A letter is known to have come either from ‘TATA NAGAR’ or from ‘CALCUTTA’. On the envelope, just two consecutive letters TA are visible. What is the probability that the letter came from ‘TATA NAGAR’?

$\frac{7}{11}$

The correct answer is Option (2) → $\frac{7}{11}$ ##

Let $E_1$ be the event that letter is from TATA NAGAR and $E_2$ be the event that letter is from CALCUTTA.

Also, let $E_3$ be the event that on the letter, two consecutive letters TA are visible.

$∴P(E_1) = \frac{1}{2} \text{ and } P(E_2) = \frac{1}{2}$

$\text{and } P(E_3 | E_1) = \frac{2}{8} \text{ and } P(E_3 | E_2) = \frac{1}{7}$

[since, if letter is from TATA NAGAR, we see that the events of two consecutive letters visible are {TA, AT, TA, AN, NA, AG, GA, AR}. So, $P(E_3 / E_1) = \frac{2}{8}$ and if letter is from CALCUTTA, we see that the events of two consecutive letters to visible are {CA, AL, LC, CU, UT, TT, TA}]

$\text{So, } P(E_3 | E_2) = \frac{1}{7}$

By Baye’s theorem using

$∴P(E_1 | E_3) = \frac{P(E_1) \cdot P(E_3 | E_1)}{P(E_1) \cdot P(E_3 | E_1) + P(E_2) \cdot P(E_3 | E_2)}$

$= \frac{\frac{1}{2} \times \frac{2}{8}}{\frac{1}{2} \times \frac{2}{8} + \frac{1}{2} \times \frac{1}{7}} = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{14}} = \frac{\frac{1}{8}}{\frac{22}{8 \times 14}} = \frac{\frac{1}{8}}{\frac{11}{56}} = \frac{7}{11}$