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

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(D) The condition $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ is satisfied if and only if the set of remainders of ${r_1, r_2, r_3}$ when divided by $3

Vedclass · Accepted Answer

$\frac{2}{9}$

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(D) The condition $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ is satisfied if and only if the set of remainders of ${r_1, r_2, r_3}$ when divided by $3$ is ${0, 1, 2}$.Let $n_0, n_1, n_2$ be the number of outcomes in ${1, 2, 3, 4, 5, 6}$ such that $r \equiv 0, 1, 2 \pmod{3}$ respectively.For a die,$n_0 = 2$ (numbers $3, 6$),$n_1 = 2$ (numbers $1, 4$),and $n_2 = 2$ (numbers $2, 5$).The probability of getting a remainder $0, 1, 2$ is $P(0) = \frac{2}{6} = \frac{1}{3}$,$P(1) = \frac{2}{6} = \frac{1}{3}$,and $P(2) = \frac{2}{6} = \frac{1}{3}$.To have $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$,the remainders must be a permutation of $(0, 1, 2)$.The number of such permutations is $3! = 6$.The probability is $6 	imes (P(0) 	imes P(1) 	imes P(2)) = 6 	imes (\frac{1}{3} 	imes \frac{1}{3} 	imes \frac{1}{3}) = 6 	imes \frac{1}{27} = \frac{6}{27} = \frac{2}{9}$.

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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