A letter is known to have come either from KOLKATA or TATANAGAR. On the envelope just two consecutive letters TA are visible. The probability that letter has come from TATANAGAR is

The letter came from either KOLKATA or TATANAGAR. The consecutive letters "TA" are visible on the envelope.

KOLKATA has 7 letters: K-O-L-K-A-T-A

TATANAGAR has 9 letters: T-A-T-A-N-A-G-A-R

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Consecutive letter pairs in KOLKATA:

K-O, O-L, L-K, K-A, A-T, T-A

Total consecutive pairs = 6

The pair "TA" appears 1 time.

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Consecutive letter pairs in TATANAGAR:

T-A, A-T, T-A, A-N, N-A, A-G, G-A, A-R

Total consecutive pairs = 8

The pair "TA" appears 2 times.

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Probability of seeing "TA" if the letter came from KOLKATA:

$P(\text{TA | KOLKATA}) = \frac{1}{6}$

Probability of seeing "TA" if the letter came from TATANAGAR:

$P(\text{TA | TATANAGAR}) = \frac{2}{8}$

$= \frac{1}{4}$

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Assuming equal prior probability for both cities:

$P(\text{KOLKATA}) = \frac{1}{2}$

$P(\text{TATANAGAR}) = \frac{1}{2}$

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Using Bayes' theorem:

$P(\text{TA visible}) = P(\text{TA | KOLKATA}) \times P(\text{KOLKATA}) + P(\text{TA | TATANAGAR}) \times P(\text{TATANAGAR})$

$= \frac{1}{6} \times \frac{1}{2} + \frac{1}{4} \times \frac{1}{2}$

$= \frac{1}{12} + \frac{1}{8}$

$= \frac{2}{24} + \frac{3}{24}$

$= \frac{5}{24}$

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$P(\text{TATANAGAR | TA}) = \frac{P(\text{TA | TATANAGAR}) \times P(\text{TATANAGAR})}{P(\text{TA visible})}$

$= \frac{\frac{1}{4} \times \frac{1}{2}}{\frac{5}{24}}$

$= \frac{\frac{1}{8}}{\frac{5}{24}}$

$= \frac{1}{8} \times \frac{24}{5}$

$= \frac{24}{40}$

$= \frac{3}{5}$

Therefore, the probability that the letter came from TATANAGAR is $\frac{3}{5}$.