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 :

The probability that the candidate travelled by bus given that they arrived late is:

$$P(E_1|L) = \frac{P(E_1) \cdot P(L|E_1)}{P(E_1) \cdot P(L|E_1) + P(E_2) \cdot P(L|E_2) + P(E_3) \cdot P(L|E_3)}$$

$$P(E_1) = \frac{2}{5}, \quad P(E_2) = \frac{1}{5}, \quad P(E_3) = \frac{2}{5}$$ (Travel probabilities)

$$P(L|E_1) = \frac{1}{5}, \quad P(L|E_2) = \frac{1}{3}, \quad P(L|E_3) = \frac{1}{4}$$ (Conditional probabilities of being late)

For Bus: $$P(E_1) \cdot P(L|E_1) = \frac{2}{5} \times \frac{1}{5} = \frac{2}{25}$$

For Scooter: $$P(E_2) \cdot P(L|E_2) = \frac{1}{5} \times \frac{1}{3} = \frac{1}{15}$$

For Car: $$P(E_3) \cdot P(L|E_3) = \frac{2}{5} \times \frac{1}{4} = \frac{2}{20} = \frac{1}{10}$$

$$P(L) = \frac{2}{25} + \frac{1}{15} + \frac{1}{10}$$

$$P(L) = \frac{24}{300} + \frac{20}{300} + \frac{30}{300} = \frac{24 + 20 + 30}{300} = \frac{74}{300}$$

$$P(E_1|L) = \frac{\frac{24}{300}}{\frac{74}{300}} = \frac{24}{74} = \frac{12}{37}$$