Question
A fair 15
-sided die, numbered 1-15
, is rolled twice.
Complete the statements by typing a value into the blank spaces.
The number of outcomes in the sample space is
. The theoretical probability of rolling a number that is divisible by 3 two times in a row is
. (Round to the nearest tenth).
-sided die, numbered 1-15
, is rolled twice.
Complete the statements by typing a value into the blank spaces.
The number of outcomes in the sample space is
. The theoretical probability of rolling a number that is divisible by 3 two times in a row is
. (Round to the nearest tenth).
Answers
GPT-4o mini
To solve this problem, we first need to determine the number of outcomes in the sample space when rolling a fair 15-sided die twice.
1. **Total Outcomes in the Sample Space:**
When a die is rolled, there are 15 possible outcomes (numbers 1 to 15). Since the die is rolled twice, we multiply the number of outcomes for each roll:
\[
\text{Total outcomes} = 15 \times 15 = 225
\]
So, the number of outcomes in the sample space is **225**.
2. **Finding the Theoretical Probability of Rolling a Number Divisible by 3:**
Next, we identify the numbers between 1 and 15 that are divisible by 3. These numbers are:
- 3
- 6
- 9
- 12
- 15
There are 5 numbers that are divisible by 3.
To find the probability of rolling a number divisible by 3 on a single roll:
\[
P(\text{divisible by 3}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{15} = \frac{1}{3}
\]
Since we want to find the probability of this event occurring two times in a row, we multiply the probabilities of the two independent events:
\[
P(\text{divisible by 3 two times}) = P(\text{divisible by 3}) \times P(\text{divisible by 3}) = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) = \frac{1}{9}
\]
3. **Rounding to the Nearest Tenth:**
To convert the probability to a decimal and round to the nearest tenth:
\[
\frac{1}{9} \approx 0.1111
\]
Rounding this to the nearest tenth gives **0.1**.
Putting it all together:
- The number of outcomes in the sample space is **225**.
- The theoretical probability of rolling a number that is divisible by 3 two times in a row is **0.1**.
1. **Total Outcomes in the Sample Space:**
When a die is rolled, there are 15 possible outcomes (numbers 1 to 15). Since the die is rolled twice, we multiply the number of outcomes for each roll:
\[
\text{Total outcomes} = 15 \times 15 = 225
\]
So, the number of outcomes in the sample space is **225**.
2. **Finding the Theoretical Probability of Rolling a Number Divisible by 3:**
Next, we identify the numbers between 1 and 15 that are divisible by 3. These numbers are:
- 3
- 6
- 9
- 12
- 15
There are 5 numbers that are divisible by 3.
To find the probability of rolling a number divisible by 3 on a single roll:
\[
P(\text{divisible by 3}) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{5}{15} = \frac{1}{3}
\]
Since we want to find the probability of this event occurring two times in a row, we multiply the probabilities of the two independent events:
\[
P(\text{divisible by 3 two times}) = P(\text{divisible by 3}) \times P(\text{divisible by 3}) = \left(\frac{1}{3}\right) \times \left(\frac{1}{3}\right) = \frac{1}{9}
\]
3. **Rounding to the Nearest Tenth:**
To convert the probability to a decimal and round to the nearest tenth:
\[
\frac{1}{9} \approx 0.1111
\]
Rounding this to the nearest tenth gives **0.1**.
Putting it all together:
- The number of outcomes in the sample space is **225**.
- The theoretical probability of rolling a number that is divisible by 3 two times in a row is **0.1**.