From 3n consecutive integers three integers are selected at random. The probability that their sum is divisible by 3 is

given 3n integers

$$\begin{gathered} 1 4 7 10 3\mathrm{n}-2 \to \left(\mathrm{n}\right) \\ 2 5 8 11 3\mathrm{n}-1 \to \left(\mathrm{n}\right) \\ 3 6 9 12 3\mathrm{n} \to \left(\mathrm{n}\right) \end{gathered}$$

$\mathrm{n}\left(\mathrm{S}\right)={}^{3\mathrm{n}}{\mathrm{C}}_3$

E=Their sum divisible by 3

The sum of the three numbers is divisible by 3 if all the three numbers are from same group or one from each group is selected

$\begin{array}{rcl}\mathrm{n}\left(\mathrm{E}\right)& =& {}^{\mathrm{n}}{\mathrm{C}}_1\times {}^{\mathrm{n}}{\mathrm{C}}_1\times {}^{\mathrm{n}}{\mathrm{C}}_1+\left({}^{\mathrm{n}}{\mathrm{C}}_3\right)3\\ \mathrm{P}\left(\mathrm{E}\right)& =& \frac{{\mathrm{n}}^3+\left({}^{\mathrm{n}}{\mathrm{C}}_3\right)3}{{}^{3\mathrm{n}}{\mathrm{C}}_3}=\end{array}$

$\mathrm{P}\left(\mathrm{E}\right)=\frac{3{\mathrm{n}}^2-3\mathrm{n}+2}{(3\mathrm{n}-1)(3\mathrm{n}-2)}$