Urn A contains 7 white and 3 black balls, um B contains 4 white and 6 black balls, urn C contains 2 white and 8 black balls. One of these urns is chosen at random with probabilities 0.2, 0.6 and 0.2 respectively. From the chosen urn, two balls are drawn at random without replacement. Both the balls happen to be white. Find the probability that the balls drawn are from the urn C.

Explanation

To solve the problem, we will use Bayes' theorem to find the probability that the balls drawn are from urn C, given that both balls drawn are white.

### Step 1: Define the Events
Let:
- \( A \): Event that urn A is chosen.
- \( B \): Event that urn B is chosen.
- \( C \): Event that urn C is chosen.
- \( E \): Event that two white balls are drawn.

### Step 2: Given Probabilities
From the problem, we have:
- \( P(A) = 0.2 \)
- \( P(B) = 0.6 \)
- \( P(C) = 0.2 \)

### Step 3: Calculate Probability of Drawing Two White Balls from Each Urn
1. **For Urn A**:
- Total balls = 10 (7 white + 3 black)
- Probability of drawing 2 white balls without replacement:
\[ P(E|A) = \frac{7}{10} \times \frac{6}{9} = \frac{7 \times 6}{10 \times 9} = \frac{42}{90} = \frac{7}{15} \]

2. **For Urn B**:
- Total balls = 10 (4 white + 6 black)
- Probability of drawing 2 white balls without replacement:
\[ P(E|B) = \frac{4}{10} \times \frac{3}{9} = \frac{4 \times 3}{10 \times 9} = \frac{12}{90} = \frac{2}{15} \]

3. **For Urn C**:
- Total balls = 10 (2 white + 8 black)
- Probability of drawing 2 white balls without replacement:
\[ P(E|C) = \frac{2}{10} \times \frac{1}{9} = \frac{2 \times 1}{10 \times 9} = \frac{2}{90} = \frac{1}{45} \]

### Step 4: Use Total Probability to Find \( P(E) \)
Using the law of total probability:
\[ P(E) = P(E|A)P(A) + P(E|B)P(B) + P(E|C)P(C) \]
Substituting the values:
\[ P(E) = \left(\frac{7}{15} \times 0.2\right) + \left(\frac{2}{15} \times 0.6\right) + \left(\frac{1}{45} \times 0.2\right) \]
Calculating each term:
- For urn A: \( \frac{7}{15} \times 0.2 = \frac{7 \times 0.2}{15} = \frac{1.4}{15} \)
- For urn B: \( \frac{2}{15} \times 0.6 = \frac{2 \times 0.6}{15} = \frac{1.2}{15} \)
- For urn C: \( \frac{1}{45} \times 0.2 = \frac{0.2}{45} \)

Now, convert \( \frac{0.2}{45} \) to a common denominator of 15:
\[ \frac{0.2}{45} = \frac{0.2 \times 15}{45 \times 15} = \frac{3}{675} = \frac{1}{225} \]
Now, we need a common denominator for all three terms:
\[ P(E) = \frac{1.4}{15} + \frac{1.2}{15} + \frac{1}{225} \]
Finding a common denominator (225):
\[ P(E) = \frac{21}{225} + \frac{18}{225} + \frac{1}{225} = \frac{40}{225} = \frac{8}{45} \]

### Step 5: Apply Bayes' Theorem
We want to find \( P(C|E) \):
\[ P(C|E) = \frac{P(E|C)P(C)}{P(E)} \]
Substituting the values:
\[ P(C|E) = \frac{\left(\frac{1}{45}\right) \times 0.2}{\frac{8}{45}} \]
Calculating:
\[ P(C|E) = \frac{\frac{0.2}{45}}{\frac{8}{45}} = \frac{0.2}{8} = \frac{1}{40} \]

### Final Answer
The probability that the balls drawn are from urn C is:
\[ \boxed{\frac{1}{40}} \]