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:
We are dealing with conditional probability where an outcome has already occurred (reaching late), and we need to find the probability of one of the prior causes (travelled by bus). This is a classic application of Bayes' Theorem.
Step 2: Key Formula or Approach:
Bayes' Theorem states:
\( P(B|L) = \frac{P(L|B) \cdot P(B)}{P(L|B) \cdot P(B) + P(L|S) \cdot P(S) + P(L|C) \cdot P(C)} \)
where \( B, S, C \) are the events of choosing bus, scooter, and car, and \( L \) is the event of reaching late.
Step 3: Detailed Explanation:
Let's list the prior probabilities:
\( P(B) = \frac{2}{5} \)
\( P(S) = \frac{1}{5} \)
\( P(C) = \frac{2}{5} \)
List the conditional probabilities of reaching late given the transport:
\( P(L|B) = \frac{1}{5} \)
\( P(L|S) = \frac{1}{3} \)
\( P(L|C) = \frac{1}{4} \)
We need to calculate the numerator (probability of taking a bus AND being late):
\( P(L \cap B) = P(L|B) \cdot P(B) = \frac{1}{5} \times \frac{2}{5} = \frac{2}{25} \).
Next, calculate the total probability of reaching late (the denominator):
\( P(L) = P(L|B)P(B) + P(L|S)P(S) + P(L|C)P(C) \)
\( P(L) = \left(\frac{1}{5} \times \frac{2}{5}\right) + \left(\frac{1}{3} \times \frac{1}{5}\right) + \left(\frac{1}{4} \times \frac{2}{5}\right) \)
\( P(L) = \frac{2}{25} + \frac{1}{15} + \frac{2}{20} \)
To add these fractions easily, find a common denominator for 25, 15, and 20, which is 300.
\( \frac{2}{25} = \frac{2 \times 12}{300} = \frac{24}{300} \)
\( \frac{1}{15} = \frac{1 \times 20}{300} = \frac{20}{300} \)
\( \frac{2}{20} = \frac{1}{10} = \frac{30}{300} \)
So, \( P(L) = \frac{24 + 20 + 30}{300} = \frac{74}{300} \).
Now, apply Bayes' Theorem:
\( P(B|L) = \frac{P(L \cap B)}{P(L)} = \frac{24/300}{74/300} = \frac{24}{74} \).
Simplifying the fraction gives:
\( \frac{24}{74} = \frac{12}{37} \).
Step 4: Final Answer:
The probability that the candidate travelled by bus is \(\frac{12}{37}\).