GMAT Focus EditionQuantitative ReasoningMedium
A quality control manager inspects a batch of 500 items. Historically, 3% of items are defective. If the manager randomly selects 20 items for inspection, what is the probability that none of the selected items are defective?
- A(0.97)^20
- B(485/500) * (484/499) * ... * (466/481)
- C(500 C 20) / (485 C 20)
- D(0.03)^20
Show answer & explanationAnswer & explanation
Correct answer: A. (0.97)^20
Since the sample size (20) is small compared to the total population (500), and the defect rate is low, we can approximate this as a binomial probability. The probability of one item not being defective is 1 - 0.03 = 0.97. For 20 items to be non-defective, it's (0.97)^20.
Why the other options are wrong
- B. Incorrect. This represents the probability if we were sampling without replacement from a known number of non-defective items, which is more complex than needed for this approximation.
- C. Incorrect. This combination setup is inverted and does not represent the desired probability.
- D. Incorrect. This would be the probability that all 20 items are defective.
Binomial Probability Approximation
When sampling without replacement from a large population, the probability of success can be approximated using binomial probability if the sample size is small relative to the population.
- Used when there are two possible outcomes (success/failure).
- Probability of success remains constant for each trial (approximately).
- Trials are independent (approximately).
Memory trick: Probability's a path, count the good, divide by all the math.