A letter is known to have come either from LONDON or CLIFTON, on the postmark only the two consecutive letters ON are eligible. The probability that it come from LONDON is

Options

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

Answer

To solve the problem of finding the probability that the letter came from LONDON given that the postmark shows the letters "ON", we can use Bayes' theorem. Here’s a step-by-step breakdown of the solution:

### Step 1: Define the Events
Let:
- \( L \) = the event that the letter is from LONDON
- \( C \) = the event that the letter is from CLIFTON
- \( E \) = the event that "ON" is visible on the postmark

We need to find \( P(L | E) \), the probability that the letter is from LONDON given that "ON" is visible.

### Step 2: Apply Bayes' Theorem
According to Bayes' theorem:
\[ P(L | E) = \frac{P(E | L) \cdot P(L)}{P(E)} \]

### Step 3: Calculate the Prior Probabilities
Since the letter can either come from LONDON or CLIFTON, and both are equally likely:
\[ P(L) = P(C) = \frac{1}{2} \]

### Step 4: Calculate \( P(E | L) \)
Next, we calculate \( P(E | L) \), the probability of seeing "ON" given that the letter is from LONDON.

In the word "LONDON", the possible pairs of consecutive letters are:
- LO
- ON
- ND
- DO

Thus, there are 4 pairs of consecutive letters in LONDON, and "ON" is one of them. Therefore:
\[ P(E | L) = \frac{1}{4} \]

### Step 5: Calculate \( P(E | C) \)
Now, we calculate \( P(E | C) \), the probability of seeing "ON" given that the letter is from CLIFTON.

In the word "CLIFTON", the possible pairs of consecutive letters are:
- CL
- LI
- IF
- FT
- TO
- ON

Thus, there are 6 pairs of consecutive letters in CLIFTON, and "ON" is one of them. Therefore:
\[ P(E | C) = \frac{1}{6} \]

### Step 6: Calculate \( P(E) \)
Now we need to calculate \( P(E) \), the total probability of seeing "ON". This can be calculated using the law of total probability:
\[ P(E) = P(E | L) \cdot P(L) + P(E | C) \cdot P(C) \]
Substituting the values we have:
\[ P(E) = \left(\frac{1}{4} \cdot \frac{1}{2}\right) + \left(\frac{1}{6} \cdot \frac{1}{2}\right) \]
\[ P(E) = \frac{1}{8} + \frac{1}{12} \]

To add these fractions, we find a common denominator (24):
\[ P(E) = \frac{3}{24} + \frac{2}{24} = \frac{5}{24} \]

### Step 7: Substitute Back into Bayes' Theorem
Now we can substitute everything back into Bayes' theorem:
\[ P(L | E) = \frac{P(E | L) \cdot P(L)}{P(E)} = \frac{\left(\frac{1}{4}\right) \cdot \left(\frac{1}{2}\right)}{\frac{5}{24}} \]
\[ P(L | E) = \frac{\frac{1}{8}}{\frac{5}{24}} = \frac{1}{8} \cdot \frac{24}{5} = \frac{3}{5} \]

### Final Answer
Thus, the probability that the letter came from LONDON given that "ON" is visible is:
\[ \boxed{\frac{3}{5}} \]