SOLUTION
Now, on a die of 6 faces, we have 3 even numbers which are 2, 4 and 6
Probability is calculated as
![\frac{expected\text{ outcome}}{total\text{ outcome }}](https://img.qammunity.org/2023/formulas/mathematics/college/13houe7ofrpyzk5i6b3m93cq0vypasgl1q.png)
So, number of even number is 3, total die faces is 6
Probability of getting an even number becomes
![(3)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/b9yjneqrj9ntkew8mpeocjxtbqn35o4sjc.png)
Also there is only 1 face with 1.
So probability of getting a 1 is
![(1)/(6)](https://img.qammunity.org/2023/formulas/mathematics/high-school/s2sdjtchslkb90he84q1tkyhaenfcyiyvp.png)
Now, probability of getting an even number or a 1, means we add both, we have
![\begin{gathered} (3)/(6)+(1)/(6) \\ (3+1)/(6) \\ =(4)/(6) \\ =(2)/(3) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/95cw41bsqdantovwmqtfc7cuz4ybhgdjdf.png)
hence the answer is 2/3, the last option nswe