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Maths Question 7 – JEE-MAIN 2026

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A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

This problem involves determining the probability of an event (letter from ANANTPUR) given that another event (visible letters "AN") has occurred. Bayes' Theorem is applicable here.

Step 1: Define Events and Prior Probabilities✦ Active

Let K be the event that the letter is from KANPUR, and A be the event that the letter is from ANANTPUR. Let E be the event that the consecutive letters "AN" are visible on the envelope. We need to find P(A|E). Since the letter is known to have arrived from either city, and no other information is given, we assume equal prior probabilities: P(K)=P(A)=12.

Step 2: Calculate Conditional Probabilities P(E|K) and P(E|A)○ Expand

For KANPUR (6 letters): The possible consecutive pairs are (KA), (AN), (NP), (PU), (UR). There are 5 such pairs. The pair "AN" appears once. Thus, P(E|K)=15.

For ANANTPUR (8 letters): The possible consecutive pairs are (AN), (NA), (AN), (NT), (TP), (PU), (UR). There are 7 such pairs. The pair "AN" appears twice. Thus, P(E|A)=27.

💡 Teacher's Secret Hint

Carefully count all possible consecutive pairs and the occurrences of 'AN' in each word.

Step 3: Apply Bayes' Theorem○ Expand

First, calculate the total probability of event E using the law of total probability:

P(E)=P(E|K)P(K)+P(E|A)P(A)=(15)(12)+(27)(12)=110+17=7+1070=1770

Now, apply Bayes' Theorem to find the probability that the letter came from ANANTPUR given that "AN" is visible:

P(A|E)=P(E|A)P(A)P(E)=(27)(12)1770=171770=17×7017=1017
💡 Teacher's Secret Hint

Ensure correct substitution of values into Bayes' Theorem and simplify the fraction carefully.

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