Six faces of a die are marked with numbers `1, -1, 0, -2, 2, 3` and the die is thrown thrice. The probability that the sum of the numbers thrown is six, is

Options

A. `(1)/(72)`
B. `(1)/(12)`
C. `(5)/(108)`
D. `(1)/(36)`

Explanation

To solve the problem, we need to calculate the probability that the sum of the numbers obtained by throwing a die marked with `1, -1, 0, -2, 2, 3` three times equals 6.

### Step-by-Step Solution:

1. **Identify the Outcomes**: The die has six faces with the following values: `1, -1, 0, -2, 2, 3`.

2. **Total Outcomes**: When the die is thrown three times, the total number of outcomes is given by:
\[ \text{Total Outcomes} = 6 \times 6 \times 6 = 216 \]

3. **Find Favorable Outcomes**: We need to find combinations of three numbers from the die that sum to 6. We will consider different combinations:

- **Combination 1**: \(1 + 2 + 3\)
- The different arrangements of these numbers can be calculated using the factorial of the number of items:
\[ \text{Ways} = 3! = 6 \]

- **Combination 2**: \(0 + 3 + 3\)
- Here, we have two 3's and one 0. The arrangements can be calculated as:
\[ \text{Ways} = \frac{3!}{2!} = 3 \]

- **Combination 3**: \(2 + 2 + 2\)
- All numbers are the same, so there is only one way to arrange them:
\[ \text{Ways} = 1 \]

4. **Total Favorable Outcomes**: Now, we sum the favorable outcomes from all combinations:
\[ \text{Total Favorable Outcomes} = 6 + 3 + 1 = 10 \]

5. **Calculate Probability**: The probability \(P\) that the sum of the numbers thrown is 6 is given by the ratio of favorable outcomes to total outcomes:
\[ P = \frac{\text{Total Favorable Outcomes}}{\text{Total Outcomes}} = \frac{10}{216} \]

6. **Simplify the Probability**: We can simplify \(\frac{10}{216}\):
\[ \frac{10}{216} = \frac{5}{108} \]

### Final Answer:
The probability that the sum of the numbers thrown is 6 is:
\[ \frac{5}{108} \]