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, $p=\frac{1}{20}, q=\frac{19}{20}$
$r \geq 1$ and $n=?$
Also, $P(X \geq 1) \geq \frac{1}{2}$
$\Rightarrow 1-P(X=0) \geq \frac{1}{2}$
$\Rightarrow 1-{ }^{n} C_{0}\left(\frac{1}{20}\right)^{0}\left(\frac{19}{20}\right)^{n-0} \geq \frac{1}{2}$
$\Rightarrow 1-\frac{1}{2} \geq\left(\frac{19}{20}\right)^{n}$
$\Rightarrow \frac{1}{2} \geq\left(\frac{19}{20}\right)^{n}$
$\Rightarrow \frac{1}{2} \geq\left(\frac{95}{100}\right)^{n}$
$\Rightarrow \log 2^{-1} \geq n[\log 95-\log 100]$
$\Rightarrow -\log 2 \geq n[\log 95-2]$
$\Rightarrow -(0.3) \geq n[1.977-2]$
$\Rightarrow n \geq \frac{0.3}{0.023}$
$\Rightarrow n \geq \frac{300}{23}$
$\Rightarrow n \geq 13.04$
$\therefore n=14,15$