JEE Main 2026 — Mathematics Probability & Statistics

Let $E_1$E1​ be the event that the letter came from KANPUR and $E_2$E2​ be the event that the letter came from ANANTPUR.

Since the letter is equally likely to come from either city, $P(E_1) = P(E_2) = \dfrac{1}{2}$P(E1​)=P(E2​)=21​.

Let $A$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, $P(A|E_1) = \dfrac{1}{5}$P(A∣E1​)=51​.

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, $P(A|E_2) = \dfrac{2}{7}$P(A∣E2​)=72​.

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

$$P(E_2|A) = \dfrac{P(E_2) P(A|E_2)}{P(E_1) P(A|E_1) + P(E_2) P(A|E_2)}$$P(E2​∣A)=P(E1​)P(A∣E1​)+P(E2​)P(A∣E2​)P(E2​)P(A∣E2​)​

$$P(E_2|A) = \dfrac{\dfrac{1}{2} \times \dfrac{2}{7}}{\dfrac{1}{2} \times \dfrac{1}{5} + \dfrac{1}{2} \times \dfrac{2}{7}}$$P(E2​∣A)=21​×51​+21​×72​21​×72​​

$$P(E_2|A) = \dfrac{\dfrac{2}{7}}{\dfrac{1}{5} + \dfrac{2}{7}}$$P(E2​∣A)=51​+72​72​​

$$P(E_2|A) = \dfrac{\dfrac{2}{7}}{\dfrac{17}{35}} = \dfrac{2}{7} \times \dfrac{35}{17} = \dfrac{10}{17}$$P(E2​∣A)=3517​72​​=72​×1735​=1710​

Answer: $\dfrac{10}{17}$1710​