Two bad eggs are mixed accidentally with 10 good ones. If three eggs are drawn at random from this lot in succession without replacement, then the variance of the probability distribution of the number of bad eggs drawn is

$P(X=0)=P\left(3\text{ good eggs }\right)$

$=\frac{{ }^{10}{\mathrm{C}}_3}{{ }^{12}{\mathrm{C}}_3}=\frac{10\times 9\times 8}{12\times 11\times 10}=\frac{12}{22}$

$P(X=1)=P\left(2\text{ good eggs and one bad egg }\right)$

$=\frac{{ }^{10}{C}_2{\times }^2{C}_1}{{ }^{12}{C}_3}=\frac{\frac{10\times 9}{1\times 2}\times \frac{2}{1}}{\frac{12\times 11\times 10}{1\times 2\times 3}}=\frac{9}{22}$

$P(X=2)=P(\text{ one good egg and two bad eggs) }$

$=\frac{{ }^{10}{C}_1{\times }^2{C}_2}{{ }^{12}{C}_3}=\frac{\frac{10}{1}\times 1}{\frac{12\times 11\times 10}{1\times 2\times 3}}=\frac{1}{22}$

$\begin{array}{r}\mu =\sum x{p}_i=\frac{1}{2}\\ {\sigma }^2=\sum x^2{p}_i-{\mu }^2=\frac{13}{22}-\frac{1}{4}=\frac{26-11}{44}=\frac{15}{44}\end{array}$