Question

Suppose a machine produces metal parts that contain some defective parts with probability $0.05$. How many parts should be produced in order that probability of at least one part being defective is $1/2$ or more? (Given ${\log }_{10}95=1.977$ and ${\log }_{10}2=0.3$)

• A. $11$
• B. $12$
• C. $15$
• D. $14$

Answer 1

Answer: (C), (D)

Given probability of defective part
$=0.05=\frac{1}{20}$
Probability of non-defective part
$=1-0.05=0.95=\frac{19}{20}$
We know that, $P(X=I)={}^n{C}_r{p}^r{q}^{n-r}$
where, $p=\frac{1}{20},q=\frac{19}{20}$
$r\ge 1$ and $n=?$
Also, $P(X\ge 1)\ge \frac{1}{2}$
$\Rightarrow 1-P(X=0)\ge \frac{1}{2}$
$\Rightarrow 1-{}^n{C}_0{\left(\frac{1}{20}\right)}^0{\left(\frac{19}{20}\right)}^{n-0}\ge \frac{1}{2}$
$\Rightarrow 1-\frac{1}{2}\ge {\left(\frac{19}{20}\right)}^n$
$\Rightarrow \frac{1}{2}\ge {\left(\frac{19}{20}\right)}^n$
$\Rightarrow \frac{1}{2}\ge {\left(\frac{95}{100}\right)}^n$
$\Rightarrow \log 2^{-1}\ge n[\log 95-\log 100]$
$\Rightarrow -\log 2\ge n[\log 95-2]$
$\Rightarrow -(0.3)\ge n[1.977-2]$
$\Rightarrow n\ge \frac{0.3}{0.023}$
$\Rightarrow n\ge \frac{300}{23}$
$\Rightarrow n\ge 13.04$
$\therefore n=14,15$