1. **State the problem:** We know that 3% of people have questionable objects in their luggage. We want to find the probability that 15 people pass through screening successfully before one person is caught with a questionable object. 2. **Identify the probability of success and failure:** The probability that a person has questionable objects is $p = 0.03$. 3. The probability that a person passes screening successfully (no questionable objects) is $q = 1 - p = 1 - 0.03 = 0.97$. 4. **Model the problem:** We want the probability that the first 15 people pass successfully, and the 16th person is caught with questionable objects. 5. This is a geometric distribution scenario where the probability of the first failure (caught) occurs on the 16th trial. 6. The probability is given by: $$P = q^{15} \times p = 0.97^{15} \times 0.03$$ 7. Calculate $0.97^{15}$: $$0.97^{15} \approx 0.642$$ 8. Multiply by $p$: $$P \approx 0.642 \times 0.03 = 0.01926$$ 9. **Final answer:** The probability that 15 people pass successfully before one is caught is approximately **0.0193** or 1.93%.