A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is :

To solve this, we use Bayes' Theorem. We need to find the probability that the letter came from ANANTPUR, given that "AN" was visible.

• $$E_1$$: The letter is from KANPUR.
• $$E_2$$: The letter is from ANANTPUR.
• $$A$$: The visible consecutive letters are "AN".

Assume $$P(E_1) = P(E_2) = \frac{1}{2}$$.

We look at the total possible pairs of consecutive letters in each word:

• KANPUR: Total pairs = 5 (KA, AN, NP, PU, UR).
• $$P(A|E_1) = \frac{1}{5}$$
• ANANTPUR: Total pairs = 7 (AN, NA, AN, NT, TP, PU, UR).
• $$P(A|E_2) = \frac{2}{7}$$

We want to find $$P(E_2|A)$$:

$$P(E_2|A) = \frac{P(E_2) \cdot P(A|E_2)}{P(E_1) \cdot P(A|E_1) + P(E_2) \cdot P(A|E_2)}$$

Substitute the values:

$$P(E_2|A) = \frac{\frac{1}{2} \cdot \frac{2}{7}}{\left(\frac{1}{2} \cdot \frac{1}{5}\right) + \left(\frac{1}{2} \cdot \frac{2}{7}\right)}$$

Cancel the common $$\frac{1}{2}$$ factor:

$$P(E_2|A) = \frac{\frac{2}{7}}{\frac{1}{5} + \frac{2}{7}} = \frac{\frac{2}{7}}{\frac{7 + 10}{35}} = \frac{2}{7} \cdot \frac{35}{17} = \frac{10}{17}$$

Correct Answer: B ($$\frac{10}{17}$$)