Probability
Bayes' Theorem
Grade 12
Question:
<p>A person goes to office either by car, scooter, bus or train, the probability of which being 1/7, 3/7, 2/7 and 1/7, respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is 2/9, 1/9, 4/9 and 1/9 respectively. Given that he reached office in time, what is the probability that he traveled by a car?</p>
<p>\(\dfrac{1}{7}\)</p>
<p>\(\dfrac{2}{7}\)</p>
<p>\(\dfrac{1}{4}\)</p>
<p>\(\dfrac{1}{49}\)</p>
Step-by-Step Solution
Key Concept: Use Bayes' theorem: P(Car|On time) = P(On time|Car) × P(Car) / P(On time). Calculate P(On time) by finding the total probability using the law of total probability across all transport modes.
<p><strong>Step 1:</strong> Identify given probabilities:</p><p>P(Car) = 1/7, P(Scooter) = 3/7, P(Bus) = 2/7, P(Train) = 1/7</p><p>P(Late|Car) = 2/9 ⟹ P(On time|Car) = 7/9</p><p>P(Late|Scooter) = 1/9 ⟹ P(On time|Scooter) = 8/9</p><p>P(Late|Bus) = 4/9 ⟹ P(On time|Bus) = 5/9</p><p>P(Late|Train) = 1/9 ⟹ P(On time|Train) = 8/9</p><p><strong>Step 2:</strong> Calculate P(On time) using law of total probability:</p><p>P(On time) = P(On time|Car)·P(Car) + P(On time|Scooter)·P(Scooter) + P(On time|Bus)·P(Bus) + P(On time|Train)·P(Train)</p><p>P(On time) = (7/9)(1/7) + (8/9)(3/7) + (5/9)(2/7) + (8/9)(1/7)</p><p>P(On time) = 1/9 + 24/63 + 10/63 + 8/63 = 7/63 + 24/63 + 10/63 + 8/63 = 49/63 = 7/9</p><p><strong>Step 3:</strong> Apply Bayes' theorem:</p><p>P(Car|On time) = [P(On time|Car) × P(Car)] / P(On time)</p><p>P(Car|On time) = [(7/9)(1/7)] / (7/9) = (1/9) / (7/9) = 1/7</p><p>∴ Answer: A (1/7)</p>
Correct Answer: A