Question

A game has a spinner with 15 equal sectors labeled 1 through 15.

What is P(multiple of 3 or multiple of 7)?

Responses

215
2 over 15

13
1 third

25
2 over 5

715

Answer 1

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).

1. **Multiples of 3**:
The multiples of 3 up to 15 are:
- 3, 6, 9, 12, 15
There are 5 multiples of 3.

2. **Multiples of 7**:
The multiples of 7 up to 15 are:
- 7, 14
There are 2 multiples of 7.

3. **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 \( \frac{7}{15} \).