_0)$
Note that we have $p(r)=0.5$ and $p(f|c_0)=0.5$ for every $r,f$ so that this simplifies:
$A=p(c_0)p(s_0|d_0,a_0)*1/2*1/2*2*\sum_r p(d_0|c_0,r)p(a_0|r)$
Putting numbers inside:
$A=0.6*0.6*0.5*(0.2+0.6*0.6)$
$B$ is a bit more cumbersome. We arrive at:
$B=\sum_{r,f} p(c_1,r,f,d_0,a_0,s_0)=p(c_1)p(s_0|d_0,a_0)1/2 \sum_{r,f}p(d_0|c_1,r)p(a_0|r)p(f|c_1)$
and the four cases $(r_0,f_0),(r_1,f_1),(r_0,f_1),(r_1,f_0)$ must be considered seperately. Up to my calculations:
$B=0.4*0.6*0.5*(0.7*0.2*0.3+0.1*0.6*0.7+0.7*0.2*0.7+0.1*0.6*0.3)$

UPDATE: If we first sum over $f$ due to normaliazation we can simplify further before plugging in the numbers:
$B=p(c_1)p(s_0|d_0,a_0)1/2 \sum_{r}p(d_0|c_1,r)p(a_0|r)$
, leading to the simplified expression for B:
$B=0.4*0.6*0.5*(0.7*0.2+0.1*0.6)$
which is equivalent to the previous one.

Finally $R=A/(A+B) \sim 0.807$, which is answer $(d)$.
A: I add an alternative approach, based on the comments of the OP, to evaluate $p(s_0,a_0,d_0)$, which is in the previous answer the denominator $A+B$.
As suggested by the comments of the OP: $p(s_0,a_0,d_0)=p(s_0|a_0,d_0)p(a_0,d_0)$ and since the Bayesian network provides $p(s_0|a_0,d_0)=0.6$ we just need the second factor, which is equal to:
$p(a_0,d_0)=\sum_{r,c} p(r)p(c)p(a_0|r)p(d_0|r,c)$
( variables $s$ and $f$ do not enter year because they have only outgoing edges from the set of variables $a,d,r,c$ )
Now this is equal to, since $p(r)=0.5$, and considering the four cases $(r,c)$:
$p(a_0,d_0)=0.5*(0.6*0.2*1.0+0.4*0.2*0.7+0.6*0.6*0.6+0.4*0.6*0.1)$
And therefore:
$p(a_0,d_0,s_0)=0.6*0.5*(0.6*0.2*1.0+0.4*0.2*0.7+0.6*0.6*0.6+0.4*0.6*0.1)$
One can numerically verify that this expression is equal to the denominator $A+B$ in the first answer. Of course one still needs to evaluate the numerator $A$ to get to the final result.
