A letter is known to have come from either TATAPUR or from CHAKRATA. On the envelope, only two letters “TA” are visible consecutively. The probability that the letter has come from CHAKRATA is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{3}{10}\)
Let \(E_1\): the letter is from TATAPUR, \(E_2\): the letter is from CHAKRATA, and A: the two consecutive letters visible are “TA”.
\(\displaystyle P(E_1) = P(E_2) = \frac{1}{2}\)
TATAPUR has 7 letters, so 6 pairs of consecutive letters: TA, AT, TA, AP, PU, UR. “TA” occurs twice.
\(\displaystyle P(A|E_1) = \frac{2}{6} = \frac{1}{3}\)
CHAKRATA has 8 letters, so 7 pairs: CH, HA, AK, KR, RA, AT, TA. “TA” occurs once.
\(\displaystyle P(A|E_2) = \frac{1}{7}\)
By Bayes’ theorem,
\(\displaystyle P(E_2|A) = \frac{P(E_2)\,P(A|E_2)}{P(E_1)\,P(A|E_1) + P(E_2)\,P(A|E_2)}\)
\(\displaystyle = \frac{\frac{1}{2} \cdot \frac{1}{7}}{\frac{1}{2} \cdot \frac{1}{3} + \frac{1}{2} \cdot \frac{1}{7}}\)
\(\displaystyle = \frac{\frac{1}{7}}{\frac{10}{21}}\)
\(\displaystyle = \frac{3}{10}\)
Shortcut: The two priors are equal, so the posterior probabilities are in the ratio of the likelihoods, \(\frac{1}{3} : \frac{1}{7} = 7 : 3\). So \(P(E_2|A) = \frac{3}{7 + 3} = \frac{3}{10}\).