Question

It is known that the probability that an item produced by a certain machine will be defective is $0.01$. Use Poisson distribution, to find the probability in a sample of $100$ items selected at random from the total output, that there are not more than one defective item.

• A. $\frac{3}{e}$
• B. $\frac{2}{e}$
• C. $\frac{1}{e}$
• D. $\frac{4}{e}$

Answer 1

The correct option is B $\frac{2}{e}$

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(parameter)
$x$: 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 $\mu$
Here, $\mu$ = $100\times 0.01=1$
$P(x;\mu )=P(x=0;1)+P(x=1;1)$
$P(x;1)=\frac{e^{-1}{1}^0}{0!}+\frac{e^{-1}{1}^1}{1!}$
$=e^{-1}+e^{-1}=2e^{-1}=\frac{2}{e}$