10 persons sit around a circular table with 10 numbered chairs. The probability that the two particular persons A and B are always together is

Since the chairs are numbered, so for counting of total number of cases it is equivalent to linear permutation. 

Hence, total number of cases $=10!$ 

If two particular persons A and B sit together then the total number of linear arrangements $=2!9!$. 

Consider one of such arrangements in which the arrangement started at chair $1\left({\mathrm{C}}_1\right)$ and ends at chair $10\left({\mathrm{C}}_{10}\right) {\mathrm{C}}_1-{\mathrm{C}}_2-{\mathrm{C}}_3-\dots \dots \dots -{\mathrm{C}}_9-{\mathrm{C}}_{10}$

If two persons sit at $C_1$ and $C_{10}$ then it will lead to $2!8!$ new arrangements. So the favourable number of 

cases $=2!9!+2!8!=2!8!\left(10\right)$

$\therefore$ Probability $=\frac{2!8!\left(10\right)}{10!}=\frac{2}{9}$