A spinner has regions numbered 1 through 15. What is the probability that the spinner will stop on an even number or a multiple of 3 ?
A. $\frac{7}{9}$
B. $\frac{1}{3}$
C. 12
D. $\frac{2}{3}$

Step 1 ：The spinner has 15 regions, so there are 15 possible outcomes.

Step 2 ：The even numbers from 1 to 15 are 2, 4, 6, 8, 10, 12, 14, and the multiples of 3 are 3, 6, 9, 12, 15.

Step 3 ：We can see that 6 and 12 are both even numbers and multiples of 3, so we should not count them twice.

Step 4 ：Therefore, the favorable outcomes are 2, 4, 6, 8, 10, 12, 14, 3, 9, 15, a total of 10 outcomes.

Step 5 ：The probability is the number of favorable outcomes divided by the total number of outcomes, which is \(\frac{10}{15} = \frac{2}{3}\).

Step 6 ：Final Answer: The probability that the spinner will stop on an even number or a multiple of 3 is \(\boxed{\frac{2}{3}}\).