Two dice are rolled with numbers 1-6. Find the probability that; (i) the sum is 13 (ii) the product is 6? (iii) the difference is 1 (iv) the pair of prime number (v) the sum is divisible by 3 or 5?

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.

Explore how to find the probabilities for different events when rolling two dice. Learn to calculate the probability of sums, products, differences, and prime pairs. This problem is suitable for students in Grades 6-8, covering key concepts in probability and combinatorics.

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