JEE Main 2026 Apr 8 Shift 2, Mathematics Q6: Total Probability Theorem And Bayes' Theorem

JEE Main2026Apr 8, Shift 2Mathematics

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 $\frac{2}{5}$ , $\frac{1}{5}$ and $\frac{2}{5}$ . The probabilities that the candidate reaches late at the examination centre are $\frac{1}{5}$ , $\frac{1}{3}$ and $\frac{1}{4}$ 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 :

A
$\frac{11}{37}$
B
$\frac{12}{37}$
C
$\frac{13}{37}$
D
$\frac{14}{37}$

Solution

By Bayes' theorem,

$P (\frac{B}{late}) = \frac{P ( B ) \cdot P ( late ∣ B )}{P ( B ) P ( late ∣ B ) + P ( S ) P ( late ∣ S ) + P ( C ) P ( late ∣ C )}$

Numerator: $\frac{2}{5} \cdot \frac{1}{5} = \frac{2}{25}$ .

Denominator: $\frac{2}{5} \cdot \frac{1}{5} + \frac{1}{5} \cdot \frac{1}{3} + \frac{2}{5} \cdot \frac{1}{4} = \frac{2}{25} + \frac{1}{15} + \frac{1}{10}$ .

LCM is $150$ : $\frac{12}{150} + \frac{10}{150} + \frac{15}{150} = \frac{37}{150}$ .

$∴$ $P (\frac{B}{late}) = \frac{2/25}{37/150} = \frac{2}{25} \cdot \frac{150}{37} = \frac{12}{37}$ .

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JEE Main 2026 Apr 8 Shift 2, Mathematics Q6: Total Probability Theorem And Bayes' Theorem

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