1. **State the problem:** We want to find the probability of having at most 2 defective items in a batch of 50, where each item has a 5% defect rate. This means we want $P(X \leq 2)$ where $X$ is the number of defective items and $X \sim \text{Binomial}(n=50, p=0.05)$. 2. **Recall the binomial probability formula:** $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=50$, $p=0.05$, and $k$ is the number of defective items. 3. **Calculate $P(X \leq 2)$:** This is the sum of probabilities for $k=0,1,2$: $$P(X \leq 2) = P(X=0) + P(X=1) + P(X=2)$$ 4. **Calculate each term:** - For $k=0$: $$P(X=0) = \binom{50}{0} (0.05)^0 (0.95)^{50} = 1 \times 1 \times 0.95^{50} = 0.95^{50}$$ - For $k=1$: $$P(X=1) = \binom{50}{1} (0.05)^1 (0.95)^{49} = 50 \times 0.05 \times 0.95^{49}$$ - For $k=2$: $$P(X=2) = \binom{50}{2} (0.05)^2 (0.95)^{48} = \frac{50 \times 49}{2} \times 0.05^2 \times 0.95^{48}$$ 5. **Evaluate the powers and combinations:** - $0.95^{50} \approx 0.0769$ - $0.95^{49} \approx 0.0809$ - $0.95^{48} \approx 0.0852$ - $\binom{50}{2} = \frac{50 \times 49}{2} = 1225$ 6. **Calculate each probability:** - $P(X=0) \approx 0.0769$ - $P(X=1) \approx 50 \times 0.05 \times 0.0809 = 2.5 \times 0.0809 = 0.2023$ - $P(X=2) \approx 1225 \times 0.0025 \times 0.0852 = 1225 \times 0.000213 = 0.2609$ 7. **Sum the probabilities:** $$P(X \leq 2) \approx 0.0769 + 0.2023 + 0.2609 = 0.5401$$ 8. **Interpretation:** There is approximately a 54% chance that a batch of 50 items will have at most 2 defective items. For quality control, this means that more than half the batches meet this defect threshold, which can guide decisions on acceptable quality levels and inspection standards.