SOLUTION
The even numbers here are 2, 4, 6 and 8. That is 4 numbers.
The numbers greater than 3 are 4, 5, 6, 7, and 8, that is 5 numbers.
And we have a total of 8 numbers.
Let P(A) be the probability of the pointer landing on an even number
Let P(B) be the probability of the pointer landing on a number greater than 3
Let P(A or B) be the probability that the pointer stops on an even number or number greater than three
From the probability formula,
![P(\text{A or B) = P(A) + P(B) - P(A}\cap B)](https://img.qammunity.org/2023/formulas/mathematics/high-school/inmcqyt2ll04wcqcc6lzobi4b82oo9f299.png)
![\text{ P(A}\cap B)\text{ means probability of A and B}](https://img.qammunity.org/2023/formulas/mathematics/college/w7ftjq51npsyr11c0dnxnikfhxvngy9ugl.png)
Hence
![\begin{gathered} P(A)=(4)/(8) \\ P(B)=(5)/(8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/7xdsg5kxabbqs4mwyuiuz672xwu9jnommv.png)
![\begin{gathered} \text{ For P(A}\cap B)\text{ we can s}ee\text{ that betwe}en\text{ } \\ \text{the even numbers 2, 4, 6, 8 and } \\ n\text{umbers greater than 3, which are 4, 5, 6, 7, 8} \\ \text{what is common is 4, }6,\text{ 8} \\ So,\text{ } \\ \text{P(A}\cap B)=(3)/(8) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/26j83xgiqzqboidi6b7skkr1x7tiwfk0e5.png)
Therefore, P(A or B) becomes
![\begin{gathered} (4)/(8)+(5)/(8)-(3)/(8) \\ (4+5-3)/(8) \\ (6)/(8) \\ =(3)/(4) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/kvx3taxlxr2jeuz0g6dq0vxfv5fzosvhfl.png)