1. **Problem statement:** We have a set of customer accounts with an error rate of 5% (probability of error $p=0.05$). The auditor samples $n$ accounts randomly. - If no errors are found in the sample, the accounts are passed. - If any error is found, all accounts are inspected. We want to solve: a. Minimum sample size $n$ so the auditor has a better than 50% chance of discovering errors. b. Probability that the accounts are passed (no errors found in sample). c. Probability that all errors in the account will be rejected (i.e., errors found in sample). 2. **Formula and explanation:** This is a binomial distribution problem where the number of errors found in the sample follows $X \sim \text{Binomial}(n, p)$. - Probability of finding exactly $k$ errors in sample of size $n$ is: $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ - Probability of finding no errors in sample ($k=0$): $$P(X=0) = (1-p)^n$$ - Probability of finding at least one error: $$P(X \geq 1) = 1 - P(X=0) = 1 - (1-p)^n$$ 3. **Part a: Minimum sample size $n$ for better than even chance to find errors** We want: $$P(X \geq 1) > 0.5$$ Substitute: $$1 - (1-p)^n > 0.5$$ Rearranged: $$(1-p)^n < 0.5$$ Take natural logarithm on both sides: $$n \ln(1-p) < \ln(0.5)$$ Since $\ln(1-p)$ is negative (because $p=0.05$), dividing reverses inequality: $$n > \frac{\ln(0.5)}{\ln(1-p)}$$ Calculate values: $$\ln(0.5) = -0.693147$$ $$\ln(1-0.05) = \ln(0.95) = -0.051293$$ So: $$n > \frac{-0.693147}{-0.051293} = 13.52$$ Minimum integer $n$ is 14. 4. **Part b: Probability that accounts are passed (no errors found in sample) for $n=10$** $$P(X=0) = (1-p)^n = 0.95^{10}$$ Calculate: $$0.95^{10} = e^{10 \ln(0.95)} = e^{10 \times (-0.051293)} = e^{-0.51293} \approx 0.5987$$ So probability accounts are passed is approximately 0.5987. 5. **Part c: Probability that all errors in the account will be rejected (errors found in sample) for $n=10$** This is the complement of part b: $$P(X \geq 1) = 1 - P(X=0) = 1 - 0.5987 = 0.4013$$ **Final answers:** a. Minimum sample size $n=14$. b. Probability accounts are passed for $n=10$ is approximately 0.5987. c. Probability all errors are rejected for $n=10$ is approximately 0.4013.