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?

This is a classic problem of conditional probability, and we can approach it using Bayes' Theorem. The question is essentially asking for the probability that you have chosen the first box (which has two gold watches) given that you have already opened one drawer and found a gold watch.

### Step 1: Define the Events
Let’s define the following events:

- \( A_1 \): The event that the first box (with two gold watches) was chosen.
- \( A_2 \): The event that the second box (with two silver watches) was chosen.
- \( A_3 \): The event that the third box (with one gold and one silver watch) was chosen.
- \( B \): The event that a gold watch was found when a drawer was opened.

We are looking for \( P(A_1 \mid B) \), the probability that we chose the first box given that a gold watch was found.

### Step 2: Use Bayes' Theorem
Bayes’ Theorem gives us:

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

Where:

- \( P(A_1) \) is the prior probability of choosing the first box, which is \( \frac{1}{3} \) since each box is equally likely to be chosen.
- \( P(B \mid A_1) \) is the probability of finding a gold watch given that we chose the first box. Since both drawers contain a gold watch, this is \( 1 \).
- \( P(B) \) is the total probability of finding a gold watch. We calculate this by considering all possible cases:

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

Now let’s calculate each term in \( P(B) \):

- \( P(B \mid A_1) = 1 \) (since both drawers in the first box have gold watches),
- \( P(B \mid A_2) = 0 \) (since the second box has only silver watches),
- \( P(B \mid A_3) = \frac{1}{2} \) (since one of the drawers in the third box has a gold watch).

Thus:

\[
P(B) = (1 \cdot \frac{1}{3}) + (0 \cdot \frac{1}{3}) + \left(\frac{1}{2} \cdot \frac{1}{3}\right) = \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{2}
\]

### Step 3: Calculate \( P(A_1 \mid B) \)
Now, we can apply Bayes' Theorem:

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

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

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Do you need more details, or have any questions on this?  
Here are 5 related questions to extend your understanding:

1. How would the probability change if there were 4 boxes instead of 3?
2. What if one box had one gold and two silver watches—how would the setup change?
3. How does Bayes’ Theorem apply to real-world decision-making situations?
4. Can you explain how conditional probability is different from regular probability?
5. What are some limitations of using Bayes' Theorem?

**Tip**: When solving probability problems, carefully define your events and apply Bayes’ Theorem step by step to simplify the process.

Explore a classic conditional probability problem involving three jewellery boxes with gold and silver watches. Learn how to use Bayes' Theorem to determine the probability that the other drawer also has a gold watch. This detailed solution is suitable for students in Grades 11-12.

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