A spinner has 8 equal sections numbered 1 through 8. What is the probability that a spin lands on a multiple of 3?

The multiples of $3$ from $1$ to $8$ are $3$ and $6$, so there are $2$ favorable sections out of $8$ equally likely sections. The probability is $\frac{2}{8}=\frac{1}{4}$, choice B. Choice A ($\frac{1}{8}$) counts only one favorable section. Choice C ($\frac{3}{8}$) counts three, perhaps by including $9$ or by counting $0$, neither of which is on the spinner. Choice D ($\frac{1}{2}$) is the probability of an even number or of an odd number, not of a multiple of $3$. List the favorable outcomes explicitly to avoid miscounting.