Question

$20\%$ of the bolts produced in a factory are found to be defective. Find the probability that in a sample of $10$ bolts chosen at random exactly $2$ will be defective using:
(i) Binomial distribution
(ii) Poisson distribution $[e^{-2}=0.1353]$

Answer 1

Given, $n=10,p=\frac{20}{100}=\frac{1}{5}$
$\therefore q=1-\frac{1}{5}=\frac{4}{5}$
Let $X$ denote the number of defective bolts chosen.
$\therefore X=2$
(i) Using Binomial distribution
$P(X-2)=10C_2$
${\left(\frac{1}{5}\right)}^2\cdot {\left(\frac{4}{5}\right)}^8=\frac{10\times 9}{1\times 2}\left(\frac{4^8}{5^{10}}\right)=45\left(\frac{4^8}{5^{10}}\right)$
(ii) Using Poisson distribution
$\lambda =np=10\cdot \left(\frac{1}{5}\right)=2$
$P(X-x)=e^{-\lambda }\frac{{\lambda }^x}{\angle x},x=0,1,2,....$
$P(X=2)=e^{-2\frac{2^2}{\angle 2}=e^{-2}\left[\frac{4}{1\times 2}\right]=e^{-2}\left(2\right)}=2\left(0.1353\right)=0.2706$.