1. **State the problem:** We want to find the probability that more than 2 parts are defective out of 7 parts produced by a machine that produces defective parts 12% of the time. 2. **Identify the distribution:** This is a binomial probability problem where the number of trials $n=7$, the probability of success (defective part) $p=0.12$, and we want $P(X>2)$ where $X$ is the number of defective parts. 3. **Formula:** The binomial probability formula is $$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$ where $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. 4. **Calculate $P(X>2)$:** $$P(X>2) = 1 - P(X \leq 2) = 1 - [P(X=0) + P(X=1) + P(X=2)]$$ 5. **Calculate each term:** - $P(X=0) = \binom{7}{0} (0.12)^0 (0.88)^7 = 1 \times 1 \times 0.88^7 = 0.88^7$ - $P(X=1) = \binom{7}{1} (0.12)^1 (0.88)^6 = 7 \times 0.12 \times 0.88^6$ - $P(X=2) = \binom{7}{2} (0.12)^2 (0.88)^5 = 21 \times 0.12^2 \times 0.88^5$ 6. **Calculate powers:** - $0.88^7 = 0.4182$ - $0.88^6 = 0.4757$ - $0.88^5 = 0.5404$ 7. **Calculate probabilities:** - $P(X=0) = 0.4182$ - $P(X=1) = 7 \times 0.12 \times 0.4757 = 7 \times 0.0571 = 0.3997$ - $P(X=2) = 21 \times 0.0144 \times 0.5404 = 21 \times 0.00778 = 0.1634$ 8. **Sum probabilities:** $$P(X \leq 2) = 0.4182 + 0.3997 + 0.1634 = 0.9813$$ 9. **Find final probability:** $$P(X > 2) = 1 - 0.9813 = 0.0187$$ 10. **Round to two decimal places:** $$\boxed{0.02}$$ So, the probability that more than 2 parts are defective is approximately 0.02.