Question
How many total different outcomes are possible when rolling two dice? Ways to roll a sum of 2 with two dice: Ways to roll a sum of 3 with two dice: Ways to roll a sum of 4 with two dice: Ways to roll a sum of 5 with two dice: Ways to roll a sum of 6 with two dice: Ways to roll a sum of 7 with two dice: Ways to roll a sum of 8 with two dice: Ways to roll a sum of 9 with two dice: Ways to roll a sum of 10 with two dice: Ways to roll a sum of 11 with two dice: Ways to roll a sum of 12 with two dice: Question 2 What is the probability of rolling a sum of 2 when rolling two dice. Show how to simplify all fractions. P(sum of 2) = ?(1 point) 10000 Word Limit Question 3 What is the probability of rolling a sum of 3 when rolling two dice. Show how to simplify all fractions. P(sum of 3) = ?(1 point) 0 / 10000 Word Limit Question 4 What is the probability of rolling a sum of 4 when rolling two dice. Show how to simplify all fractions. P(sum of 4) = ?(1 point) 0 / 10000 Word Limit Question 5 What is the probability of rolling a sum of 5 when rolling two dice. Show how to simplify all fractions. P(sum of 5) = ?(1 point) 0 / 10000 Word Limit Question 6 What is the probability of rolling a sum of 6 when rolling two dice. Show how to simplify all fractions. P(sum of 6) = ?(1 point) 0 / 10000 Word Limit Question 7 What is the probability of rolling a sum of 7 when rolling two dice. Show how to simplify all fractions. P(sum of 7) = ?(1 point) 0 / 10000 Word Limit Question 8 What is the probability of rolling a sum of 8 when rolling two dice. Show how to simplify all fractions. P(sum of 8) = ?(1 point) 0 / 10000 Word Limit Question 9 What is the probability of rolling a sum of 9 when rolling two dice. Show how to simplify all fractions. P(sum of 9) = ?(1 point) 0 / 10000 Word Limit Question 10 What is the probability of rolling a sum of 10 when rolling two dice. Show how to simplify all fractions. P(sum of 10) = ?(1 point) 0 / 10000 Word Limit Question 11 What is the probability of rolling a sum of 11 when rolling two dice. Show how to simplify all fractions. P(sum of 11) = ?(1 point) 0 / 10000 Word Limit Question 12 What is the probability of rolling a sum of 12 when rolling two dice. Show how to simplify all fractions. P(sum of 12) = ?(1 point) 0 / 10000 Word Limit Skip to navigation
Answers
GPT-4o mini
### Question 1: Total Outcomes and Sums
When rolling two six-sided dice, the total number of different outcomes is calculated by multiplying the number of outcomes for each die. Since each die has 6 faces, the total combinations are:
\[
\text{Total outcomes} = 6 \times 6 = 36
\]
Here’s the breakdown of the ways to roll sums from 2 to 12 with two dice:
- **Ways to roll a sum of 2**: (1, 1) → **1 way**
- **Ways to roll a sum of 3**: (1, 2), (2, 1) → **2 ways**
- **Ways to roll a sum of 4**: (1, 3), (2, 2), (3, 1) → **3 ways**
- **Ways to roll a sum of 5**: (1, 4), (2, 3), (3, 2), (4, 1) → **4 ways**
- **Ways to roll a sum of 6**: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → **5 ways**
- **Ways to roll a sum of 7**: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → **6 ways**
- **Ways to roll a sum of 8**: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → **5 ways**
- **Ways to roll a sum of 9**: (3, 6), (4, 5), (5, 4), (6, 3) → **4 ways**
- **Ways to roll a sum of 10**: (4, 6), (5, 5), (6, 4) → **3 ways**
- **Ways to roll a sum of 11**: (5, 6), (6, 5) → **2 ways**
- **Ways to roll a sum of 12**: (6, 6) → **1 way**
### Summarized Ways of Rolling Each Sum:
- Sum of 2: 1
- Sum of 3: 2
- Sum of 4: 3
- Sum of 5: 4
- Sum of 6: 5
- Sum of 7: 6
- Sum of 8: 5
- Sum of 9: 4
- Sum of 10: 3
- Sum of 11: 2
- Sum of 12: 1
### Question 2: Probability of Rolling a Sum of 2
The probability \( P \) of rolling a sum of 2 is calculated as follows:
\[
P(\text{sum of 2}) = \frac{\text{Number of ways to roll a sum of 2}}{\text{Total outcomes}} = \frac{1}{36}
\]
### Question 3: Probability of Rolling a Sum of 3
\[
P(\text{sum of 3}) = \frac{\text{Number of ways to roll a sum of 3}}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18}
\]
### Question 4: Probability of Rolling a Sum of 4
\[
P(\text{sum of 4}) = \frac{\text{Number of ways to roll a sum of 4}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}
\]
### Question 5: Probability of Rolling a Sum of 5
\[
P(\text{sum of 5}) = \frac{\text{Number of ways to roll a sum of 5}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}
\]
### Question 6: Probability of Rolling a Sum of 6
\[
P(\text{sum of 6}) = \frac{\text{Number of ways to roll a sum of 6}}{\text{Total outcomes}} = \frac{5}{36}
\]
### Question 7: Probability of Rolling a Sum of 7
\[
P(\text{sum of 7}) = \frac{\text{Number of ways to roll a sum of 7}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}
\]
### Question 8: Probability of Rolling a Sum of 8
\[
P(\text{sum of 8}) = \frac{\text{Number of ways to roll a sum of 8}}{\text{Total outcomes}} = \frac{5}{36}
\]
### Question 9: Probability of Rolling a Sum of 9
\[
P(\text{sum of 9}) = \frac{\text{Number of ways to roll a sum of 9}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}
\]
### Question 10: Probability of Rolling a Sum of 10
\[
P(\text{sum of 10}) = \frac{\text{Number of ways to roll a sum of 10}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}
\]
### Question 11: Probability of Rolling a Sum of 11
\[
P(\text{sum of 11}) = \frac{\text{Number of ways to roll a sum of 11}}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18}
\]
### Question 12: Probability of Rolling a Sum of 12
\[
P(\text{sum of 12}) = \frac{\text{Number of ways to roll a sum of 12}}{\text{Total outcomes}} = \frac{1}{36}
\]
### Summary of Probabilities
- \( P(\text{sum of 2}) = \frac{1}{36} \)
- \( P(\text{sum of 3}) = \frac{1}{18} \)
- \( P(\text{sum of 4}) = \frac{1}{12} \)
- \( P(\text{sum of 5}) = \frac{1}{9} \)
- \( P(\text{sum of 6}) = \frac{5}{36} \)
- \( P(\text{sum of 7}) = \frac{1}{6} \)
- \( P(\text{sum of 8}) = \frac{5}{36} \)
- \( P(\text{sum of 9}) = \frac{1}{9} \)
- \( P(\text{sum of 10}) = \frac{1}{12} \)
- \( P(\text{sum of 11}) = \frac{1}{18} \)
- \( P(\text{sum of 12}) = \frac{1}{36} \)
When rolling two six-sided dice, the total number of different outcomes is calculated by multiplying the number of outcomes for each die. Since each die has 6 faces, the total combinations are:
\[
\text{Total outcomes} = 6 \times 6 = 36
\]
Here’s the breakdown of the ways to roll sums from 2 to 12 with two dice:
- **Ways to roll a sum of 2**: (1, 1) → **1 way**
- **Ways to roll a sum of 3**: (1, 2), (2, 1) → **2 ways**
- **Ways to roll a sum of 4**: (1, 3), (2, 2), (3, 1) → **3 ways**
- **Ways to roll a sum of 5**: (1, 4), (2, 3), (3, 2), (4, 1) → **4 ways**
- **Ways to roll a sum of 6**: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → **5 ways**
- **Ways to roll a sum of 7**: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1) → **6 ways**
- **Ways to roll a sum of 8**: (2, 6), (3, 5), (4, 4), (5, 3), (6, 2) → **5 ways**
- **Ways to roll a sum of 9**: (3, 6), (4, 5), (5, 4), (6, 3) → **4 ways**
- **Ways to roll a sum of 10**: (4, 6), (5, 5), (6, 4) → **3 ways**
- **Ways to roll a sum of 11**: (5, 6), (6, 5) → **2 ways**
- **Ways to roll a sum of 12**: (6, 6) → **1 way**
### Summarized Ways of Rolling Each Sum:
- Sum of 2: 1
- Sum of 3: 2
- Sum of 4: 3
- Sum of 5: 4
- Sum of 6: 5
- Sum of 7: 6
- Sum of 8: 5
- Sum of 9: 4
- Sum of 10: 3
- Sum of 11: 2
- Sum of 12: 1
### Question 2: Probability of Rolling a Sum of 2
The probability \( P \) of rolling a sum of 2 is calculated as follows:
\[
P(\text{sum of 2}) = \frac{\text{Number of ways to roll a sum of 2}}{\text{Total outcomes}} = \frac{1}{36}
\]
### Question 3: Probability of Rolling a Sum of 3
\[
P(\text{sum of 3}) = \frac{\text{Number of ways to roll a sum of 3}}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18}
\]
### Question 4: Probability of Rolling a Sum of 4
\[
P(\text{sum of 4}) = \frac{\text{Number of ways to roll a sum of 4}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}
\]
### Question 5: Probability of Rolling a Sum of 5
\[
P(\text{sum of 5}) = \frac{\text{Number of ways to roll a sum of 5}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}
\]
### Question 6: Probability of Rolling a Sum of 6
\[
P(\text{sum of 6}) = \frac{\text{Number of ways to roll a sum of 6}}{\text{Total outcomes}} = \frac{5}{36}
\]
### Question 7: Probability of Rolling a Sum of 7
\[
P(\text{sum of 7}) = \frac{\text{Number of ways to roll a sum of 7}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}
\]
### Question 8: Probability of Rolling a Sum of 8
\[
P(\text{sum of 8}) = \frac{\text{Number of ways to roll a sum of 8}}{\text{Total outcomes}} = \frac{5}{36}
\]
### Question 9: Probability of Rolling a Sum of 9
\[
P(\text{sum of 9}) = \frac{\text{Number of ways to roll a sum of 9}}{\text{Total outcomes}} = \frac{4}{36} = \frac{1}{9}
\]
### Question 10: Probability of Rolling a Sum of 10
\[
P(\text{sum of 10}) = \frac{\text{Number of ways to roll a sum of 10}}{\text{Total outcomes}} = \frac{3}{36} = \frac{1}{12}
\]
### Question 11: Probability of Rolling a Sum of 11
\[
P(\text{sum of 11}) = \frac{\text{Number of ways to roll a sum of 11}}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18}
\]
### Question 12: Probability of Rolling a Sum of 12
\[
P(\text{sum of 12}) = \frac{\text{Number of ways to roll a sum of 12}}{\text{Total outcomes}} = \frac{1}{36}
\]
### Summary of Probabilities
- \( P(\text{sum of 2}) = \frac{1}{36} \)
- \( P(\text{sum of 3}) = \frac{1}{18} \)
- \( P(\text{sum of 4}) = \frac{1}{12} \)
- \( P(\text{sum of 5}) = \frac{1}{9} \)
- \( P(\text{sum of 6}) = \frac{5}{36} \)
- \( P(\text{sum of 7}) = \frac{1}{6} \)
- \( P(\text{sum of 8}) = \frac{5}{36} \)
- \( P(\text{sum of 9}) = \frac{1}{9} \)
- \( P(\text{sum of 10}) = \frac{1}{12} \)
- \( P(\text{sum of 11}) = \frac{1}{18} \)
- \( P(\text{sum of 12}) = \frac{1}{36} \)