Question

A manufacturer of cotter pins knows that $5$% of his product is defective. If he sells cotter pins in boxes of $100$ and guarantees that not more than $10$ pins will be defective, the approximate probability that a box will fail to meet the guaranteed quality is

• A. $\frac{e^{-5}{5}^{10}}{10!}$
• B. $1-\sum _{x=0}^{10}\frac{e^{-5}{5}^x}{x!}$
• C. $1-\sum _{x=0}^{\infty }\frac{e^{-5}{5}^x}{x!}$
• D. $\sum _{x=0}^{\infty }\frac{e^{-5}{5}^x}{x!}$

Answer 1

Answer: (B)

$1-\sum _{x=0}^{10}\frac{e^{-5}{5}^x}{x!}$

we are given $n=100$
let $p=$ probability of a defective bulb $=5$%$=0.05$
$\therefore m=$ mean number of defective bulbs in a box of $100=np=100\times 0.05=5$
Since p is small , we can use poison's distribution.
Probability of $x$ defective bulbs in a box of $100$ is
$P(X=x)=\frac{e^{-m}{m}^x}{x!}=\frac{e^{-5}{5}^x}{x!},x=0,1,\mathrm{2...}$
Probability that is box will fail to meet the guarented quality is $P(X>10)=1-P(X\le 10)=1-\sum _{x=0}^{10}\frac{e^{-5}{5}^x}{x!}=1-e^{-5}\sum _{x=0}^{10}\frac{5^x}{x!}$