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Maths Question 6 – JEE-MAIN 2026

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A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are 25, 15 and 25. The probabilities that the candidate reaches late at the examination centre are 15, 13 and 14 if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is:

This is a conditional probability problem where you are given prior probabilities of events and conditional probabilities of an outcome given those events.

Step 1: Identify Given Probabilities and Goal✦ Active

Let B, S, C be the events of traveling by bus, scooter, and car, respectively. Let L be the event of reaching late. We are given the following probabilities:

P(B)=25,P(S)=15,P(C)=25

The conditional probabilities of reaching late are:

P(L|B)=15,P(L|S)=13,P(L|C)=14

We need to find the probability that the candidate travelled by bus given that they reached late, i.e., P(B|L).

Step 2: Calculate the Total Probability of Reaching Late○ Expand

Using the Law of Total Probability, P(L)=P(L|B)P(B)+P(L|S)P(S)+P(L|C)P(C).

P(L)=(15)(25)+(13)(15)+(14)(25) P(L)=225+115+220=225+115+110

The least common multiple (LCM) of 25, 15, and 10 is 150. Convert the fractions to have a common denominator:

P(L)=2⋅6150+1⋅10150+1⋅15150 P(L)=12150+10150+15150=12+10+15150=37150
Step 3: Apply Bayes' Theorem○ Expand

Now, apply Bayes' Theorem to find P(B|L):

P(B|L)=P(L|B)P(B)P(L) P(B|L)=(15)(25)37150=22537150 P(B|L)=225⋅15037=2⋅637=1237

Thus, the probability that the candidate travelled by bus given that they reached late is 1237.

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