To solve the problem, we need to find the probability that \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \), where \( \omega \) is a complex cube root of unity and \( r_1, r_2, r_3 \) are the outcomes of rolling a fair die three times.
### Step 1: Understanding Cube Roots of Unity
The complex cube roots of unity are given by:
- \( 1 \)
- \( \omega \)
- \( \omega^2 \)
where \( \omega = e^{2\pi i / 3} \) and \( \omega^2 = e^{4\pi i / 3} \). The important property we will use is:
\[
1 + \omega + \omega^2 = 0
\]
### Step 2: Total Outcomes
When a fair die is thrown three times, each throw has 6 possible outcomes (1 to 6). Therefore, the total number of outcomes when throwing the die three times is:
\[
6 \times 6 \times 6 = 216
\]
### Step 3: Finding Favorable Outcomes
We need to find the combinations of \( r_1, r_2, r_3 \) such that \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \). This can happen if \( r_1, r_2, r_3 \) take values from the sets corresponding to \( 0 \mod 3 \), \( 1 \mod 3 \), and \( 2 \mod 3 \).
- **Values for \( r \equiv 0 \mod 3 \)**: 3, 6
- **Values for \( r \equiv 1 \mod 3 \)**: 1, 4
- **Values for \( r \equiv 2 \mod 3 \)**: 2, 5
### Step 4: Counting Favorable Outcomes
To satisfy \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \), we need one number from each of the three categories:
1. One number from \( \{3, 6\} \) (2 choices)
2. One number from \( \{1, 4\} \) (2 choices)
3. One number from \( \{2, 5\} \) (2 choices)
The number of ways to choose one number from each category is:
\[
2 \times 2 \times 2 = 8
\]
### Step 5: Arranging the Outcomes
Since the three numbers can be arranged in any order, we need to multiply by the number of permutations of the three chosen numbers, which is \( 3! = 6 \).
Thus, the total number of favorable outcomes is:
\[
8 \times 6 = 48
\]
### Step 6: Calculating the Probability
The probability \( P \) that \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \) is given by the ratio of favorable outcomes to total outcomes:
\[
P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{48}{216}
\]
### Step 7: Simplifying the Probability
We can simplify \( \frac{48}{216} \):
\[
P = \frac{48 \div 24}{216 \div 24} = \frac{2}{9}
\]
### Final Answer
Thus, the probability that \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \) is \( \frac{2}{9} \).
