areas, numbered 1 through 13 . If the spinner is spun one time, $$ \{2,4,6,8,10, ? $$ 2. There is a spinner with 15 equal areas, numbered 1 through 15 . If the spinner is spun one time, what is the probability that the result is a multiple of 2 or a multiple of 5 ? $$ 2,4,6,8,16,1214,1,53,55,10,68 $$ = 3. There is a spinner with 12 equal areas, numbered 1 through 12 . If the spinner is spun one time, what is the probability that the result is a multiple of 6 and a multiple of 4 ? What is the probability that the result is a m $$ 5,6,4,12\} $$ 5. There is a spinner with 5. There is a spinner with 15 equal areas, numbered 1 through 15 . If the spinner is spun one time, What is the probability that the result is a multiple of 6 or a multiple of 4 ? is the probability that the result is a multiple of 5 and a multiple of 2 ? Independent Events:

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UpStudy AI · Accepted Answer

Let's break down each part of the problem step by step.

### Problem 1
**Spinner with 13 equal areas, numbered 1 through 13.**

We need to find the probability of landing on a specific set of numbers: $$ \{2, 4, 6, 8, 10\} $$.

1. **Total outcomes**: There are 13 areas, so the total number of outcomes is 13.
2. **Favorable outcomes**: The numbers in the set are 2, 4, 6, 8, and 10. There are 5 favorable outcomes.

The probability $$ P $$ is given by the formula:
$$
P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{13}
$$

### Problem 2
**Spinner with 15 equal areas, numbered 1 through 15.**

We need to find the probability that the result is a multiple of 2 or a multiple of 5.

1. **Multiples of 2**: The multiples of 2 from 1 to 15 are $$ \{2, 4, 6, 8, 10, 12, 14\} $$. There are 7 multiples of 2.
2. **Multiples of 5**: The multiples of 5 from 1 to 15 are $$ \{5, 10, 15\} $$. There are 3 multiples of 5.
3. **Common multiples (Multiples of 10)**: The common multiples of 2 and 5 from 1 to 15 are $$ \{10\} $$. There is 1 common multiple.

Using the principle of inclusion-exclusion:
$$
P(\text{multiple of 2 or 5}) = P(\text{multiple of 2}) + P(\text{multiple of 5}) - P(\text{common multiples})
$$
Calculating:
$$
P(\text{multiple of 2}) = \frac{7}{15}, \quad P(\text{multiple of 5}) = \frac{3}{15}, \quad P(\text{common multiples}) = \frac{1}{15}
$$
Thus,
$$
P(\text{multiple of 2 or 5}) = \frac{7}{15} + \frac{3}{15} - \frac{1}{15} = \frac{9}{15} = \frac{3}{5}
$$

### Problem 3
**Spinner with 12 equal areas, numbered 1 through 12.**

We need to find the probability that the result is a multiple of 6 and a multiple of 4.

1. **Multiples of 6**: The multiples of 6 from 1 to 12 are $$ \{6, 12\} $$. There are 2 multiples of 6.
2. **Multiples of 4**: The multiples of 4 from 1 to 12 are $$ \{4, 8, 12\} $$. There are 3 multiples of 4.
3. **Common multiples (Multiples of 12)**: The common multiples of 6 and 4 from 1 to 12 are $$ \{12\} $$. There is 1 common multiple.

Since we are looking for the probability of landing on a number that is both a multiple of 6 and a multiple of 4, we only consider the common multiples:
$$
P(\text{multiple of 6 and 4}) = \frac{\text{Number of common multiples}}{\text{Total outcomes}} = \frac{1}{12}
$$

### Problem 4
**Spinner with 15 equal areas, numbered 1 through 15.**

We need to find the probability that the result is a multiple of 6 or a multiple of 4.

1. **Multiples of 6**: The multiples of 6 from 1 to 15 are $$ \{6, 12\} $$. There are 2 multiples of 6.
2. **Multiples of 4**: The multiples of 4 from 1 to 15 are $$ \{4, 8, 12\} $$. There are 3 multiples of 4.
3. **Common multiples (Multiples of 12)**: The common multiples of 6 and 4 from 1 to 15 are $$ \{12\} $$. There is 1 common multiple.

Using the principle of inclusion-exclusion:
$$
P(\text{multiple of 6 or 4}) = P(\text{multiple of 6}) + P(\text{multiple of 4}) - P(\text{common multiples})
$$
Calculating:
$$
P(\text{multiple of 6}) = \frac{2}{15}, \quad P(\text{multiple of 4}) = \frac{3}{15}, \quad P(\text{common multiples}) = \frac{1}{15}
$$
Thus,
$$
P(\text{multiple of 6 or 4}) = \frac{2}{15} + \frac{3}{15} - \frac{1}{15} = \frac{4}{15}
$$

### Summary of Probabilities
1. Probability of landing on $$ \{2, 4, 6, 8, 10\} $$: $$ \frac{5}{13} $$
2. Probability of landing on a multiple of 2 or 5: $$ \frac{3}{5} $$
3. Probability of landing on a multiple of 6 and 4: $$ \frac{1}{12} $$
4. Probability of landing on a multiple of 6 or 4: $$ \frac{4}{15} $$
- **Problem 1**: Probability of landing on $$ \{2, 4, 6, 8, 10\} $$ is $$ \frac{5}{13} $$. - **Problem 2**: Probability of landing on a multiple of 2 or 5 is $$ \frac{3}{5} $$. - **Problem 3**: Probability of landing on a multiple of 6 and 4 is $$ \frac{1}{12} $$. - **Problem 4**: Probability of landing on a multiple of 6 or 4 is $$ \frac{4}{15} $$.

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