A letter is known to have come 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.

Answer

To solve the problem, we will find the probability that the letter has come from LONDON or CLIFTON given that the visible letters are "ON".

### Step-by-Step Solution:

1. **Identify the Sources**:
The letter can either come from LONDON or CLIFTON.

2. **Count the Total Outcomes**:
- For LONDON, the letters are: L, O, N, D, O, N.
- The consecutive pairs are: LO, ON, ND, DO, ON.
- So, there are 5 pairs in total.
- For CLIFTON, the letters are: C, L, I, F, T, O, N.
- The consecutive pairs are: CL, LI, IF, FT, TO, ON.
- So, there are 6 pairs in total.

3. **Count the Favorable Outcomes**:
- From LONDON, the pairs that contain "ON" are: ON (appears twice).
- Thus, there are 2 favorable outcomes for LONDON.
- From CLIFTON, the pairs that contain "ON" are: ON (appears once).
- Thus, there is 1 favorable outcome for CLIFTON.

4. **Calculate the Probability for Each Source**:
- The probability of the letter coming from LONDON given "ON":
\[ P(L|ON) = \frac{P(ON|L) \cdot P(L)}{P(ON|L) \cdot P(L) + P(ON|C) \cdot P(C)} \]
Where:
- \( P(ON|L) = \frac{2}{5} \) (favorable outcomes from LONDON)
- \( P(ON|C) = \frac{1}{6} \) (favorable outcomes from CLIFTON)
- \( P(L) = P(C) = \frac{1}{2} \) (equal probabilities for LONDON and CLIFTON)

- Plugging in the values:
\[ P(L|ON) = \frac{\frac{2}{5} \cdot \frac{1}{2}}{\frac{2}{5} \cdot \frac{1}{2} + \frac{1}{6} \cdot \frac{1}{2}} \]
\[ = \frac{\frac{2}{10}}{\frac{2}{10} + \frac{1}{12}} \]

5. **Finding a Common Denominator**:
- The common denominator for \( \frac{2}{10} \) and \( \frac{1}{12} \) is 60.
- Convert the fractions:
\[ \frac{2}{10} = \frac{12}{60}, \quad \frac{1}{12} = \frac{5}{60} \]
- Thus:
\[ P(L|ON) = \frac{\frac{12}{60}}{\frac{12}{60} + \frac{5}{60}} = \frac{12}{17} \]

6. **Calculate the Probability for CLIFTON**:
- Similarly, we can find the probability for CLIFTON:
\[ P(C|ON) = \frac{P(ON|C) \cdot P(C)}{P(ON|L) \cdot P(L) + P(ON|C) \cdot P(C)} \]
- Plugging in the values:
\[ P(C|ON) = \frac{\frac{1}{6} \cdot \frac{1}{2}}{\frac{2}{10} + \frac{1}{12}} = \frac{\frac{1}{12}}{\frac{12}{60} + \frac{5}{60}} = \frac{\frac{1}{12}}{\frac{17}{60}} = \frac{5}{17} \]

### Final Answers:
- The probability that the letter has come from LONDON is \( \frac{12}{17} \).
- The probability that the letter has come from CLIFTON is \( \frac{5}{17} \).