Probability of a letter coming from a city I got this question:

A letter is known to have come either from $\text{TATANAGAR}$ or from $\text{CALCUTTA}$. On the envelope just two consecutive Letters $\text{TA}$ are visible. What is the probability that the letters came from $\text{TATANAGAR}$?

My attempt:
Total number of cases $=3$, as there are three such pairs.
Probability $={{\text{Favourable}}\over{\text{Total}}}=\frac23$
However, the answer is given to be $7\over11$.
How is this possible? Please help.
 A: Let $T$ be the event that the letter with two consecutive letters preserved came from Tatanagar, $C$ the event that the letter came from Calcutta, and $X$ the event that the letter came with exactly the pair of letters "TA" on the address.
First, note that we have:
$$P(X\mid T)=\frac{2}{8}=\frac{1}{4}$$
because there are $8$ pairs of consecutive letters in the 9-letter word TATANAGAR, and $2$ of them are "TA". Here, the assumption is that, in a word, all consecutive pairs of letters are equally likely to be preserved.
Similarly, $P(X\mid C)=\frac{1}{7}$ because there is only one pair "TA" in the 8-letter word CALCUTTA.
Now, applying Bayes' formula, we have:
$$P(T\mid X)=\frac{P(X\cap T)}{P(X)}=\frac{P(X\mid T)P(T)}{P(X\mid T)P(T)+P(X\mid C)P(C)}$$
which, by replacing above calculated $P(X\mid T)$ and $P(X\mid C)$ gives:
$$P(T\mid X)=\frac{\frac{1}{4}P(T)}{\frac{1}{4}P(T)+\frac{1}{7}P(C)}$$
Now, they apparently also assume that $P(T)=P(C)=\frac{1}{2}$, i.e. that a priori you have equally many letters with just two consecutive letters preserved coming from Tatanagar as from Calcutta (is that true?- doesn't matter...), so we finally have:
$$P(T\mid X)=\frac{\frac{1}{4}\frac{1}{2}}{\frac{1}{4}\frac{1}{2}+\frac{1}{7}\frac{1}{2}}=\frac{7}{11}$$
Note: I've been careful to write, in the above answer, the hidden assumptions from the problem in italic. I'd be much happier with the problem statement if they were explicitly spelled out...
A: Let $E_1$ be the event of the letter coming from TATANAGAR, $E_2$ be the event it came from CALCUTTA, and $A $ be the event that denotes two consecutive alphabets visible are TA.
Given: $$P (E_1)=P (E_2)=\frac12$$ $$ P (A \mid E_1) = \frac28, P (A \mid E_2)=\frac17$$ because there are two occurrences of TA in TATANAGAR in eight pairs (letters should be consecutive, hence TR isn't counted), and $1$ in CALCUTTA with seven pairs..
Conclude using Bayes' theorem.
A: The reason you cannot use 2/3 directly is because occurrence of TA from both city names in not equally likely.
TATANAGAR- Probability of having TA is 2/8
CALCUTTA - Probability of having TA is 1/7
Probability it is from TATANAGAR= $\frac{1/4}{1/4+1/7}$=$7/11$
A: $$T\equiv \text{comes from Tatanagar}\quad \text{and}\quad C\equiv \text{Comes from Calcutta}\rightarrow \left\{ \begin{array}{lcc}
             p(T)=\dfrac{1}{2} \\
             \\ p(C)=\dfrac{1}{2}
             \end{array}
   \right.$$
$$\text{Possible choices of two consecutive letters:}\left\{ \begin{array}{lcc}
             \text{In TATANAGAR}: 8 \\
             \\ \text{In CALCUTTA}: 7
             \end{array}
   \right.$$
$$\text{Event} D\equiv \text{The chosen couple of letters is TA}\rightarrow \left\{ \begin{array}{lcc}
             p(D|T)=\dfrac{2}{8} \\
             \\ p(D|C)=\dfrac{1}{7}
             \end{array}
   \right.$$
$$p(T|D)=\dfrac{p(T\cap D)}{p(D)}=\dfrac{\frac{1}{2}\cdot \frac{2}{8}}{\frac{1}{2}\cdot \frac{2}{8}+\frac{1}{2}\cdot \frac{1}{7}}=\dfrac{\frac{1}{8}}{\frac{7+4}{56}}=\dfrac{7}{11}$$
