Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. A fair die is thrown three times. If $r_1…
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
$\frac{1}{36}$
$\frac{1}{8}$
$\frac{1}{9}$
$\frac{2}{9}$
Solution
Solution:
$\begin{aligned} & \because \omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0 \\ & \Rightarrow \text { each of } r_1, r_2, r_3 \text { belongs to each of the categories } 3 k, 3 k+1, \\ & 3 k+2\end{aligned}$
So the required probability
$3 \times \frac{1}{3} \times \frac{1}{3} \times \frac{1}{3}=\frac{2}{9}$