Question:
A person goes to college either by bus, scooter or car. The probability that he goes by bus is \(\frac{2}{5}\), by scooter is \(\frac{1}{5}\) and by car is \(\frac{3}{5}\). The probability that he entered late in college if he goes by bus is \(\frac{1}{7}\), by scooter is \(\frac{3}{7}\) and by car is \(\frac{1}{7}\). If it is given that he entered late in college, then the probability that he goes to college by car is
A person goes to college either by bus, scooter or car. The probability that he goes by bus is \(\frac{2}{5}\), by scooter is \(\frac{1}{5}\) and by car is \(\frac{3}{5}\). The probability that he entered late in college if he goes by bus is \(\frac{1}{7}\), by scooter is \(\frac{3}{7}\) and by car is \(\frac{1}{7}\). If it is given that he entered late in college, then the probability that he goes to college by car is
Updated On: Apr 13, 2026
- \(\frac{3}{7}\)
- \(\frac{3}{8}\)
- \(\frac{4}{7}\)
- \(\frac{5}{8}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept:
This problem uses Bayes' Theorem. We are given the probabilities of different modes of transport (prior probabilities) and the conditional probabilities of being late given the mode of transport. We need to find the posterior probability of using a car given the person is late.
Step 2: Key Formula or Approach:
Let \(B, S, C\) be the events of going by Bus, Scooter, and Car respectively. Let \(L\) be the event of being Late. \[ P(C|L) = \frac{P(C) \cdot P(L|C)}{P(B) \cdot P(L|B) + P(S) \cdot P(L|S) + P(C) \cdot P(L|C)} \] *Note: The total probability in the problem statement (\(2/5 + 1/5 + 3/5 = 6/5\)) exceeds 1, suggesting a typo in the car probability. In standard versions of this problem, the car probability is \(2/5\).*
Step 3: Detailed Explanation:
1. Assume \(P(B) = 2/5\), \(P(S) = 1/5\), and \(P(C) = 2/5\) (so total = 1). 2. Given: \(P(L|B) = 1/7\), \(P(L|S) = 3/7\), \(P(L|C) = 1/7\). 3. Calculate Total Probability of being late \(P(L)\): \[ P(L) = \left(\frac{2}{5} \cdot \frac{1}{7}\right) + \left(\frac{1}{5} \cdot \frac{3}{7}\right) + \left(\frac{2}{5} \cdot \frac{1}{7}\right) = \frac{2}{35} + \frac{3}{35} + \frac{2}{35} = \frac{7}{35} \] 4. Calculate \(P(C|L)\): \[ P(C|L) = \frac{P(C) \cdot P(L|C)}{P(L)} = \frac{\frac{2}{35}}{\frac{7}{35}} = \frac{2}{7} \] *If using the user's provided car probability (\(3/5\)):* \[ P(L) = \frac{2}{35} + \frac{3}{35} + \frac{3}{35} = \frac{8}{35} \] \[ P(C|L) = \frac{3/35}{8/35} = \frac{3}{8} \]
Step 4: Final Answer:
The probability that he goes by car given he is late is \(\frac{3}{8}\).
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