To find the probability that the spinner will stop on an even number or a multiple of 3, let's break down the problem step by step:
1. Total Numbers:
The spinner has regions numbered from 1 to 18. So, the total number of outcomes is 18.
2. Even Numbers:
We first need to count the even numbers between 1 and 18.
The even numbers are: 2, 4, 6, 8, 10, 12, 14, 16, 18.
There are 9 even numbers.
3. Multiples of 3:
Next, we need to count the multiples of 3 between 1 and 18.
The multiples of 3 are: 3, 6, 9, 12, 15, 18.
There are 6 multiples of 3.
4. Numbers that are both Even and Multiples of 3:
We also need to subtract the numbers that are both even and multiples of 3, as we have counted them twice.
These numbers are: 6, 12, 18.
There are 3 numbers that are both even and multiples of 3.
5. Inclusion-Exclusion Principle:
To find the total number of favorable outcomes (numbers that are either even or multiples of 3), we use the inclusion-exclusion principle:
[tex]\[
\text{Favorable outcomes} = \text{(Number of even numbers)} + \text{(Number of multiples of 3)} - \text{(Number of numbers that are both even and multiples of 3)}
\][/tex]
Substituting the numbers, we get:
[tex]\[
\text{Favorable outcomes} = 9 + 6 - 3 = 12
\][/tex]
6. Probability:
The probability that the spinner will stop on an even number or a multiple of 3 is given by the ratio of favorable outcomes to the total number of outcomes.
[tex]\[
\text{Probability} = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{12}{18} = \frac{2}{3}
\][/tex]
Therefore, the probability that the spinner will stop on an even number or a multiple of 3 is [tex]\(\frac{2}{3}\)[/tex].
The correct answer is:
[tex]\[
\boxed{\frac{2}{3}}
\][/tex]
