An example about independence of three discrete random variables Let A, B and C be random variables with discrete probability distributions. Consider the following two joint probability tables:

Can we know that whether A and C are independent or not? Can we compute the conditional probability $P(B=b_1|A=a_2,C=c_3)$?
I am truly confused about the independence. I think this is a good example to figure it out. Any help is appreiciated. Thank you!
 A: We compute the marginal distributions of $A$, $B$, and $C$ as follows:
\begin{align}
\mathbb P(A=a_1) &= \sum_{j=1}^3 \mathbb P(A=a_1,B=b_j) = \frac1{10}+\frac1{20}+\frac3{20} = \frac3{10}\\
\mathbb P(A=a_2) &= \sum_{j=1}^3 \mathbb P(A=a_2,B=b_j) = \frac1{10}+\frac1{20}+\frac3{10} = \frac9{20}\\
\mathbb P(A=a_3) &= \sum_{j=1}^3 \mathbb P(A=a_3,B=b_j) = \frac1{20}+\frac3{20}+\frac1{20} = \frac14\\
\mathbb P(B=b_1) &= \sum_{j=1}^3 \mathbb P(B=b_1,A=a_j) = \frac1{10}+\frac1{10}+\frac1{20}=\frac14\\
\mathbb P(B=b_2) &= \sum_{j=1}^3 \mathbb P(B=b_2,A=a_j) = \frac1{20}+\frac1{20}+\frac3{20}=\frac14\\
\mathbb P(B=b_3) &= \sum_{j=1}^3 \mathbb P(B=b_3,A=a_j) = \frac3{20}+\frac3{10}+\frac1{20}=\frac12\\
\mathbb P(C=c_1) &= \sum_{j=1}^3 \mathbb P(C=c_1,B=b_j) = \frac1{50}+\frac3{100}+\frac7{20} = \frac25\\
\mathbb P(C=c_2) &= \sum_{j=1}^3 \mathbb P(C=c_2,B=b_j) = \frac7{50}+\frac1{20}+\frac1{25}=\frac{23}{100}\\
\mathbb P(C=c_3) &= \sum_{j=1}^3 \mathbb P(C=c_3,B=b_j) = \frac3{50}\\
\mathbb P(C=c_4) &= \sum_{j=1}^3 \mathbb P(C=c_4,B=b_j) = \frac3{100} + \frac{17}{100} + \frac{11}{100} = \frac{31}{100}.
\end{align}
$A$ and $C$ are independent if and only if $\mathbb P(A=a_i,C=c_j) = \mathbb P(A=a_i)\mathbb P(C=c_j)$ for $i=1,2,3$ and $j=1,2,3,4$. I will leave this to you to compute.
The conditional probability $\mathbb P(B=b_1\mid A=a_2,C=c_3)$ is given by
$$
\frac{\mathbb P(B=b_1,A=a_2,C=c_3)}{\mathbb P(A=a_2,C=c_3}.
$$
Again, I will leave this to you to compute.
