SAT Math · Probability and conditional probability · hard

A factory's Machine A produces 60% of all units with a 3% defect rate, and Machine B produces the remaining 40% with a 5% defect rate.…

A factory's Machine A produces 60% of all units with a 3% defect rate, and Machine B produces the remaining 40% with a 5% defect rate. What is the overall probability, as a percent, that a randomly selected unit from this factory is defective?

Answer and explanation

Answer: 3.8

Use the law of total probability: weight each machine's defect rate by its share of production. Machine A contributes 0.60×0.03=0.0180.60 \times 0.03 = 0.018, and Machine B contributes 0.40×0.05=0.0200.40 \times 0.05 = 0.020. Together, 0.018+0.020=0.0380.018 + 0.020 = 0.038, which is 3.8%. A common wrong answer is 4, from simply averaging the two defect rates, 3+52\frac{3 + 5}{2}, which ignores that Machine A makes more of the units. Another is 8, from adding 3% and 5%. A third is 1.8, which counts only Machine A. The overall rate must lie between 3% and 5%, closer to 3%.

More Probability and conditional probability questions

All Probability and conditional probability questions

Learn the method behind questions like this one

The SAT Math course teaches each tested skill through lessons, step-by-step solutions to sample problems, the complete study guide and full-length practice tests.