This problem involves conditional probability and can be solved using Bayes' Theorem. We need to find the probability that the candidate travelled by bus, given that they reached the examination centre late.

Defining Events and Probabilities

Let:

• B be the event the candidate travels by bus.
• S be the event the candidate travels by scooter.
• C be the event the candidate travels by car.
• L be the event the candidate reaches late at the examination centre.

Given probabilities are:

• $P(B) = \frac{2}{5}$
• $P(S) = \frac{1}{5}$
• $P(C) = \frac{2}{5}$
• $P(L|B) = \frac{1}{5}$ (Probability of being late given travel by bus)
• $P(L|S) = \frac{1}{3}$ (Probability of being late given travel by scooter)
• $P(L|C) = \frac{1}{4}$ (Probability of being late given travel by car)

Calculating Total Probability of Reaching Late

To use Bayes' Theorem, we first need the overall probability of the candidate reaching late, $P(L)$. We can calculate this using the law of total probability:

$P(L) = P(L|B)P(B) + P(L|S)P(S) + P(L|C)P(C)$

Substitute the given values:

• $P(L|B)P(B) = \frac{1}{5} \times \frac{2}{5} = \frac{2}{25}$
• $P(L|S)P(S) = \frac{1}{3} \times \frac{1}{5} = \frac{1}{15}$
• $P(L|C)P(C) = \frac{1}{4} \times \frac{2}{5} = \frac{2}{20} = \frac{1}{10}$

Now, sum these probabilities:

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

To add these fractions, find a common denominator, which is 150:

$P(L) = \frac{2 \times 6}{150} + \frac{1 \times 10}{150} + \frac{1 \times 15}{150}$

$P(L) = \frac{12}{150} + \frac{10}{150} + \frac{15}{150} = \frac{12 + 10 + 15}{150} = \frac{37}{150}$

Applying Bayes' Theorem

We want to find the probability that the candidate travelled by bus given they reached late, which is $P(B|L)$. Bayes' Theorem states:

$P(B|L) = \frac{P(L|B)P(B)}{P(L)}$

We already calculated the numerator $P(L|B)P(B) = \frac{2}{25}$ and the denominator $P(L) = \frac{37}{150}$.

Substitute these values into Bayes' Theorem:

$P(B|L) = \frac{\frac{2}{25}}{\frac{37}{150}}$

To divide fractions, multiply by the reciprocal of the denominator:

$P(B|L) = \frac{2}{25} \times \frac{150}{37}$

Simplify the expression:

$P(B|L) = \frac{2 \times 150}{25 \times 37} = \frac{300}{925}$

Simplify the fraction by dividing the numerator and denominator by their greatest common divisor, which is 25:

$P(B|L) = \frac{300 \div 25}{925 \div 25} = \frac{12}{37}$