JEE Main 2026 — Probability Question with Solution

Let $E_1$ be the event that the letter came from KANPUR and $E_2$ be the event that the letter came from ANANTPUR.

Since the letter is equally likely to come from either city, $P(E_1)=P(E_2)=\frac{1}{2}$.

Let $A$ be the event that the two consecutive visible letters are AN.

In the word KANPUR, there are 6 letters, so there are 5 pairs of consecutive letters: KA, AN, NP, PU, UR. The pair AN appears exactly once. Thus, $P(A|E_1)=\frac{1}{5}$.

In the word ANANTPUR, there are 8 letters, so there are 7 pairs of consecutive letters: AN, NA, AN, NT, TP, PU, UR. The pair AN appears exactly twice. Thus, $P(A|E_2)=\frac{2}{7}$.

By Bayes' theorem, the probability that the letter came from ANANTPUR given that AN is visible is:

$P(E_2|A)=\frac{P(E_2)P(A|E_2)}{P(E_1)P(A|E_1)+P(E_2)P(A|E_2)}$

$P(E_2|A)=\frac{\frac{1}{2}\times \frac{2}{7}}{\frac{1}{2}\times \frac{1}{5}+\frac{1}{2}\times \frac{2}{7}}$

$P(E_2|A)=\frac{\frac{2}{7}}{\frac{1}{5}+\frac{2}{7}}$

$P(E_2|A)=\frac{\frac{2}{7}}{\frac{17}{35}}=\frac{2}{7}\times \frac{35}{17}=\frac{10}{17}$

Answer: $\frac{10}{17}$