Since the events are mutually exclusive (you can't hit a yellow section and 1 section at the same time), we can add the probabilities of both A (hitting a yellow section) and B (hitting a section with a 1).
[tex]P(A) = \frac{2}{8}[/tex]
[tex]P(B) = \frac{4}{8}[/tex]
Thus, the probability of hitting a yellow section or a section with a 1 is [tex]\frac{2}{8} + \frac{4}{8} = \boxed{\frac{6}{8}}[/tex].
