Let $\omega$ = 1 be a complex cube root of unity. A fair die is thrown three times. If r₁, r₂ and r₃ are the numbers obtained on the die, then the probability that $\omega$^r₁ + $\omega$^r₂ + $\omega$^r₃ = 0 is

${\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\in \{1,2,\dots .6\}\Rightarrow \mathrm{n}\left(\mathrm{s}\right)=6^3$
${\mathrm{w}}^{{\mathrm{r}}_1}+{\mathrm{w}}^{{\mathrm{r}}_2}+{\mathrm{w}}^{{\mathrm{r}}_3}=0\text{ iff }{\mathrm{r}}_1,{\mathrm{r}}_2,{\mathrm{r}}_3\text{ belongs to sets }\mathrm{A},\mathrm{B},\mathrm{C}$
(one each in any order where
A={3,6}, B={1,4}, C={2,5} $\left[\because 1+\mathrm{w}+{\mathrm{w}}^2=0\right]$
$\therefore$No. of favourable cases $=\left({ }^2{\mathrm{C}}_1{\times }^2{\mathrm{C}}_1{\times }^2{\mathrm{C}}_1\right)\times \overline{)3}$ and 
Required probability$=\left(\frac{8\times 6}{6^3}\right)=\frac{2}{9}$