1. Let's state the first problem: We have four inspectors who stamp film packages with different probabilities of failure. We want the probability that a package missing the expiration date was stamped by John. 2. Define events: - Let $J, T, U, P$ be the events that John, Tom, Jeư, and Pat stamped the package respectively. - Their probabilities are $P(J)=0.20$, $P(T)=0.60$, $P(U)=0.15$, $P(P)=0.05$. - Failure rates (no stamp) are: $P(F|J) = \frac{1}{200} = 0.005$, $P(F|T) = \frac{1}{100} = 0.01$, $P(F|U) = \frac{1}{90} \approx 0.0111$, $P(F|P) = \frac{1}{200} = 0.005$. 3. Use total probability to find $P(F)$, the probability that a package is not stamped: $$P(F) = P(F|J)P(J) + P(F|T)P(T) + P(F|U)P(U) + P(F|P)P(P)$$ Calculate: $$P(F) = 0.005 \times 0.20 + 0.01 \times 0.60 + 0.0111 \times 0.15 + 0.005 \times 0.05$$ $$= 0.001 + 0.006 + 0.001665 + 0.00025 = 0.008915$$ 4. We want $P(J|F)$, the probability the package was stamped by John given failure: $$P(J|F) = \frac{P(F|J)P(J)}{P(F)} = \frac{0.005 \times 0.20}{0.008915} = \frac{0.001}{0.008915} \approx 0.1121$$ 5. Answer: The probability that the failed package was inspected by John is approximately $0.112$ or 11.2%. 6. Now, problem 2a: Given 20% defect rate ($p=0.2$), probability all three items are defective is $$P(3 \text{ defective}) = p^3 = 0.2^3 = 0.008$$ 7. Problem 2b: Probability that in 4 items, exactly 3 are defective follows the binomial formula: $$P(X=3) = \binom{4}{3} p^3 (1-p)^1 = 4 \times 0.2^3 \times 0.8 = 4 \times 0.008 \times 0.8 = 0.0256$$ 8. Answer: a) Probability all three defective is 0.008 or 0.8%. b) Probability exactly three defective out of four is 0.0256 or 2.56%.