A factory produces 500 units of a product per day. Each unit is inspected by three different inspectors. If each inspector finds defects in 5% of the units they inspect, what is the probability that a randomly chosen unit will have no defects found by any of the inspectors?
Let's analyze the situation and calculate the probability:
1. We know that there are three inspectors, and each one finds defects in 5% of the units they inspect.
2. This means that each inspector approves 95% of the units they inspect (100% - 5%).
3. Since all three inspectors must approve a unit for it to be considered defect-free, we need to find the probability that all three inspectors approve a randomly chosen unit.
4. The probability of independent events occurring together is calculated by multiplying their individual probabilities: 0.95 * 0.95 * 0.95 ≈ 0.857375.
5. To express this as a percentage, we multiply by 100: 0.857375 * 100 ≈ 85.74%.

The probability that a randomly chosen unit will have no defects found by any of the inspectors is approximately 85.74%.