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) → $\frac{4}{e^3}$
Defective rate = $3\% = 0.03$
Sample size = $100$
Mean of Poisson distribution: $\lambda = np = 100 \times 0.03 = 3$
Required probability: $P(X \leq 1) = P(0) + P(1)$
$P(0) = \frac{e^{-3} 3^0}{0!} = e^{-3}$
$P(1) = \frac{e^{-3} 3^1}{1!} = 3e^{-3}$
$P(X \leq 1) = e^{-3} + 3e^{-3} = 4e^{-3}$