Let omega be a complex cube root of unity wit | vedclass

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(C) The condition $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$ is satisfied if the remainders of $r_1, r_2, r_3$ when divided by $3$ are ${0, 1, 2

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$2/9$

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(C) The condition $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$ is satisfied if the remainders of $r_1, r_2, r_3$ when divided by $3$ are ${0, 1, 2}$ in some order.For a die,the possible values are ${1, 2, 3, 4, 5, 6}$.Remainders modulo $3$ are:$1 \equiv 1, 2 \equiv 2, 3 \equiv 0, 4 \equiv 1, 5 \equiv 2, 6 \equiv 0$.There are $2$ numbers for each remainder: $0$ (from ${3, 6}$),$1$ (from ${1, 4}$),and $2$ (from ${2, 5}$).Number of favorable outcomes $= (2 	imes 2 	imes 2) 	imes 3! = 8 	imes 6 = 48$.Total outcomes $= 6^3 = 216$.Probability $= \frac{48}{216} = \frac{2}{9}$.

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,what is the probability that $\omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0$?

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