Given three identical boxes I, II and III, each containing two coins. In box I both coins are gold coins, in box II both are silver coins and in box III there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the first coin is gold, what is the probability that the other coin in the box is gold.

Answer

To solve the problem, we need to determine the probability that the other coin in the box is gold given that the first coin drawn is gold. We will use Bayes' theorem for this calculation.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( B_1 \): the event of selecting Box I (two gold coins).
- Let \( B_2 \): the event of selecting Box II (two silver coins).
- Let \( B_3 \): the event of selecting Box III (one gold and one silver coin).
- Let \( G \): the event that the first coin drawn is gold.

2. **Calculate Prior Probabilities**:
Since the boxes are chosen at random:
\[ P(B_1) = P(B_2) = P(B_3) = \frac{1}{3} \]

3. **Calculate Conditional Probabilities**:
- If Box I is chosen, the probability of drawing a gold coin:
\[ P(G | B_1) = 1 \quad \text{(both coins are gold)} \]
- If Box II is chosen, the probability of drawing a gold coin:
\[ P(G | B_2) = 0 \quad \text{(both coins are silver)} \]
- If Box III is chosen, the probability of drawing a gold coin:
\[ P(G | B_3) = \frac{1}{2} \quad \text{(one gold and one silver)} \]

4. **Apply Total Probability Theorem**:
We need to find \( P(G) \):
\[ P(G) = P(G | B_1) P(B_1) + P(G | B_2) P(B_2) + P(G | B_3) P(B_3) \]
Substituting the values:
\[ P(G) = 1 \cdot \frac{1}{3} + 0 \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{3} \]
\[ P(G) = \frac{1}{3} + 0 + \frac{1}{6} = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

5. **Use Bayes' Theorem**:
We want to find \( P(B_1 | G) \):
\[ P(B_1 | G) = \frac{P(G | B_1) P(B_1)}{P(G)} \]
Substituting the known values:
\[ P(B_1 | G) = \frac{1 \cdot \frac{1}{3}}{\frac{1}{2}} = \frac{\frac{1}{3}}{\frac{1}{2}} = \frac{1}{3} \cdot \frac{2}{1} = \frac{2}{3} \]

6. **Conclusion**:
The probability that the other coin in the box is gold, given that the first coin drawn is gold, is:
\[ \boxed{\frac{2}{3}} \]