A letter is known to have come either from TATA NAGAR

Let E1 be the event that the letter comes from TATA NAGAR,

E2 be the event that the letter comes from CALCUTTA

And, E3 be the event that on the letter, two consecutive letters TA are visible

Now,

P(E1) = ½, P(E2) = ½, P(E3/E1) = 2/8 and P(E3/E2) = 1/7

For TATA NAGAR, the two consecutive letters visible are TA, AT, TA, AN, NA, AG, GA and AR

So, P(E3/E1) = 2/8

And, for CALCULTTA the two consecutive letters visible are CA, AL, LC, CU, UT, TT and TA

So, P(E3/E2) = 1/7

Now, using Baye’s Theorem we have

Therefore, the required probability is 7/11.