Respuesta :
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,
[tex]P(\text{A or B) = P(A) + P(B) - P(A}\cap B)[/tex][tex]\text{ P(A}\cap B)\text{ means probability of A and B}[/tex]Hence
[tex]\begin{gathered} P(A)=\frac{4}{8} \\ P(B)=\frac{5}{8} \end{gathered}[/tex][tex]\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)=\frac{3}{8} \end{gathered}[/tex]Therefore, P(A or B) becomes
[tex]\begin{gathered} \frac{4}{8}+\frac{5}{8}-\frac{3}{8} \\ \frac{4+5-3}{8} \\ \frac{6}{8} \\ =\frac{3}{4} \end{gathered}[/tex]