A letter is known to have come either from 'KRISHNAGIRI' or 'DHARMAPURI'. On the post mark only the two consecutive letters 'RI' are visible. Then, the chance that it came from 'KRISHNAGIRI' is

Let £1 be the event that the letter came from 'DHARMAPURI' and £2 be the event that the letter came from 'KRISHNAGIRI'. Let A be the event that two consecutive letters RI are visible as post mark. Since, letters have come either from KRISHNAGIRI or 'DHARMAPURI' . 

$\therefore$      $P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$

  If £1 has occurred, then it means that letters came from 'DHARAMPURI'. 

In the word 'DHARMAPURI' there are 1 0 letters in which RI occurs in the end. 

Considering RI as one letter there are 9 letters out of which one can be in 9 ways.  

$\therefore$        $P\left(\frac{A}{E_1}\right)=\frac{1}{9}$

If   $E_2$has occurred, then the letters came from 'KRISHNAGIRI'. In the word 'KRISHNAGIRI' there are 11 letters in which RI occurs twice. Considering one of the two RI as one letter there are 1 0 letters. 

 $\therefore$      $P\left(\frac{A}{E_2}\right)=\frac{2}{10}$

By Baye's theorem , we have 

  $\begin{array}{l}P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right)P\left(\frac{A}{E_2}\right)}{P\left(E_1\right)P\left(\frac{A}{E_1}\right)+P\left(E_2\right)P\left(\frac{A}{E_2}\right)}\\ =\frac{\frac{1}{2}\times \frac{2}{10}}{\frac{1}{2}\times \frac{1}{9}+\frac{1}{2}\times \frac{2}{10}}=\frac{\frac{2}{10}}{\frac{1}{9}+\frac{2}{10}}=\frac{2}{10}\times \frac{90}{28}=\frac{9}{14}\end{array}$