To solve this, we analyze the given scenarios based on the outcomes of rolling two dice. Each die has 6 faces numbered 1 through 6, and the total number of outcomes when rolling two dice is $6 \times 6 = 36$.

Now let's address each part of the problem.

(i) Probability that the sum is 13

• The highest possible sum from rolling two dice is 12 (6 + 6).
• Since 13 is not achievable, the probability of getting a sum of 13 is $\boxed{0}$.

(ii) Probability that the product is 6

We need to identify the pairs of numbers from the two dice whose product equals 6. The pairs are:

• (1, 6)
• (2, 3)
• (3, 2)
• (6, 1)

These are 4 favorable outcomes out of 36 total outcomes, so the probability is: $P(\text{product = 6}) = \frac{4}{36} = \frac{1}{9}$

(iii) Probability that the difference is 1

We want the difference between the two dice to be 1. The possible pairs are:

• (2, 1), (3, 2), (4, 3), (5, 4), (6, 5)
• (1, 2), (2, 3), (3, 4), (4, 5), (5, 6)

There are 10 favorable outcomes, so the probability is: $P(\text{difference = 1}) = \frac{10}{36} = \frac{5}{18}$

(iv) Probability of rolling a pair of prime numbers

Prime numbers between 1 and 6 are 2, 3, and 5. We need to find the pairs where both dice show a prime number. The possible pairs are:

• (2, 2), (2, 3), (2, 5)
• (3, 2), (3, 3), (3, 5)
• (5, 2), (5, 3), (5, 5)

There are 9 favorable outcomes, so the probability is: $P(\text{pair of prime numbers}) = \frac{9}{36} = \frac{1}{4}$

(v) Probability that the sum is divisible by 3 or 5

We need to check for sums that are divisible by either 3 or 5. The sums of two dice can range from 2 to 12.

• Sums divisible by 3: 3, 6, 9, 12
• Sums divisible by 5: 5, 10

Now, let's count how many outcomes give these sums:

• For sum = 3: (1, 2), (2, 1) → 2 outcomes
• For sum = 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1) → 5 outcomes
• For sum = 9: (3, 6), (4, 5), (5, 4), (6, 3) → 4 outcomes
• For sum = 12: (6, 6) → 1 outcome
• For sum = 5: (1, 4), (2, 3), (3, 2), (4, 1) → 4 outcomes
• For sum = 10: (4, 6), (5, 5), (6, 4) → 3 outcomes

Adding them all together, we get $2 + 5 + 4 + 1 + 4 + 3 = 19$ favorable outcomes. Therefore, the probability is: $P(\text{sum divisible by 3 or 5}) = \frac{19}{36}$

Final Results:

1. Probability the sum is 13: $\boxed{0}$
2. Probability the product is 6: $\boxed{\frac{1}{9}}$
3. Probability the difference is 1: $\boxed{\frac{5}{18}}$
4. Probability of a pair of prime numbers: $\boxed{\frac{1}{4}}$
5. Probability the sum is divisible by 3 or 5: $\boxed{\frac{19}{36}}$

Would you like more details or have any questions?

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Here are 5 related questions to think about:

1. What is the probability of getting an even product from two dice rolls?
2. What is the probability that the sum of two dice is a prime number?
3. How would the probabilities change if the dice had numbers 1-8 instead of 1-6?
4. What is the probability of rolling a double (the same number on both dice)?
5. How does the probability of rolling an odd sum compare to rolling an even sum?

Tip: Always break down complex probability questions by first identifying all possible outcomes and then counting favorable ones.