Given three identical Boxes A, B and C, Box A contains 2 gold and 1 silver coins, Box B contains 1 gold and 2 silver coins and Box C contains 3 silver coins. A person chooses a Box at random and takes out a coin. If the coin drawn is of silver, find the probability that it has been drawn from the Box which has the remaining two coins also of silver.

Answer

To solve the problem, we need to find the probability that a silver coin drawn came from Box C, which contains only silver coins, given that a silver coin was drawn. We can use Bayes' theorem to find this probability.

### Step-by-Step Solution:

1. **Identify the Boxes and Their Contents:**
- Box A: 2 Gold coins, 1 Silver coin
- Box B: 1 Gold coin, 2 Silver coins
- Box C: 0 Gold coins, 3 Silver coins

2. **Calculate the Total Probability of Drawing a Silver Coin (P(S)):**
- Probability of choosing any box (P(A), P(B), P(C)) = 1/3
- Probability of drawing a silver coin from each box:
- From Box A: P(S|A) = 1/3 (1 silver out of 3 total coins)
- From Box B: P(S|B) = 2/3 (2 silver out of 3 total coins)
- From Box C: P(S|C) = 3/3 = 1 (3 silver out of 3 total coins)

Now, we calculate the total probability of drawing a silver coin:
\[ P(S) = P(A) \cdot P(S|A) + P(B) \cdot P(S|B) + P(C) \cdot P(S|C) \]
\[ P(S) = \frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} \cdot \frac{2}{3} + \frac{1}{3} \cdot 1 \]
\[ P(S) = \frac{1}{9} + \frac{2}{9} + \frac{3}{9} = \frac{6}{9} = \frac{2}{3} \]

3. **Calculate the Probability of Drawing from Box C Given a Silver Coin (P(C|S)):**
Using Bayes' theorem:
\[ P(C|S) = \frac{P(C) \cdot P(S|C)}{P(S)} \]
Substituting the values:
\[ P(C|S) = \frac{\frac{1}{3} \cdot 1}{\frac{2}{3}} = \frac{1/3}{2/3} = \frac{1}{2} \]

4. **Conclusion:**
The probability that the silver coin drawn came from Box C, which contains only silver coins, is \( \frac{1}{2} \).