A letter is known to have come for 'RAVINDRA' or 'DEVINDRAN'. On the envelope just two consecutive letters 'RA' are visible. The probability that the letter has come for 'RAVINDRA' is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{16}{23} \)
Let \( E_1 \): the letter is for RAVINDRA, \( E_2 \): the letter is for DEVINDRAN, and A: the visible pair is “RA”. With no other information, \( P(E_1)=P(E_2)=\frac{1}{2} \).
RAVINDRA has 8 letters, so 7 pairs of consecutive letters: RA, AV, VI, IN, ND, DR, RA. Two of them are RA.
DEVINDRAN has 9 letters, so 8 pairs: DE, EV, VI, IN, ND, DR, RA, AN. One of them is RA.
\( \displaystyle P(A|E_1)=\frac{2}{7} \)
\( \displaystyle P(A|E_2)=\frac{1}{8} \)
By Bayes' theorem:
\( \displaystyle P(E_1|A)=\frac{\frac{1}{2}\cdot\frac{2}{7}}{\frac{1}{2}\cdot\frac{2}{7}+\frac{1}{2}\cdot\frac{1}{8}} \)
\( \displaystyle =\frac{\frac{2}{7}}{\frac{2}{7}+\frac{1}{8}} \)
\( \displaystyle =\frac{\frac{16}{56}}{\frac{16}{56}+\frac{7}{56}} \)
\( \displaystyle =\frac{16}{23} \)
Shortcut: With equal priors, Bayes' theorem is just the ratio of the likelihoods:
\( \displaystyle \frac{2}{7}:\frac{1}{8}=16:7 \)
\( \displaystyle P(E_1|A)=\frac{16}{16+7}=\frac{16}{23} \)
Why the other options are incorrect
Option (2) \( \displaystyle\frac{7}{11} \) — Incorrect. The likelihoods are in the ratio 16 : 7, which gives \( \frac{16}{23} \), not \( \frac{7}{11} \).
Option (3) \( \displaystyle\frac{4}{11} \) — Incorrect. The likelihoods are in the ratio 16 : 7, which gives \( \frac{16}{23} \), not \( \frac{4}{11} \).
Option (4) \( \displaystyle\frac{7}{23} \) — Incorrect. \( \frac{7}{23} \) is the probability that the letter came for DEVINDRAN, the other name.