JEE Main 2026 Apr 5 Shift 1, Mathematics Q7: Total Probability Theorem And Bayes' Theorem

Let $E_1$ denote the letter being from KANPUR and $E_2$ from ANANTPUR, each with prior $\frac{1}{2}$.

KANPUR has $6$ letters and so $5$ pairs of consecutive letters: KA, AN, NP, PU, UR. Exactly one pair is AN, giving $P(\text{AN}\mid E_1)=\frac{1}{5}$.

ANANTPUR has $8$ letters and $7$ consecutive pairs: AN, NA, AN, NT, TP, PU, UR. Two of these are AN, giving $P(\text{AN}\mid E_2)=\frac{2}{7}$.

By Bayes' theorem,

$P(E_2\mid \text{AN})=\frac{\frac{1}{2}\cdot \frac{2}{7}}{\frac{1}{2}\cdot \frac{1}{5}+\frac{1}{2}\cdot \frac{2}{7}}=\frac{2/7}{1/5+2/7}=\frac{2/7}{17/35}=\frac{10}{17}$.