Subtopic: Poisson, normal and binomial distributions
Question Statement:
In a seed quality testing trial, the number of defective seeds per unit volume follows a Poisson distribution with mean parameter $\lambda = 2$. The probability that a randomly drawn sample contains at least $2$ defective seeds is ________ (round off to three decimal places).
Answer Key:
0.594
Step-by-Step Solution & Explanation
For a Poisson distribution with mean $\lambda = 2$:
$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!} = \frac{e^{-2} 2^k}{k!}$
We seek $P(X \ge 2) = 1 - [P(X = 0) + P(X = 1)]$.
$P(X = 0) = \frac{e^{-2} 2^0}{0!} = e^{-2}$
$P(X = 1) = \frac{e^{-2} 2^1}{1!} = 2 e^{-2}$
Sum of probabilities for $X = 0$ and $X = 1$:
$P(X = 0) + P(X = 1) = 3 e^{-2}$
Given $e^{-2} \approx 0.135335$:
$3 e^{-2} \approx 3 \times 0.135335 = 0.406006$
$P(X \ge 2) = 1 - 0.406006 = 0.593994 \approx 0.594$