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Use the complement rule. “At least one defective” = 1 − P(no defective). “No defective” means all 3 drawn from the 16 good ones. The number of 3-component samples with no defective is (163)=560\binom{16}{3} = 560. The total number of 3-component samples from 20 is (203)=1140\binom{20}{3} = 1140. Thus

P(no defective)=(163)(203)=5601140=56114=2857.P(\text{no defective})=\frac{\binom{16}{3}}{\binom{20}{3}}=\frac{560}{1140}=\frac{56}{114}=\frac{28}{57}.

Therefore

P(at least one defective)=1−2857=57−2857=2957.P(\text{at least one defective}) = 1 - \frac{28}{57} = \frac{57-28}{57} = \frac{29}{57}.

Choice C. This structure—“at least one”—nearly always favors complementary counting because the direct approach would require summing multiple cases (exactly 1 defective + exactly 2 + exactly 3). With combinations, ensure you use the same denominator for all outcomes and simplify fractions at the end to avoid arithmetic clutter. A quick plausibility check: since 20% of the lot is defective, drawing three without replacement should yield a probability modestly above 1−0.83=0.4881 - 0.8^3 = 0.488. Our exact result 29/57≈0.50929/57 \approx 0.509 is consistent, because without replacement slightly increases the chance of hitting a defective compared with independent draws.