A letter is known to have come either from LONDON or CLIFTON. On the envelope just two consecutive letters ON are visible. What is the probability that the letter has come from (i) LONDON (ii) CLIFTON?

Options

A. `1//7`
B. `12//17`
C. `17//30`
D. `3//5`

Explanation

To solve the problem, we need to determine the probability that the letter came from LONDON or CLIFTON given that the visible letters are "ON."

### Step-by-Step Solution:

1. **Identify the Events**:
- Let \( A_1 \) be the event that the letter is from LONDON.
- Let \( A_2 \) be the event that the letter is from CLIFTON.
- Let \( E \) be the event that the visible letters are "ON."

2. **Determine the Possible Outcomes**:
- The word LONDON contains the consecutive letters "ON" in the positions 2 and 3.
- The word CLIFTON also contains the consecutive letters "ON" in the positions 5 and 6.

3. **Calculate the Probabilities**:
- The total number of pairs of consecutive letters in LONDON:
- LONDON has 5 pairs: (L, O), (O, N), (N, D), (D, O), (O, N).
- The pairs containing "ON" are: (O, N) (which appears twice).
- Therefore, \( P(E | A_1) = \frac{2}{5} \).

- The total number of pairs of consecutive letters in CLIFTON:
- CLIFTON has 6 pairs: (C, L), (L, I), (I, F), (F, T), (T, O), (O, N).
- The pairs containing "ON" are: (O, N) (which appears once).
- Therefore, \( P(E | A_2) = \frac{1}{6} \).

4. **Use Bayes' Theorem**:
- We want to find \( P(A_1 | E) \) and \( P(A_2 | E) \).
- According to Bayes' theorem:
\[ P(A_1 | E) = \frac{P(E | A_1) \cdot P(A_1)}{P(E)} \]
\[ P(A_2 | E) = \frac{P(E | A_2) \cdot P(A_2)}{P(E)} \]
- Assuming the letter is equally likely to come from LONDON or CLIFTON, we have:
- \( P(A_1) = P(A_2) = \frac{1}{2} \).

5. **Calculate \( P(E) \)**:
- Using the law of total probability:
\[ P(E) = P(E | A_1) \cdot P(A_1) + P(E | A_2) \cdot P(A_2) \]
\[ P(E) = \left(\frac{2}{5} \cdot \frac{1}{2}\right) + \left(\frac{1}{6} \cdot \frac{1}{2}\right) \]
\[ P(E) = \frac{2}{10} + \frac{1}{12} \]
- Finding a common denominator (60):
\[ P(E) = \frac{12}{60} + \frac{5}{60} = \frac{17}{60} \]

6. **Calculate \( P(A_1 | E) \)**:
- Substitute into Bayes' theorem:
\[ P(A_1 | E) = \frac{P(E | A_1) \cdot P(A_1)}{P(E)} = \frac{\left(\frac{2}{5}\right) \cdot \left(\frac{1}{2}\right)}{\frac{17}{60}} \]
\[ P(A_1 | E) = \frac{\frac{2}{10}}{\frac{17}{60}} = \frac{2}{10} \cdot \frac{60}{17} = \frac{12}{17} \]

7. **Calculate \( P(A_2 | E) \)**:
- Similarly,
\[ P(A_2 | E) = \frac{P(E | A_2) \cdot P(A_2)}{P(E)} = \frac{\left(\frac{1}{6}\right) \cdot \left(\frac{1}{2}\right)}{\frac{17}{60}} \]
\[ P(A_2 | E) = \frac{\frac{1}{12}}{\frac{17}{60}} = \frac{1}{12} \cdot \frac{60}{17} = \frac{5}{17} \]

### Final Answers:
- The probability that the letter is from LONDON: \( P(A_1 | E) = \frac{12}{17} \)
- The probability that the letter is from CLIFTON: \( P(A_2 | E) = \frac{5}{17} \)