(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?
Best 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}
$$Pr =\dfrac{\frac37\frac36}{\frac37\frac18 + \frac78\frac47 + \frac37\frac36 +\frac26\frac47\ + \frac78\frac36 + \frac26\frac18 }=\frac{24}{161}$$