Independent defects and a noisy inspection signal
A hypothetical production model assigns every item an independent defect probability of0.10. A researcher selects five items under those stated assumptions and records the defect count X. Separately, an inspection signal is positive with probability0.90 for a defective item and with probability0.05 for a nondefective item. The researcher draws a tree diagram before interpreting a positive signal. A manager treats the inspection sensitivity as the probability that a positive item is defective, overlooking the rate of defects in the population. Another manager says that an expected defect count of one-half means that exactly half an item must be defective in every sample. All rates are supplied invented exercise data, not claims about an actual device, laboratory assay or population.
1. Specify X and calculate the probability that at least one of five items is defective.
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Under the independent constant-probability assumptions, X~Binomial(5,0.1). P(X≥1)=1-P(X=0)=1-0.9⁵=0.40951. At least one is not the same event as exactly one.
2. Calculate P(X=2) and E(X), explaining what the expectation means.
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P(X=2)=C(5,2)(0.1)²(0.9)³=0.0729. E(X)=5×0.1=0.5, a long-run average count over repeated samples, not a possible observed half-item count.
3. Find the probability that an item is defective given a positive inspection signal.
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The defective-positive probability is0.1×0.9=0.09. The total positive probability is0.09+0.9×0.05=0.135. Bayes gives0.09/0.135=2/3. Sensitivity0.90 conditions on a different event.