1. **Problem statement:** A company has three plants producing 50%, 30%, and 20% of relays. The probabilities of a relay being defective from these plants are 0.02, 0.05, and 0.01 respectively. We want to find: (a) The probability a randomly selected relay is defective. (b) Given a relay is defective, the probability it was made by plant 2. 2. **Formula and rules:** - Total probability of defectiveness: $$P(B) = \sum P(B|A_i)P(A_i)$$ where $A_i$ is the event relay is from plant $i$. - Bayes' theorem for conditional probability: $$P(A_2|B) = \frac{P(B|A_2)P(A_2)}{P(B)}$$ 3. **Calculations for (a):** $$P(B) = (0.02)(0.5) + (0.05)(0.3) + (0.01)(0.2)$$ $$= 0.01 + 0.015 + 0.002 = 0.027$$ 4. **Calculations for (b):** $$P(A_2|B) = \frac{(0.05)(0.3)}{0.027} = \frac{0.015}{0.027}$$ Showing cancellation: $$= \frac{\cancel{0.015}}{\cancel{0.027}} = 0.556$$ 5. **Final answers:** - (a) Probability relay is defective: $0.027$ - (b) Probability defective relay is from plant 2: $0.556$ These results use the law of total probability and Bayes' theorem to combine and invert conditional probabilities.