Question

Two bad eggs are accidently mixed with ten good ones. Two eggs are drawn at random one by one with replacement from this lot. Find the mean and variance for for the number of bad eggs.

Answer 1

Let $x$ be no. of bad eggs $x=0,1$

$p(x=0)=p$ ($2$ good eggs ) $=\frac{{}^{10}{C}_2}{{}^{12}{C}_2}=\frac{10\times 9}{12\times 11}=\frac{90}{131}$

$p(x=1)=p$ ($1$ good egg and $1$ bad egg) $=\frac{{}^{10}{C}_1{\times }^2{C}_1}{{}^{12}{C}_2}=\frac{10\times 2}{6\times 11}=\frac{10}{33}$

$p(x=2)=p$ (two bad eggs ) $=\frac{{}^2{C}_2}{{}^{12}{C}_2}=\frac{1}{\frac{612\times 11\times 10}{10!\times 2}}=\frac{1}{66}$

$x$	$pi$	$xpi$	$x^{2}pi$
$0$	$\frac{90}{131}$	$\theta$	$0$
$1$	$\frac{10}{33}$	$\frac{10}{33}$	$\frac{10}{33}$
$2$	$\frac{1}{66}$	$\frac{1}{33}$	$\frac{2}{33}$
	$\sum$	$\frac{11}{33}$	$\frac{12}{33}$

$\mu =\sum xpi=\frac{11}{33}$

${\sigma }^2=\sum x^{2}p;-{\mu }^2=\frac{12}{33}-{\left(\frac{11}{33}\right)}^2$