Answer to Question #215933 in Statistics and Probability for Ray

Question

A company manufactures fuses. The percentage of non-defective fuses is 95.4%. A sample of 9 fuse was selected. Calculate the probability of selecting at least 3 defective fuses.   

Answer 1

The Poisson distribution is

$P\left(X=x\right)=\frac{e^{-\lambda }\lambda ^x}{x!}$

Let $x$ denote the defective fuse

$P=100\%-95.4\%=4.6\%=0.046$

$n=9$

$mean=\lambda =np=9\times 0.046=0.414$

$P\left(at\,least\:3\:defective\:fuses\right)=\sum_{x \ge 3} \frac{e^{-0.414}0.414^3}{3}=0.0156$