A letter is expected to come either from city 'SURAT' or from city 'RAMPUR' through post office. If on the way, envelope containing the letter is damaged and only two consecutive alphabets RA are visible on it, then the probability that letter comes from the city 'SURAT' is : 

$\frac{5}{7}$

Let E₁​ denote the event that the letter came from 'SURAT' and E₂​ denote the event that the letter came from 'RAMPUR'.

Let A be the event that the two consecutive alphabets visible on the envelope are RA.

Since the letters have come from either SURAT or RAMPUR.

∴ $P(E_1​)=\frac{1}{2}​,P(E_2​)=\frac{1}{2}$

If E_1​ has occurred then the letter has come from SURAT. In the word SURAT there are {SU, UR, RA, AT} 4 consecutive alphabets and RA occurs two times.

$∴P(A/E_1)=\frac{2}{4}$

If E_2​ has occurred then the letter has come from RAMPUR. In the word RAMPUR there are {RA, AM, MP, PU, UR} 5 consecutive alphabets and RA occurs once.

$∴P(A/E_2)=\frac{1}{5}$

The probability that the letter has come from SURAT is $P(E_1​/A)$

By Bayes theorem, $P(E_1​/A)=\frac{P(E_1)P(A/E_1)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)}$

$=\frac{\frac{1}{2}×\frac{2}{4}}{\frac{1}{2}×\frac{2}{4}+\frac{1}{2}×\frac{1}{5}}$

$=\frac{\frac{2}{8×10}}{\frac{2}{8×10}+\frac{1}{10×8}}=\frac{\frac{20}{80}}{\frac{20}{80}}=\frac{20}{80}×\frac{80}{28}=\frac{5}{7}$

Hence the probability that the letter has come from SURAT is $\frac{5}{7}$​.