Question

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. \(\frac{14}{37} \)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This problem is an application of Bayes' Theorem, which calculates the probability of an event based on prior knowledge of conditions that might be related to the event.
Step 2: Key Formula or Approach:
Bayes' Theorem: \[ P(A|B) = \frac{P(B|A)P(A)}{P(B)} \] Where $P(B)$ is the total probability of being late.
Step 3: Detailed Explanation:
Let $B, S, C$ be the events of choosing Bus, Scooter, and Car, and $L$ be the event of being late. $P(B) = 2/5, P(S) = 1/5, P(C) = 2/5$. $P(L|B) = 1/5, P(L|S) = 1/3, P(L|C) = 1/4$. First, calculate the total probability of being late $P(L)$: \[ P(L) = P(B)P(L|B) + P(S)P(L|S) + P(C)P(L|C) \] \[ P(L) = \left(\frac{2}{5} \times \frac{1}{5}\right) + \left(\frac{1}{5} \times \frac{1}{3}\right) + \left(\frac{2}{5} \times \frac{1}{4}\right) \] \[ P(L) = \frac{2}{25} + \frac{1}{15} + \frac{1}{10} = \frac{12 + 10 + 15}{150} = \frac{37}{150} \] Now, use Bayes' Theorem for $P(B|L)$: \[ P(B|L) = \frac{P(B)P(L|B)}{P(L)} = \frac{2/25}{37/150} \] \[ P(B|L) = \frac{12/150}{37/150} = \frac{12}{37} \]
Step 4: Final Answer:
The probability that the candidate travelled by bus, given they were late, is $\frac{12}{37}$.