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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
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Verified Answer
The correct answer is:
$\frac{2}{9}$
$\frac{2}{9}$
Sample space A dice is thrown thrice, $n(s)=6 \times 6 \times 6$.
Favorable events $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ ie, $\left(r_1, r_2, r_3\right)$ are ordered 3-triples which can take values, $\left.\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array}\right\}$ ie, 8 ordered pairs and each can be arranged in 3 ! ways $=6$
$$
\begin{array}{ll}
\therefore & n(E)=8 \times 6 \\
\Rightarrow & P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}
\end{array}
$$
Favorable events $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ ie, $\left(r_1, r_2, r_3\right)$ are ordered 3-triples which can take values, $\left.\begin{array}{llll}(1,2,3), & (1,5,3), & (4,2,3), & (4,5,3) \\ (1,2,6), & (1,5,6), & (4,2,6), & (4,5,6)\end{array}\right\}$ ie, 8 ordered pairs and each can be arranged in 3 ! ways $=6$
$$
\begin{array}{ll}
\therefore & n(E)=8 \times 6 \\
\Rightarrow & P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}
\end{array}
$$
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