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(D) The total number of outcomes when a die is thrown $3$ times is $6 	imes 6 	imes 6 = 216$. Given the condition $\omega^{\beta_1} + \omega^{\beta_

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$\frac{2}{9}$

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(D) The total number of outcomes when a die is thrown $3$ times is $6 	imes 6 	imes 6 = 216$. Given the condition $\omega^{\beta_1} + \omega^{\beta_2} = -\omega^{\beta_3}$, we can rewrite this as $\omega^{\beta_1} + \omega^{\beta_2} + \omega^{\beta_3} = 0$. Since $\omega$ is a complex cube root of unity, $\omega^n$ can take values $\omega, \omega^2, 1$ depending on whether $n \equiv 1, 2, 0 \pmod{3}$. For the sum to be zero, the values ${\omega^{\beta_1}, \omega^{\beta_2}, \omega^{\beta_3}}$ must be a permutation of ${1, \omega, \omega^2}$. This means one of $\beta_1, \beta_2, \beta_3$ must be of the form $3k$, one of the form $3k+1$, and one of the form $3k+2$. In the set ${1, 2, 3, 4, 5, 6}$, there are two numbers of each type: Type $0$ $(n \equiv 0 \pmod{3})$: ${3, 6}$ Type $1$ $(n \equiv 1 \pmod{3})$: ${1, 4}$ Type $2$ $(n \equiv 2 \pmod{3})$: ${2, 5}$ The number of ways to choose one number from each set is $2 	imes 2 	imes 2 = 8$. Since the order of $\beta_1, \beta_2, \beta_3$ matters, we multiply by $3! = 6$. Favourable outcomes $= 8 	imes 6 = 48$. Probability $= \frac{48}{216} = \frac{2}{9}$.

$\omega$ is a complex cube root of unity. When an unbiased die is thrown $3$ times, if $\beta_1, \beta_2, \beta_3$ are the numbers appeared on the die, then the probability that $\beta_1, \beta_2$ and $\beta_3$ satisfy $\omega^{\beta_1}+\omega^{\beta_2}=-\omega^{\beta_3}$ is

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