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Problem 611

AMC 10/12, early questions

Combinatorics Difficulty 3.7 Multiple choice CEMC Fermat · Canada · 2016

Box 1 contains one gold marble and one black marble. Box 2 contains one gold marble and two black marbles. Box 3 contains one gold marble and three black marbles. Whenever a marble is chosen randomly from one of the boxes, each marble in that box is equally likely to be chosen. A marble is randomly chosen from Box 1 and placed in Box 2. Then a marble is randomly chosen from Box 2 and placed in Box 3. Finally, a marble is randomly chosen from Box 3. What is the probability that the marble chosen from Box 3 is gold?

Pick one

Official solution

When a marble is chosen from Box 1, the probability is $\frac{1}{2}$ that it will be gold and the probability is $\frac{1}{2}$ that it will be black.

Thus, after this choice is made, there is a probability of $\frac{1}{2}$ that Box 2 contains 2 gold marbles and 2 black marbles, and there is a probability of $\frac{1}{2}$ that Box 2 contains 1 gold marble and 3 black marbles.

In the first case (which occurs with probability $\frac{1}{2}$ ), the probability that a gold marble is chosen from Box 2 is $\frac{2}{4}=\frac{1}{2}$ and the probability that a black marble is chosen from Box 2 is $\frac{2}{4}=\frac{1}{2}$ .

In the second case (which occurs with probability $\frac{1}{2}$ ), the probability that a gold marble is chosen from Box 2 is $\frac{1}{4}$ and the probability that a black marble is chosen from Box 2 is $\frac{3}{4}$ .

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Therefore, the probability that a gold marble is chosen from Box 2 is $\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{1}{4} = \frac{3}{8}$ and the probability that a black marble is chosen from Box 2 is $\frac{1}{2}\cdot \frac{1}{2}+\frac{1}{2}\cdot \frac{3}{4} = \frac{5}{8}$ .

Thus, after this choice is made, there is a probability of $\frac{3}{8}$ that Box 3 contains 2 gold marbles and 3 black marbles, and a probability of $\frac{5}{8}$ that Box 3 contains 1 gold marble and 4 black marbles.

Finally, the probability that a gold marble is chosen from Box 3 equals the probability that Box 3 contains 2 gold marbles and 3 black marbles times the probability of choosing a gold marble in this situation (that is, $\frac{3}{8}\cdot \frac{2}{5}$ ) plus the probability that Box 3 contains 1 gold marble and 4 black marbles times the probability of choosing a gold marble in this situation.

In other words, this probability is $\frac{3}{8}\cdot \frac{2}{5} + \frac{5}{8}\cdot \frac{1}{5} = \frac{11}{40}$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.