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:

$\frac{4}{e^3}$

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: $P(X \leq 1) = P(0) + P(1)$

$P(0) = \frac{e^{-3} 3^0}{0!} = e^{-3}$

$P(1) = \frac{e^{-3} 3^1}{1!} = 3e^{-3}$

$P(X \leq 1) = e^{-3} + 3e^{-3} = 4e^{-3}$