Question

The probability that atmost $5$ defective fuses will be found in a box of $200$ fuses, if experience shows that $20\%$ of such fuses are defective, is

• A. $\frac{e^{-40}{40}^5}{5!}$
• B. $\sum _{x=0}^5\frac{e^{-40}{40}^x}{x!}$
• C. $\sum _{x=6}^{\infty }\frac{e^{-40}{40}^x}{x!}$
• D. $1-\sum _{x=6}^{\infty }\frac{e^{-40}{40}^x}{x!}$

Answer 1

The correct option is B $\sum _{x=0}^5\frac{e^{-40}{40}^x}{x!}$

The poisson distribution is
$P(X=x)=\frac{e^{-\lambda }{\lambda }^x}{x!}$ , $x=0,1,2,3,..$
Let, $X$ denote the defective fuse

$p=\frac{20}{100}$
$n=200$
mean$=\lambda =np=200\times \frac{20}{100}=40$
$=>P($atmost $5$ defective fuses)${=}_{x=0}^5\sum \frac{e^{-40}{40}^x}{x!}$