Listing All Possible Outcomes:

When rolling two six-faced dice, each die can show a number between 1 and 6. The total outcomes are pairs $(x, y)$, where $x$ is the result of the first die and $y$ is the result of the second die. The 36 outcomes are:

$\{(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), \ldots, (6, 6)\}$

1. Probability of Getting a Sum of 11:

To find the probability, count the outcomes where the sum of the two dice equals 11. These are:

$(5, 6), (6, 5)$

Thus, there are 2 outcomes where the sum is 11. The probability is:

$P(\text{Sum} = 11) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18}.$

2. Probability of Getting a Sum Less Than or Equal to 4:

The possible outcomes with sums less than or equal to 4 are:

• Sum = 2: $(1, 1)$ — 1 outcome
• Sum = 3: $(1, 2), (2, 1)$ — 2 outcomes
• Sum = 4: $(1, 3), (2, 2), (3, 1)$ — 3 outcomes

Thus, there are $1 + 2 + 3 = 6$ favorable outcomes. The probability is:

$P(\text{Sum} \leq 4) = \frac{6}{36} = \frac{1}{6}.$

3. Probability of Getting a Sum of 13 or More:

The maximum sum from rolling two dice is 12 ($6 + 6$). Therefore, it's impossible to get a sum of 13 or more. The probability is:

$P(\text{Sum} \geq 13) = 0.$

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Summary of Probabilities:

1. $P(\text{Sum} = 11) = \frac{1}{18}$
2. $P(\text{Sum} \leq 4) = \frac{1}{6}$
3. $P(\text{Sum} \geq 13) = 0$

Would you like further details or visualizations for this problem?

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Follow-Up Questions:

1. How can you generalize these probabilities for dice with more or fewer faces?
2. What is the probability of getting an odd sum from two dice rolls?
3. What is the probability of rolling doubles (e.g., $(1,1), (2,2), \ldots$)?
4. If one die is biased to favor a 6, how does that affect the probabilities?
5. What is the expected sum of two dice rolls?

Tip:

Always ensure the total number of outcomes is consistent when calculating probabilities to avoid errors.