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 the probability of atleast one part being defective is $\frac{1}{2}$ or more?

(Given that, ${\log }_{10}95=1.977$ and ${\log }_{10}2=0.3$)

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

Answer 1

Answer: (C), (D)

The correct options are
C $14$
D $15$

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, $P(X=r){=}^n{C}_r{P}^r{q}^{n-r}$

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}$

$\Rightarrow 1-\frac{1}{2}\ge {\left(\frac{19}{20}\right)}^n$

$\Rightarrow \frac{1}{2}\ge {\left(\frac{19}{20}\right)}^n$

$\Rightarrow \frac{1}{2}\ge {\left(\frac{95}{100}\right)}^n$

$\Rightarrow log{2}^{-1}\ge n[log95-2]$

$\Rightarrow -\left(0.3\right)\ge n[1.977-2]$

$\Rightarrow n\ge \frac{0.3}{0.023}$

$\Rightarrow n\ge \frac{300}{23}$

$\Rightarrow n\ge 13.04$

$\therefore n=14,15$