A letter is known to have come either from LONDON or CLIFTON. On the envelope, just two consecutive letter ‘ON’ are visible then the probability that the letter has come from CLIFTON is:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{5}{17} \)
Let \( E_1 \): the letter is from LONDON, \( E_2 \): from CLIFTON, each with probability \( \frac{1}{2} \); A: the visible pair is ON.
LONDON has 5 pairs of consecutive letters: LO, ON, ND, DO, ON; two of them are ON.
CLIFTON has 6 pairs: CL, LI, IF, FT, TO, ON; one of them is ON.
\( \displaystyle P(A|E_1)=\frac{2}{5} \)
\( \displaystyle P(A|E_2)=\frac{1}{6} \)
By Bayes’ theorem:
\( \displaystyle P(E_2|A)=\frac{\frac{1}{2}\cdot\frac{1}{6}}{\frac{1}{2}\cdot\frac{2}{5}+\frac{1}{2}\cdot\frac{1}{6}} \)
\( \displaystyle =\frac{\frac{1}{6}}{\frac{2}{5}+\frac{1}{6}} \)
\( \displaystyle =\frac{\frac{1}{6}}{\frac{17}{30}} \)
\( \displaystyle =\frac{30}{102} \)
\( \displaystyle =\frac{5}{17} \)
Shortcut: The priors are equal, so they cancel: the answer is the ratio \( \frac{1}{6}:\frac{2}{5}=5:12 \), giving \( \frac{5}{5+12}=\frac{5}{17} \).
Tip: A word of n letters has n − 1 pairs of consecutive letters; count repeated pairs such as the two ON in LONDON. The probability for LONDON is \( \frac{12}{17} \), Option (4).