A letter is known to have come either from KANPUR or from ANANTNAG. On the envelope, only two consecutive letters AN of the place name are visible. The probability that the letter came from ANANTNAG is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{10}{17}\)
Let \(E_1\) be the event that the letter came from KANPUR, \(E_2\) the event that it came from ANANTNAG, and V the event that the visible pair is AN. Each place is equally likely:
\( \displaystyle P(E_1) = P(E_2) = \frac{1}{2} \)
KANPUR has 5 pairs of consecutive letters: KA, AN, NP, PU, UR. One of them is AN:
\( \displaystyle P(V|E_1) = \frac{1}{5} \)
ANANTNAG has 7 pairs: AN, NA, AN, NT, TN, NA, AG. Two of them are AN:
\( \displaystyle P(V|E_2) = \frac{2}{7} \)
By Bayes’ theorem:
\( \displaystyle P(E_2|V) = \frac{\frac{1}{2}\times\frac{2}{7}}{\frac{1}{2}\times\frac{1}{5}+\frac{1}{2}\times\frac{2}{7}} \)
\( \displaystyle = \frac{\frac{2}{7}}{\frac{1}{5}+\frac{2}{7}} = \frac{\frac{10}{35}}{\frac{17}{35}} = \frac{10}{17} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{7}{17}\) — Incorrect. This is \(P(E_1|V)\), the probability that the letter came from KANPUR.
Option (2) \(\displaystyle\frac{2}{7}\) — Incorrect. This is \(P(V|E_2)\), the chance of seeing AN on a letter from ANANTNAG, not the other way round.
Option (3) \(\displaystyle\frac{1}{2}\) — Incorrect. This is the chance before the visible letters are taken into account.