A letter is known to have come either from KOLKATA or TATANAGAR. On the envelope just two consecutive letters TA are visible. The probability that letter has come from TATANAGAR is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3}{5}\)
Let \(E_1\): the letter is from KOLKATA, \(E_2\): the letter is from TATANAGAR, and A: the pair TA is visible. Then \(P(E_1) = P(E_2) = \frac{1}{2}\).
A word of n letters has n − 1 pairs of consecutive letters.
KOLKATA (7 letters) has 6 pairs: KO, OL, LK, KA, AT, TA. TA occurs once:
\( \displaystyle P(A|E_1) = \frac{1}{6} \)
TATANAGAR (9 letters) has 8 pairs: TA, AT, TA, AN, NA, AG, GA, AR. TA occurs twice:
\( \displaystyle P(A|E_2) = \frac{2}{8} = \frac{1}{4} \)
By Bayes' theorem:
\( \displaystyle P(E_2|A) = \frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2)} \)
\( \displaystyle = \frac{\frac{1}{2} \cdot \frac{1}{4}}{\frac{1}{2} \cdot \frac{1}{6} + \frac{1}{2} \cdot \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{1}{6} + \frac{1}{4}} \)
\( \displaystyle = \frac{\frac{3}{12}}{\frac{5}{12}} = \frac{3}{5} \)
Shortcut: With equal priors, the posterior probabilities are in the ratio of the likelihoods:
\( \displaystyle \frac{1}{6} : \frac{1}{4} = 2 : 3 \)
So \(P(E_2|A) = \frac{3}{2 + 3} = \frac{3}{5}\).