Bayes' theorem problem for class 12 mathematics 
A letter is known to come either from TATANAGAR or CALCUTTA. One the envelope just two consecutive letters TA are visible. What is the probability that the letter comes from CALCUTTA, and what is the probability that it comes from TATANAGAR?

 A: Assuming that both locations are equally likely, and that each pair of letters is equally likely to be visible, we can use Bayes' theorem to solve this problem. If the letter comes from Calcutta, there are seven possible letter combinations, one of them which is $TA$. If the letter comes from Tatanagar, there are eight possible combinations, two of them which are $TA$. Let $C$ and $T$ denote the events that the letter comes from Calcutta en Tatanagar, respectively. We then find:
$$P(C) = \frac{\frac{1}{2}\frac{1}{7}}{\frac{1}{2}\frac{1}{7} + \frac{1}{2}\frac{2}{8}} = \frac{\frac{1}{14}}{\frac{1}{14} + \frac{1}{8}} = \frac{4}{4 + 7} = \frac{4}{11} \approx 0.364$$
$$P(T) = 1 - P(C) = \frac{7}{11} \approx 0.636$$
A: Am I missing something? Since there are three occurrences of TA between the two letters, it seems to me that the chances are 1:3 for CALCUTTA and 2:3 for TATANAGAR. I hope this helps.
A: E1:  Letter (TA) came from Tatanagar
E2 :  Letter (TA) came from Calcutta
E : Event that 2 consecutive times the letter (TA) is visible.
P(E1) = P(E2) = 1/2
TATANAGAR can be broken down to : TA,AT,TA,AN,NA,AG,GA,AR = 8 pairs pairs in which two pairs contain the letter TA .
Therefore,
$P(\frac{\text{Number of times TA is present}}{\text{letter came from Tatanagar}}) = P(\frac{E}{_{E1}}) = \frac{2}{8}$
CALCUTTA can be broken down to : CA,AL,LC,CU,UT,TT,TA = 7 pairs in which ONE pair contains the letter TA .
$P(\frac{\text{Number of times TA is present}}{\text{letter came from Calcutta}}) = P(\frac{E}{_{E2}}) = \frac{1}{7}$
Then the Probability that the letter TA came from Calcutta when Just 2 consecutive letter(TA) are visible is given  = $P(\frac{E2}{E}) = \frac{P(E2) * P(\frac{E}{E2})}{{P(E2) * P(\frac{E}{E2}) + P(E1) * P(\frac{E}{E1})}} = \frac{\frac{1}{2} * \frac{1}{7}}{\frac{1}{2} * [ \frac{1}{7} + \frac{2}{8}]} $
