It is known that 3% of plastic bags manufactured in a factory are defective. Using the Poisson distribution on a sample of 100 bags, the probability of at most one defective bag is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{4}{e^3}\)
The number X of defective bags in the sample follows a Poisson distribution with mean
\(\lambda = np = 100 \times 0.03 = 3\)
\( \displaystyle P(X = r) = \frac{e^{-\lambda}\lambda^r}{r!} \)
“At most one” means X = 0 or X = 1:
\(P(X \le 1) = P(X = 0) + P(X = 1)\)
\(= e^{-3} + 3e^{-3}\)
\(= 4e^{-3}\)
\( \displaystyle = \frac{4}{e^3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{4}{e^2}\) — Incorrect. The mean is \(\lambda = 3\), so the factor is \(e^{-3}\), not \(e^{-2}\).
Option (3) \(e^{-3}\) — Incorrect. This is P(X = 0) alone, the probability of no defective bag; it leaves out P(X = 1).
Option (4) \(\displaystyle\frac{1}{3}\) — Incorrect. \(4e^{-3} \approx 0.199\), not 0.333.