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,
\[P({{E}_{1}})\text{ }=\text{ 1/2},\text{ }P({{E}_{2}})\text{ }=\text{ 1/2},\text{ }P({{E}_{3}}/{{E}_{1}})\text{ }=\text{ }2/8\] and \[P({{E}_{3}}/{{E}_{2}})\text{ }=\text{ }1/7\]
For TATA NAGAR, the two consecutive letters visible are TA, AT, TA, AN, NA, AG, GA and AR
So, \[P({{E}_{3}}/{{E}_{1}})\text{ }=\text{ }2/8\]
And, for CALCULTTA the two consecutive letters visible are CA, AL, LC, CU, UT, TT and TA
So, \[P({{E}_{3}}/{{E}_{2}})\text{ }=\text{ }1/7\]
Now, using Baye’s Theorem we have
Therefore, the required probability is \[7/11\].