Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. A fair die is thrown three times. If $r_1$, $r_2$ and $r_3$ are the numbers obtained on the die, then the probability that $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$ is

$\omega^3 = 1, \quad \omega \neq 1 \ \Rightarrow \ 1 + \omega + \omega^2 = 0$

$\text{Die thrown } 3 \text{ times} \ \Rightarrow \ 6^3 = 216 \text{ total outcomes}$

$\text{To find: } P\!\left(\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0\right)$

$\omega^{k} \in \{1,\ \omega,\ \omega^{2}\} \quad \text{(depends on } k \bmod 3)$

$k \equiv 0:\ \{3,6\} \rightarrow \omega^{k} = 1$

$k \equiv 1:\ \{1,4\} \rightarrow \omega^{k} = \omega$

$k \equiv 2:\ \{2,5\} \rightarrow \omega^{k} = \omega^{2}$

$1 + \omega + \omega^{2} = 0$

$\text{Sum} = 0 \iff \{\omega^{r_1},\omega^{r_2},\omega^{r_3}\} = \{1,\omega,\omega^{2}\}$

$3,\ 3\omega,\ 3\omega^{2} \neq 0$

$2+\omega,\ 2+\omega^{2},\ 1+2\omega \neq 0$

$\Rightarrow \text{ one outcome from each class}$

$\text{Favourable}: r_1,\ r_2,\ r_3 \text{ from three different classes}$

$\text{Assign the 3 classes to } r_1,r_2,r_3:\ 3! = 6$

$\text{Each class offers } 2 \text{ outcomes}:\ 2^{3} = 8$

$\text{Favourable} = 6 \times 8 = 48$

$P = \frac{48}{216} = \frac{2}{9}$

$1 \times \frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$

$P = \frac{2}{9} \ \Rightarrow \ \text{option (3)}$