Each of the three jewellery boxes has two drawers. In each drawer of the first box there is a gold watch. In each drawer of the second box there is a silver watch. In one drawer of the third box there is a gold watch while in the other there is a silver watch. One box is selected at random, opened one of the drawers and it contains a gold watch. What is the probability that the other drawer also has a gold watch?

Let's define the problem in terms of probability. We have three boxes, and each box has two drawers. Here are the contents of each box:

1. **Box 1**: Gold watch in both drawers (GG).
2. **Box 2**: Silver watch in both drawers (SS).
3. **Box 3**: One Gold watch and one Silver watch (GS).

We are told that a box is selected at random, and then a drawer is opened, revealing a gold watch. We need to find the probability that the other drawer in the same box also contains a gold watch.

### Step-by-Step Solution:

1. **Probability of selecting each box**: Since the box is selected at random, the probability of choosing each box is \(\frac{1}{3}\).

2. **Probability of finding a gold watch in each box**:
   - For **Box 1 (GG)**: Probability of choosing a drawer with a gold watch is 1 (both drawers have gold).
   - For **Box 2 (SS)**: Probability of choosing a drawer with a gold watch is 0 (no gold watch).
   - For **Box 3 (GS)**: Probability of choosing a drawer with a gold watch is \(\frac{1}{2}\) (one of the two drawers has a gold watch).

3. **Calculating the probability of selecting a box given that a gold watch is found**:
   We use Bayes' theorem to find the probability that it is Box 1 (GG) given that we found a gold watch.

   Let \( A_1 \) be the event that Box 1 is chosen, \( A_3 \) be the event that Box 3 is chosen, and \( G \) be the event that a gold watch is found.

   \[
   P(A_1 \mid G) = \frac{P(G \mid A_1) \cdot P(A_1)}{P(G)}
   \]

   - \( P(G \mid A_1) = 1 \) (Probability of finding a gold watch in Box 1)
   - \( P(G \mid A_3) = \frac{1}{2} \) (Probability of finding a gold watch in Box 3)
   - \( P(G \mid A_2) = 0 \) (Probability of finding a gold watch in Box 2)

   - \( P(A_1) = \frac{1}{3} \)
   - \( P(A_2) = \frac{1}{3} \)
   - \( P(A_3) = \frac{1}{3} \)

   So, the total probability of finding a gold watch \( P(G) \) is:

   \[
   P(G) = P(G \mid A_1) \cdot P(A_1) + P(G \mid A_3) \cdot P(A_3) + P(G \mid A_2) \cdot P(A_2)
   \]

   \[
   P(G) = 1 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} + 0 \cdot \frac{1}{3}
   \]

   \[
   P(G) = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}
   \]

4. **Finding \( P(A_1 \mid G) \)**:

   \[
   P(A_1 \mid G) = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{2}{3}
   \]

5. **Final Answer**:
The probability that the other drawer also has a gold watch, given that one drawer has a gold watch, is \(\boxed{\frac{2}{3}}\).

Do you want more details or have any questions? Here are five related questions:

1. How would the probability change if the third box had two silver watches?
2. What is the probability of choosing Box 3 given that a gold watch is found?
3. How does Bayes' theorem help in determining conditional probabilities in such problems?
4. If there were four boxes with different combinations, how would you set up the problem?
5. Can you explain how independent and dependent events influence probability in this scenario?

**Tip:** When solving probability problems, clearly define events and conditions to avoid confusion and ensure accurate results.

Explore how to calculate the probability that a randomly selected jewellery box contains two gold watches, given that one drawer reveals a gold watch. This problem uses Bayes' Theorem and conditional probability, making it ideal for students in Grades 9-12.

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