Question

Let $\omega$ be a complex cube root of unity with $\omega \ne 1$. 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 ?

• A. $\frac{1}{18}$
• B. $\frac{1}{9}$
• C. $\frac{2}{9}$
• D. $\frac{1}{36}$

Answer 1

The correct option is C $\frac{2}{9}$

Sample space: A die is thrown thrice, $n\left(S\right)=6\times 6\times 6.$
Favourable events ${\omega }^{r_1}+{\omega }^{r_2}+{\omega }^{r_3}=0$
i.e. $(r_1,r_2,r_3)$ are ordered triplets which can take values,
$\begin{array}{cccc}(1,2,3),& (1,5,3),& (4,2,3),& (4,5,3),\\ (1,2,6)& (1,5,6)& (4,2,6,)& (4,5,6)\end{array}\}$ i.e. 8 ordered triplets and each can be arranged in 3! ways = 6
$\therefore n\left(E\right)=8\times 6\Rightarrow P\left(E\right)=\frac{8\times 6}{6\times 6\times 6}=\frac{2}{9}$