To find the probability \( P(\text{multiple of 3 or multiple of 7}) \), we need to determine how many of the 15 sectors are multiples of 3, multiples of 7, and those that are multiples of both (to avoid double counting).
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Multiples of 3: The multiples of 3 up to 15 are:
- 3, 6, 9, 12, 15 There are 5 multiples of 3.
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Multiples of 7: The multiples of 7 up to 15 are:
- 7, 14 There are 2 multiples of 7.
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Multiples of both 3 and 7 (i.e., multiples of 21): Since 21 is greater than 15, there are no multiples of both 3 and 7 within the range from 1 to 15.
Now we can apply the principle of inclusion-exclusion to find the total count of sectors that are either multiples of 3 or multiples of 7:
\[ N(3 \text{ or } 7) = N(3) + N(7) - N(3 \text{ and } 7) \] \[ N(3 \text{ or } 7) = 5 + 2 - 0 = 7 \]
So, there are 7 favorable outcomes (the numbers labeled 3, 6, 7, 9, 12, 14, 15).
Now, the total number of sectors on the spinner is 15. Therefore, the probability is:
\[ P(\text{multiple of 3 or multiple of 7}) = \frac{N(3 \text{ or } 7)}{\text{Total outcomes}} = \frac{7}{15} \]
So, the final answer is \( \frac{7}{15} \). Checking the available responses:
- \( \frac{7}{15} \) would be the closest match.
Thus, the correct answer is \( \frac{7}{15} \).