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 and only if the remainders of $r_1, r_2, r_3$ when divided by $3$ are $0,

Vedclass · Accepted Answer

$\frac{2}{9}$

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(C) The condition $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ is satisfied if and only if the remainders of $r_1, r_2, r_3$ when divided by $3$ are $0, 1, 2$ in some order.For each $r_i \in \{1, 2, 3, 4, 5, 6\}$,the remainders modulo $3$ are:$r_i \equiv 1 \pmod{3} \implies r_i \in \{1, 4\}$ ($2$ values)$r_i \equiv 2 \pmod{3} \implies r_i \in \{2, 5\}$ ($2$ values)$r_i \equiv 0 \pmod{3} \implies r_i \in \{3, 6\}$ ($2$ values)Let $n_0, n_1, n_2$ be the number of outcomes with remainder $0, 1, 2$ respectively. We need one of each,so the number of favorable outcomes is $3! 	imes (n_0 	imes n_1 	imes n_2) = 6 	imes (2 	imes 2 	imes 2) = 48$.The total number of outcomes is $6^3 = 216$.Thus,the probability is $\frac{48}{216} = \frac{2}{9}$.

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

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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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