A letter is known to have come from either TATAPUR or from CHAKRATA. On the envelope, only two letters "TA" are visible consecutively. The probability that the letter has come from CHAKRATA is:

$\frac{3}{10}$

The correct answer is Option (2) → $\frac{3}{10}$

Total possible consecutive letter pairs in a word of length $n$ is $n-1$.

For TATAPUR ($n=7$), total pairs = $6$. Pairs: TA, AT, TA, AP, PU, UR → "TA" appears $2$ times.

For CHAKRATA ($n=8$), total pairs = $7$. Pairs: CH, HA, AK, KR, RA, AT, TA → "TA" appears $1$ time.

Let $E$ = event that "TA" is seen, $A_1$ = TATAPUR, $A_2$ = CHAKRATA.

$P(A_1) = P(A_2) = \frac12$

$P(E|A_1) = \frac{2}{6} = \frac13$, $P(E|A_2) = \frac{1}{7}$

By Bayes' theorem: $P(A_2|E) = \frac{P(E|A_2)P(A_2)}{P(E|A_1)P(A_1)+P(E|A_2)P(A_2)}$

$P(A_2|E) = \frac{\frac{1}{7} \cdot \frac12}{\frac{1}{3} \cdot \frac12 + \frac{1}{7} \cdot \frac12} = \frac{\frac{1}{14}}{\frac{1}{6} + \frac{1}{14}} = \frac{\frac{1}{14}}{\frac{7+3}{42}} = \frac{\frac{1}{14}}{\frac{10}{42}} = \frac{3}{10}$