JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Probability

Each origin is equally likely, $P(K)=P(A)=\frac{1}{2}$. KANPUR has the five consecutive pairs KA, AN, NP, PU, UR, among which AN occurs once, so $P(\text{AN}\mid K)=\frac{1}{5}$. ANANTPUR has the seven pairs AN, NA, AN, NT, TP, PU, UR, in which AN occurs twice, so $P(\text{AN}\mid A)=\frac{2}{7}$. By Bayes' theorem, $P(A\mid \text{AN})=\frac{\frac{1}{2}\cdot \frac{2}{7}}{\frac{1}{2}\cdot \frac{1}{5}+\frac{1}{2}\cdot \frac{2}{7}}=\frac{\frac{2}{7}}{\frac{1}{5}+\frac{2}{7}}=\frac{\frac{2}{7}}{\frac{17}{35}}=\frac{10}{17}$. The probability that the letter came from ANANTPUR is $\frac{10}{17}$.