Q.7 In a factory, two machines M1 and M2 manufacture 60% and 40% of the autocomponents respectively. Out of the total production, 2% of M1 and 3% of M2 are found to be defective. If a randomly drawn autocomponent from the combined lot is found defective, what is the probability that it was manufactured by M2?
Introduction
Problems based on conditional probability and
Bayes’ theorem are frequently asked in board exams and
competitive examinations such as JEE and NEET.
In this article, we solve a probability problem involving two machines
manufacturing auto-components and determine the probability that a
defective component was produced by machine M2.
Key Concept Used
This problem is solved using Bayes’ Theorem, which helps
determine the probability of a cause given that an event has already occurred.
Bayes’ Theorem:
P(M2 | D) =P(M2) · P(D | M2)⁄P(D)
Step-by-Step Solution
Step 1: Given Probabilities
P(M1) = 0.60
P(M2) = 0.40
P(D | M1) = 0.02
P(D | M2) = 0.03
Step 2: Total Probability of a Defective Component
P(D) =P(M1)P(D | M1) +P(M2)P(D | M2)
P(D) = (0.60 × 0.02) + (0.40 × 0.03)
P(D) = 0.012 + 0.012 = 0.024
Step 3: Apply Bayes’ Theorem
P(M2 | D) =0.40 × 0.03⁄0.024
P(M2 | D) =0.012⁄0.024 = 0.5
Final Answer
Probability that the defective component was manufactured by M2 = 0.5
Correct Option: (C)
Quick Exam Tip
Whenever a question states “given that an item is defective”
and involves multiple sources, always apply Bayes’ Theorem.
Conclusion
This problem highlights the importance of conditional probability in real-life
situations. Even though M2 produces fewer components, its higher defect rate
makes it equally likely to be responsible for a defective item.
Understanding Bayes’ theorem ensures accuracy and confidence in probability-based
exam questions.
