Question

A manufacturer makes condensers which on an average are 1% defective. He packs them in boxes of 100. Calculate the probability that a box picked at random will contain 3 or more faulty condensers?

• A. 0.08

Answer 1

The correct option is A 0.08
$P=$1 % = 0.01

$n=100$

$m=nP=100\times 0.01=1$

$P\left(r\right)=\frac{e^{-m}{m}^r}{r!}$

P(3 or more faulty condensers)

$=P\left(3\right)+P\left(4\right)+......+P\left(100\right)$

$=1-\left[P\right(0)+P(1)+P(2\left)\right]$

$=1-[\frac{e^{-1}{1}^0}{0!}+\frac{e^{-1}{1}^1}{1!}+\frac{e^{-1}{1}^2}{2!}]=1-[e^{-1}+e^{-1}+\frac{e^{-1}}{2}]$
$=1-e^{-1}\left[\frac{5}{2}\right]=0.0803$