Question

If $2$% of pipes manufactured by a company are defective, the probability that in a sample of $1000$ pipes, exactly $6$ pipes are defective is (Assume Poisson distribution for the result)

• A. $\frac{e^{-2}{2}^6}{6!}$
• B. $\frac{e^{-20}(20{)}^6}{6!}$
• C. $e^{-20}$
• D. $e^{-2}$

Answer 1

Answer: (B)

$\frac{e^{-20}(20{)}^6}{6!}$

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 $\mu$.
$\mu =1000\times 0.02=20$
$x=6$
$P(6;20)=\frac{e^{-20}{20}^6}{6!}$