A letter is known to have come either from LONDON or CLIFTON. On the envelop just two consecutive letter \'ON\' are visible. The probability that the letter has come from LONDON is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \displaystyle\frac{12}{17} \)
Let \( E_1 \) = letter from LONDON, \( E_2 \) = letter from CLIFTON, A = the pair ON is visible. Before seeing the envelope, \( P(E_1)=P(E_2)=\frac{1}{2} \).
LONDON has 6 letters, so 5 pairs of consecutive letters: LO, ON, ND, DO, ON. Two of them are ON:
\( \displaystyle P(A|E_1)=\frac{2}{5} \)
CLIFTON has 7 letters, so 6 pairs: CL, LI, IF, FT, TO, ON. One of them is ON:
\( \displaystyle P(A|E_2)=\frac{1}{6} \)
By Bayes' theorem,
\( \displaystyle P(E_1|A)=\frac{\frac{1}{2}\cdot\frac{2}{5}}{\frac{1}{2}\cdot\frac{2}{5}+\frac{1}{2}\cdot\frac{1}{6}} \)
\( \displaystyle =\frac{\frac{2}{5}}{\frac{2}{5}+\frac{1}{6}} \)
\( \displaystyle =\frac{\frac{12}{30}}{\frac{17}{30}} \)
\( \displaystyle =\frac{12}{17} \)
Shortcut: With equal priors, Bayes reduces to a ratio of the likelihoods: \( \frac{2}{5}:\frac{1}{6}=12:5 \), so \( P(E_1|A)=\frac{12}{12+5}=\frac{12}{17} \).
Why the other options are incorrect
Option (1) \( \displaystyle\frac{5}{17} \) — Incorrect. \( \frac{5}{17} \) is the probability that the letter came from CLIFTON.
Option (3) \( \displaystyle\frac{1}{17} \) — Incorrect. The probability is \( \frac{12}{17} \), not \( \frac{1}{17} \).
Option (4) \( \displaystyle\frac{2}{17} \) — Incorrect. The probability is \( \frac{12}{17} \), not \( \frac{2}{17} \).