1. **State the problem:** We want to find the probability of exactly 3 defective bulbs in a sample of 50, where each bulb has a 7% chance of being defective. 2. **Identify the distribution:** This is a binomial probability problem because each bulb is either defective or not, independently. 3. **Formula:** The binomial probability formula is: $$P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $n=50$, $k=3$, and $p=0.07$. 4. **Calculate the binomial coefficient:** $$\binom{50}{3} = \frac{50!}{3! \times (50-3)!} = \frac{50 \times 49 \times 48}{3 \times 2 \times 1} = 19600$$ 5. **Calculate the probability:** $$P(X=3) = 19600 \times (0.07)^3 \times (0.93)^{47}$$ 6. **Evaluate powers:** $$0.07^3 = 0.000343$$ $$0.93^{47} \approx 0.049$$ 7. **Multiply all parts:** $$P(X=3) = 19600 \times 0.000343 \times 0.049 \approx 0.329$$ **Final answer:** The probability of finding exactly 3 defective bulbs in a sample of 50 is approximately **0.329**.