Probability That Envelope Came From ANANTPUR Given Letters AN

To find the probability that the letter came from ANANTPUR given that two consecutive visible letters are AN, we use Bayes' Theorem.

Let the events be defined as:

• $E_1$: The letter came from KANPUR.
• $E_2$: The letter came from ANANTPUR.
• $A$: Two consecutive letters visible on the envelope are AN.

Since the letter is equally likely to come from either location, we have: $P(E_1) = P(E_2) = \frac{1}{2}$

Step 1: Calculate $P(A | E_1)$

The word KANPUR consists of 6 letters. The total number of consecutive two-letter pairs is: $6 - 1 = 5$

The consecutive letter pairs in KANPUR are:

1. KA
2. AN
3. NP
4. PU
5. UR

Number of times AN appears = $1$. Thus, the conditional probability is: $P(A | E_1) = \frac{1}{5}$

Step 2: Calculate $P(A | E_2)$

The word ANANTPUR consists of 8 letters. The total number of consecutive two-letter pairs is: $8 - 1 = 7$

The consecutive letter pairs in ANANTPUR are:

1. AN
2. NA
3. AN
4. NT
5. TP
6. PU
7. UR

Number of times AN appears = $2$. Thus, the conditional probability is: $P(A | E_2) = \frac{2}{7}$

Step 3: Apply Bayes' Theorem

We need to find $P(E_2 | A)$, the probability that the letter came from ANANTPUR given that the visible letters are AN:

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

Substituting the known values: $P(E_2 | A) = \frac{\frac{1}{2} \times \frac{2}{7}}{\left(\frac{1}{2} \times \frac{1}{5}\right) + \left(\frac{1}{2} \times \frac{2}{7}\right)}$

Canceling out $\frac{1}{2}$ from the numerator and denominator: $P(E_2 | A) = \frac{\frac{2}{7}}{\frac{1}{5} + \frac{2}{7}}$

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

$P(E_2 | A) = \frac{2}{7} \times \frac{35}{17} = \frac{10}{17}$

Thus, the required probability is $\frac{10}{17}$, which corresponds to option B.