Probability problem - Fermat Grade 11 contest

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 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 $\frac{3}{4}$ .

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

Ergo, 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, $\frac{3}{8}⋅\frac{2}{5}$ ) plus the probability that Box 3 contains $1$ gold marble and $4$ black marbles multiplied by the probability of choosing a gold marble in this situation.

In other words, the probability is $\frac{3}{8}⋅\frac{2}{5}+\frac{5}{8}⋅\frac{1}{5}=\frac{11}{40}=27.5\%$ percent.