Let `omega` be a cube of unity . If `r_1,r_2,and r_3` be the numbers obtained on the die. Then probability of `omega^(r_1)+omega^(r_2)+omega^(r_3)=0` is

Options

A. `1/18`
B. `1/9`
C. `2/9`
D. `1/36`

Answer

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 cube root of unity and \( r_1, r_2, r_3 \) are the results of rolling a die.

### Step-by-step Solution:

1. **Understanding Cube Roots of Unity**:
The cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3}, \quad \omega^2 = e^{4\pi i / 3}, \quad \text{and} \quad 1 \]
They satisfy the equation:
\[ 1 + \omega + \omega^2 = 0 \]
This means that for \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \), we need one of the values to be \( 1 \) and the other two to be \( \omega \) and \( \omega^2 \).

2. **Possible Values of \( r_1, r_2, r_3 \)**:
The numbers \( r_1, r_2, r_3 \) can take values from \( 1 \) to \( 6 \) since they are the outcomes of rolling a die.

3. **Identifying Conditions**:
We need to find the cases where:
- One of \( r_1, r_2, r_3 \) is a multiple of \( 3 \) (which gives \( \omega^0 = 1 \)).
- The other two must not be multiples of \( 3 \) and must correspond to \( \omega \) and \( \omega^2 \).

4. **Finding Multiples of 3**:
The multiples of \( 3 \) in the range \( 1 \) to \( 6 \) are \( 3 \) and \( 6 \). Thus, we can have:
- Case 1: \( r_3 = 3 \)
- Case 2: \( r_3 = 6 \)

5. **Case Analysis**:
- **Case 1**: \( r_3 = 3 \)
- The remaining values \( r_1 \) and \( r_2 \) can be \( 1 \) and \( 2 \) or \( 4 \) and \( 5 \). The valid combinations are:
- \( (1, 2) \)
- \( (2, 1) \)
- \( (4, 5) \)
- \( (5, 4) \)
- Total combinations for this case: \( 4 \).

- **Case 2**: \( r_3 = 6 \)
- The remaining values \( r_1 \) and \( r_2 \) can again be \( 1 \) and \( 2 \) or \( 4 \) and \( 5 \). The valid combinations are:
- \( (1, 2) \)
- \( (2, 1) \)
- \( (4, 5) \)
- \( (5, 4) \)
- Total combinations for this case: \( 4 \).

6. **Total Favorable Outcomes**:
Adding both cases gives us \( 4 + 4 = 8 \) favorable outcomes.

7. **Total Outcomes**:
The total number of outcomes when rolling a die three times is:
\[ 6 \times 6 \times 6 = 216 \]

8. **Calculating Probability**:
The probability \( P \) is given by:
\[ P = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{8}{216} = \frac{1}{27} \]

### Final Answer:
The probability that \( \omega^{r_1} + \omega^{r_2} + \omega^{r_3} = 0 \) is \( \frac{1}{27} \).