Q31 / 75 · JEE Main 2026 (08 April Shift 2) · Mathematics
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 $\dfrac{2}{5}$, $\dfrac{1}{5}$ and $\dfrac{2}{5}$. The probabilities that the candidate reaches late at the examination centre are $\dfrac{1}{5}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$ 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$\dfrac{11}{37}$
- Correct answerB$\dfrac{12}{37}$
- C$\dfrac{13}{37}$
- D$\dfrac{14}{37}$
Solution
Let $E_1$, $E_2$, and $E_3$ be the events that the candidate travels by bus, scooter, and car, respectively.
The probabilities of choosing these modes of transport are:
$P(E_1) = \dfrac{2}{5}$
$P(E_2) = \dfrac{1}{5}$
$P(E_3) = \dfrac{2}{5}$
Let $A$ be the event that the candidate reaches the examination centre late. The conditional probabilities of reaching late are:
$P(A|E_1) = \dfrac{1}{5}$
$P(A|E_2) = \dfrac{1}{3}$
$P(A|E_3) = \dfrac{1}{4}$
We need to find the probability that the candidate travelled by bus given that they reached late, which is $P(E_1|A)$.
Using Bayes' theorem:
$P(E_1|A) = \dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$
Substituting the given values:
$P(E_1|A) = \dfrac{\dfrac{2}{5} \times \dfrac{1}{5}}{\left(\dfrac{2}{5} \times \dfrac{1}{5}\right) + \left(\dfrac{1}{5} \times \dfrac{1}{3}\right) + \left(\dfrac{2}{5} \times \dfrac{1}{4}\right)}$
$P(E_1|A) = \dfrac{\dfrac{2}{25}}{\dfrac{2}{25} + \dfrac{1}{15} + \dfrac{1}{10}}$
Taking the LCM of the denominators $25$, $15$, and $10$, which is $150$:
$P(E_1|A) = \dfrac{\dfrac{12}{150}}{\dfrac{12}{150} + \dfrac{10}{150} + \dfrac{15}{150}}$
$P(E_1|A) = \dfrac{12}{12 + 10 + 15} = \dfrac{12}{37}$
Answer: $\dfrac{12}{37}$
The probabilities of choosing these modes of transport are:
$P(E_1) = \dfrac{2}{5}$
$P(E_2) = \dfrac{1}{5}$
$P(E_3) = \dfrac{2}{5}$
Let $A$ be the event that the candidate reaches the examination centre late. The conditional probabilities of reaching late are:
$P(A|E_1) = \dfrac{1}{5}$
$P(A|E_2) = \dfrac{1}{3}$
$P(A|E_3) = \dfrac{1}{4}$
We need to find the probability that the candidate travelled by bus given that they reached late, which is $P(E_1|A)$.
Using Bayes' theorem:
$P(E_1|A) = \dfrac{P(E_1)P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$
Substituting the given values:
$P(E_1|A) = \dfrac{\dfrac{2}{5} \times \dfrac{1}{5}}{\left(\dfrac{2}{5} \times \dfrac{1}{5}\right) + \left(\dfrac{1}{5} \times \dfrac{1}{3}\right) + \left(\dfrac{2}{5} \times \dfrac{1}{4}\right)}$
$P(E_1|A) = \dfrac{\dfrac{2}{25}}{\dfrac{2}{25} + \dfrac{1}{15} + \dfrac{1}{10}}$
Taking the LCM of the denominators $25$, $15$, and $10$, which is $150$:
$P(E_1|A) = \dfrac{\dfrac{12}{150}}{\dfrac{12}{150} + \dfrac{10}{150} + \dfrac{15}{150}}$
$P(E_1|A) = \dfrac{12}{12 + 10 + 15} = \dfrac{12}{37}$
Answer: $\dfrac{12}{37}$
Hence, the correct answer is option: 'B'.