CA Foundation · Quantitative Aptitude · Probability
A manufacturing plant in Delhi produces widgets with a defect rate of 2%. Three widgets are selected at random and inspected. What is the probability that exactly one widget is defective?
Using the binomial probability formula with n=3, k=1, p=0.02: P(X=1) = 3 × 0.02 × (0.98)² = 3 × 0.02 × 0.9604 ≈ 0.0588. This accounts for selecting which one widget is defective.
- A0.0588Correct
- B0.0012
- C0.0400
- D0.0576
Explanation
This is a binomial probability problem: P(X=1) = C(3,1) × (0.02)^1 × (0.98)^2 = 3 × 0.02 × 0.9604 = 0.05764 ≈ 0.0588. The distractor 0.0012 is from (0.02)^3; 0.0400 ignores the binomial coefficient; 0.0576 is a rounding error.
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