Question

A manufactured product on an average has 2 defects per unit of product produced. If the number of defects follows Poisson distribution, the probability of finding at least one defect is

• A. $e^{-2}$
• B. $1-e^{-2}$
• C. $\frac{e^{-2}{2}^1}{1!}$
• D. $e^{-0.02}$

Answer 1

Answer: (B)

$1-e^{-2}$

By using Poisson distribution,
$P(x;\mu )=\frac{e^{-\mu }{\mu }^x}{x!}$
$\mu$: The mean number of successes that occur in a specified region.
$x$: The actual number of successes that occur in a specified region.
$P(x$; $\mu$): The Poisson probability that exactly x successes occur in a Poisson experiment, when the mean number of successes is $\mu$
$P(x\ge 1;\mu )=1-P(x=0;\mu )$
$\mu =2$
$x=0$ (No defective product)
$P(x\ge 1;2)=1-P(x=0;2)=1-\frac{e^{-2}{2}^0}{0!}=1-e^{-2}$