To determine the probability that the sum of two rolled six-sided dice is a prime number, let's work step by step:

Step 1: Identify all possible outcomes

Each die has 6 faces, so the total number of outcomes is: $6 \times 6 = 36$

Step 2: List possible sums and identify prime sums

The possible sums range from $2$ (1+1) to $12$ (6+6). The prime numbers within this range are: $2, 3, 5, 7, 11$

Step 3: Count the combinations for each prime sum

We will count the number of outcomes (pairs of dice rolls) that result in each prime sum:

• Sum = 2: $(1, 1)$ → $1$ outcome
• Sum = 3: $(1, 2), (2, 1)$ → $2$ outcomes
• Sum = 5: $(1, 4), (2, 3), (3, 2), (4, 1)$ → $4$ outcomes
• Sum = 7: $(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)$ → $6$ outcomes
• Sum = 11: $(5, 6), (6, 5)$ → $2$ outcomes

Total prime outcomes: $1 + 2 + 4 + 6 + 2 = 15$

Step 4: Calculate the probability

The probability of rolling a prime sum is the number of favorable outcomes divided by the total number of outcomes: $P(\text{prime sum}) = \frac{15}{36} = \frac{5}{12}$

Final Answer:

The probability that the sum is a prime number is: $\boxed{\frac{5}{12}}$

Would you like further clarification or details? Here are some related questions to expand this concept:

1. What is the probability of rolling an even sum with two dice?
2. Can we compute the probability of rolling a sum greater than 8?
3. How does the outcome change if one die is biased?
4. What is the distribution of sums when rolling three dice instead of two?
5. How can we calculate probabilities for non-prime or composite sums?

Tip: Prime numbers are only divisible by 1 and themselves. This property is key in analyzing dice outcomes systematically!