find the probability the black marble is from box A and the blue one is from C 
(HMMT 2000 Guts Round #43) Box A contains 3 black marbles and 4 blue marbles. Box B has 7 black marbles and 1 blue marble. Box C has 2 black marbles, 3 blue marbles, and 1 green marble. Person A closes their eyes and picks two marbles from 2 different boxes. If it turns out that A gets $1$ black and 1 blue marble, what is the probability that the black marble is from box A and the blue one is from C?


My question is at the bottom.

We assume the process is equivalent to uniformly choosing 2 distinct boxes and then independently and randomly choosing a marble from each box (!)
Let the colours black, blue, and green be denoted by the numbers 1,2,3. A has 3 black and 4 blue marbles, B has 7 black and 1 blue, and C has 2 black, 3 blue, and 1 green. Consider the sample space $\Omega := \{\{(m_1,b_1),(m_2,b_2)\} : b_i \in \{A,B,C\}, m_i \in \{1,2,3\}, i = 1,2, b_1\neq b_2\}$. Here, the $b_i$'s in a pair refer to the boxes chosen. Let $X = \{\{(m_1,b_1),(m_2,b_2)\} \in \Omega : (m_1 = 1 \wedge m_2 = 2)\},Y = \{\{(m_1,b_1),(m_2,b_2)\} \in \Omega : (m_1 = 1 \wedge m_2 = 2 \wedge b_1 = A \wedge b_2 = C)  \}.$
For $R\neq S \in \{A,B,C\},$ let $P_{R,S} = \{\{(m_1,b_1),(m_2,b_2)\} \in \Omega : b_1 = R \wedge m_2 = S\}.$ We want to find $P(Y | X) = P(Y\cap X)/P(X)$. To find $P(X),$ we first find $P(X\cap P_{A,B}) , P(X\cap P_{A,C}) , P(X\cap P_{B,C})$. Now $P(X\cap P_{A,B}) = P(X | P_{A,B}) \cdot P(P_{A,B}). P(P_{A,B}) = P(P_{A,C}) = P(P_{B,C}) = 1/3,$ since every pair of two distinct boxes is selected with equal probability (by (!)). Now $P(X | P_{A,B}) = 3/7\cdot 1/8 + 4/7\cdot 7/8 = 31/56.$ Similarly, $P(X | P_{A,C}) = 3/7\cdot 3/6 + 4/7\cdot 2/6 = 17/42, P(X|P_{B,C}) = 7/8\cdot 3/6 + 1/8\cdot 2/6 = 23/48.$ Thus $P(X) = 1/3(31/56+17/42+23/48) = (483/1008).$ To find $P(Y\cap X),$ we find $P(Y\cap X \cap P_{A,B}), P(Y\cap X \cap P_{A,C}), P(Y\cap X \cap P_{B,C})$. $P(Y\cap X \cap P_{A,B}) = P(Y\cap X \cap P_{B,C}) = P(\emptyset) = 0.$ Now  $P(Y\cap X \cap P_{A,C}) = P(Y\cap X | P_{A,C}) P(P_{A,C}) = 1/3 \cdot (3/7\cdot 3/6) = 1/14.$ So the desired probability is $\dfrac{24}{161}$. This is different from the solution on HMMT.org, which is $120/1147$.

Did I make a mistake somewhere, and if so, where and how can I get the correct answer?


The simplest way is to make an array.
Since the probabilities of choosing each box is equal, we can tabulate conditional probabilities of drawing black-blue in that order given we have chosen the indicated boxes
\begin{array}{| c | c | c | c | c | c |}\hline \\Combo &AB & BA & AC & CA & BC & CB \\\hline Black-Blue &\frac37\frac18 & \frac78\frac47 & \frac37\frac36 &\frac26\frac47 & \frac78\frac36 & \frac26\frac18 \\ \hline \end{array}
$$Pr =\dfrac{\frac37\frac36}{\frac37\frac18 + \frac78\frac47 + \frac37\frac36 +\frac26\frac47\ + \frac78\frac36 + \frac26\frac18 }=\frac{24}{161}$$
