Question

If on an average ,5 percent of the output in a factory making certain parts, is defective and that 200 units are in a package then the probability that atmost 4 defective parts may be found in that package is

• A. $e^{-10}[1+\frac{100}{1!}+\frac{{100}^2}{2!}+\frac{{100}^3}{3!}+\frac{{100}^4}{4!}]$
• B. $e^{-10}[1+\frac{10}{1!}+\frac{{10}^2}{2!}+\frac{{10}^3}{3!}+\frac{{10}^4}{4!}]$
• C. $e^{-10}[1-\frac{10}{1!}+\frac{{10}^2}{2!}+\frac{{10}^3}{3!}+\frac{{10}^4}{4!}]$
• D. $e^{-10}[1-\frac{100}{1!}+\frac{{100}^2}{2!}+\frac{{100}^3}{3!}+\frac{{100}^4}{4!}]$

Answer 1

The correct option is B $e^{-10}[1+\frac{10}{1!}+\frac{{10}^2}{2!}+\frac{{10}^3}{3!}+\frac{{10}^4}{4!}]$

Here $\lambda$
$=\frac{5}{100}.200$
$=10$.
Hence by applying Poisson distribution, we get that the probability that atmost 4 defective part are found is
$={\sum }_{k=0}^{k=4}\frac{e^{-10}{.10}^k}{k!}$
$=e^{-10}[1+\frac{10}{1!}+\frac{{10}^2}{2!}+\frac{{10}^3}{3!}+\frac{{10}^4}{4!}]$.