Find the probability that the letter came from MAHARASTRA A mail is known to have come from MAHARASTRAor MADRAS. On the post mark only consecutive letters RA can be read clearly. What is the chance that the mail came from MAHARASTRA? 
Please assume that equal number of mails per year are posted from Maharastra and Madras.
My try:-
As the number of RA in MAHARASTRA is 2 and in Madras is 1, the chance is 2/3.
But I think the answer is incorrect. Please help me to identify my mistake.
 A: The question cannot be answered without further information. What is the volume of post from Maharashtra compared to Madras? For example, if only one letter per year comes from Maharashtra, and ten million letters come from Madras, then chances are that it's from Madras.
Also, Maharashtra is a state while Madras is a city, so depending on the format of post marks, we might be able to deduce the answer right away (e.g., if the typeface is larger for states than for cities, or if the position is an indication).
Finally, note that Madras was officially renamed Chennai in 1996, so if the letter was sent within the past 20 years, chances are it's from Maharashtra.
A: By Bayes' theorem, 
$P$(letter from Maharashtra|we see 'RA') = $\frac{P(RA| Maha)P(Maha)}{P(RA| Maha)P(Maha) + P(RA|Mad)P(Mad)}$
and use $P(Maha) = P(Mad)$
A: I think I can explain your mistake - on the assumption that in a mangled postmark the pair of visible letters is chosen uniformly from all the pairs. 
Here is an analogy that might help. Imagine two bags, one with $1000$ blue marbles and $2$ red marbles, one with $1$ blue marble and $1$ red marble. Someone says they picked a bag at random and then a marble at random and it was red. Which bag is more likely? Can you see intuitively why the probability that it was the second bag is quite large, and not just the $1/3$ you get by looking only at the three red marbles?
To get the precise probability you use Bayes' theorem, as in the answer from @ab123 .
A: E1:  Letter (RA) came from MADRAS
E2 :  Letter (RA) came from MAHARASHTRA
E : Event that 2 consecutive times the letter (RA) is visible.
P(E1) = P(E2) = 1/2
MADRAS can be broken down to : MA,AD,DR,RA,AS = 5 pairs in which ONE pair contains letter RA .
Therefore,
$P(\frac{\text{Number of times RA is present}}{\text{letter came from MADRAS}}) = P(\frac{E}{_{E1}}) = \frac{1}{5}$
MAHARASHTRA can be broken down to : MA,AH,HA,AR,RA,AS,SH,HT,TR,RA = 10 pairs in which TWO pairs cRAtain the letter RA .
$P(\frac{\text{Number of times RA is present}}{\text{letter came from MAHARASHTRA}}) = P(\frac{E}{_{E2}}) = \frac{2}{10}$
Then the Probability that the letter RA came from MAHARASHTRA when Just 2 consecutive letter(RA) 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{2}{10}}{\frac{1}{2} * [ \frac{2}{10} + \frac{1}{5}]} $
