Given that:
- A fair die is rolled three times.
- A 1 is considered "success".
- All other outcomes are "failures".
You know that a die has 6 faces numbered from 1 to 6. In this case, these outcomes are considered "failures":
![\begin{gathered} 2 \\ 3 \\ 4 \\ 5 \\ 6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/wx9nenm76uc0lz03iqhgllun0ruc0mhx4h.png)
That is a total of 5 numbers considered "failures".
Since the events are independent, you can determine that the probability of 0 successes is:
![P=(5)/(6)\cdot(5)/(6)\cdot(5)/(6)](https://img.qammunity.org/2023/formulas/mathematics/college/h17fqskc4cjhqu8cef3czndumhfldw70g9.png)
Solving the Multiplications, you get:
![P=(5\cdot5\cdot5)/(6\cdot6\cdot6)](https://img.qammunity.org/2023/formulas/mathematics/college/uydruw058dq1awmq39go6024mku0ypac9f.png)
![P=(125)/(216)](https://img.qammunity.org/2023/formulas/mathematics/college/afrwo527zfxm1a626wivgdqgjeh06p1r6c.png)
Hence, the answer is:
![P=(125)/(216)](https://img.qammunity.org/2023/formulas/mathematics/college/afrwo527zfxm1a626wivgdqgjeh06p1r6c.png)