Defective Bulbs and the Warranty Rule
A bulb manufacturer in Assam claims that 5% of bulbs in a large consignment are defective. A retailer receives a consignment and checks 20 bulbs selected at random. The warranty policy says the retailer may return the consignment if at least one defective bulb is found in the sample. The manufacturer argues that since 5% of 20 is 1, finding one defective is “normal” and the return rule is unfair. The retailer wants to know the probability of at least one defective, the probability of exactly one defective, and the expected number and standard deviation of defectives in a sample of 20. For a larger order of 100 bulbs, the retailer plans to use a Poisson approximation with λ = 5, but is unsure whether that is justified. Assume independence and a constant defect rate.
1. Using the binomial distribution, compute the probability of at least one defective and the probability of exactly one defective in 20 bulbs.
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Let X be the number of defectives, with n = 20 and p = 0.05. Then
[ P(X=0)=0.95^{20}\approx0.3585 ] [ P(X\ge1)=1-P(X=0)\approx1-0.3585=0.6415 ]
For exactly one defective:
[ P(X=1)=\binom{20}{1}(0.05)^1(0.95)^{19} =20(0.05)(0.3774)\approx0.3774 ]
So there is about a 64.2% chance that the return rule is triggered, and about a 37.7% chance of exactly one defective. The manufacturer's phrase “5% of 20 is 1” describes an expected value, not the probability of the event. In fact, getting at least one defective is more likely than not.
2. Compute the expected number and standard deviation of defectives in the sample of 20, and comment on the manufacturer's argument.
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For a binomial distribution:
[ E(X)=np=20(0.05)=1 ] [ SD(X)=\sqrt{np(1-p)}=\sqrt{20(0.05)(0.95)}=\sqrt{0.95}\approx0.975 ]
The expected number is 1, but expectation is a long-run average, not a prediction that exactly one defective is the most likely outcome. The distribution is spread out: P(X = 0) ≈ 0.359, P(X = 1) ≈ 0.377 and P(X = 2) ≈ 0.189. The manufacturer's argument that one defective is “normal” is therefore misleading. The return rule triggers in almost two-thirds of samples of this size, so if the rule is considered too strict, the policy should be changed explicitly rather than defended by confusing an expected value with a probability.
3. For a sample of 100 bulbs with p = 0.05, assess whether the Poisson approximation with λ = 5 is justified.
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Here n = 100, p = 0.05 and λ = np = 5. The Poisson approximation to the binomial is generally reasonable when n is large and p is small, with np moderate. The rule-of-thumb conditions n ≥ 20 and p ≤ 0.05 are satisfied, so the approximation is acceptable for quick work. For example, the Poisson gives
[ P(X=0)=e^{-5}\approx0.0067 ]
compared with the binomial value ((0.95)^{100}\approx0.0059). The approximation is close but not exact; with software or tables, the binomial is preferable. Also, the Poisson model assumes independent bulbs and a constant defect rate, so a batch effect or machine drift would violate the model.