1. Understanding the Problem Setup

Let $\omega$ be a complex cube root of unity other than 1. This means it satisfies the properties $\omega^3 = 1$ and the equation $x^3 - 1 = (x-1)(x^2+x+1) = 0$. Since $\omega \ne 1$, it must be a root of $x^2+x+1=0$, which implies $1 + \omega + \omega^2 = 0$.

A fair die is thrown three times, and the outcomes are $r_1, r_2, r_3$. Each of these outcomes can be any integer from the set $S = \{1, 2, 3, 4, 5, 6\}$.

The total number of possible outcomes for the three throws is $6 \times 6 \times 6 = 6^3 = 216$. Each outcome is equally likely.

We want to find the probability of the event ${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$.

2. Analyzing the Condition for Favorable Outcomes

The condition is ${\omega ^{{r_1}}} + {\omega ^{{r_2}}} + {\omega ^{{r_3}}} = 0$. From the properties of cube roots of unity, we know that $1 + \omega + \omega^2 = 0$. This is the only way a sum of three powers of $\omega$ can be zero. Therefore, the set of values $\{\omega^{r_1}, \omega^{r_2}, \omega^{r_3}\}$ must be a permutation of the set $\{1, \omega, \omega^2\}$.

3. Categorizing Die Outcomes

The value of $\omega^r$ depends on the remainder of $r$ when divided by 3. Let's categorize the possible outcomes of a single die roll, $r

in {1, 2, 3, 4, 5, 6}$, based on this remainder:

• Category 0 ($r \equiv 0 \pmod 3$): If $r$ is a multiple of 3, then $\omega^r = (\omega^3)^k = 1^k = 1$. The numbers in $S$ are $\{3, 6\}$. Let's call this set $S_0$. The number of outcomes in this category is $|S_0|=2$.

• Category 1 ($r \equiv 1 \pmod 3$): If $r$ leaves a remainder of 1 when divided by 3, then $\omega^r = \omega^{3k+1} = (\omega^3)^k \omega^1 = \omega$. The numbers in $S$ are $\{1, 4\}$. Let's call this set $S_1$. The number of outcomes in this category is $|S_1|=2$.

• Category 2 ($r \equiv 2 \pmod 3$): If $r$ leaves a remainder of 2 when divided by 3, then $\omega^r = \omega^{3k+2} = (\omega^3)^k \omega^2 = \omega^2$. The numbers in $S$ are $\{2, 5\}$. Let's call this set $S_2$. The number of outcomes in this category is $|S_2|=2$.

4. Counting the Favorable Outcomes

For the sum to be zero, the three outcomes $(r_1, r_2, r_3)$ must be such that one number is from Category 0, one is from Category 1, and one is from Category 2. This ensures that the terms $\omega^{r_1}, \omega^{r_2}, \omega^{r_3}$ are $1, \omega, \omega^2$ in some order.

Let's count the number of ways this can happen:

• We need to choose one number from $S_0$, one from $S_1$, and one from $S_2$. The number of ways to choose one number from each set is $|S_0| \times |S_1| \times |S_2| = 2 \times 2 \times 2 = 8$.
• The chosen numbers must be assigned to the throws $r_1, r_2, r_3$. For example, $r_1$ could be from $S_0$, $r_2$ from $S_1$, and $r_3$ from $S_2$. Or $r_1$ from $S_1$, $r_2$ from $S_0$, and $r_3$ from $S_2$, etc. The number of ways to arrange which throw comes from which category is the number of permutations of the three categories, which is $3! = 6$.

So, the total number of favorable outcomes is: $N_{\text{favorable}} = (\text{Number of ways to assign categories to } r_1, r_2, r_3) \times (\text{Number of ways to choose numbers from each category})$ $N_{\text{favorable}} = 3! \times (|S_0| \times |S_1| \times |S_2|) = 6 \times (2 \times 2 \times 2) = 6 \times 8 = 48$

5. Calculating the Probability

The probability of the event is the ratio of the number of favorable outcomes to the total number of outcomes.

Total number of outcomes, $N_{\text{total}} = 6^3 = 216$. Number of favorable outcomes, $N_{\text{favorable}} = 48$.

The required probability is: $P = \frac{N_{\text{favorable}}}{N_{\text{total}}} = \frac{48}{216}$

Simplifying the fraction: $P = \frac{48 \div 24}{216 \div 24} = \frac{2}{9}$

Alternatively, divide by common factors: $P = \frac{48}{216} = \frac{24}{108} = \frac{12}{54} = \frac{6}{27} = \frac{2}{9}$

Comparing this result with the options, it matches option C.

Final Answer: The probability is $2/9$.