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Let's examine the table and the question you provided.

The table shows the sums of two dice: one red die and one black die, each numbered from 1 to 6. The sums of each possible roll are already calculated in the table. Now let's address the questions one by one.

1. **(i) What is the probability that the score on the two dice will add up to an even number?**

To solve this, we need to count the number of even sums in the table and divide by the total number of possible outcomes.

### Total number of outcomes: 
Since both dice are 6-sided, there are \(6 \times 6 = 36\) possible outcomes.

### Even sums:
From the table, the even sums are: 2, 4, 6, 8, 10, and 12. Let's count how many even sums are present:

- Row 1 (Red die = 1): 2, 4, 6 → 3 even sums
- Row 2 (Red die = 2): 4, 6, 8, 10 → 4 even sums
- Row 3 (Red die = 3): 6, 8, 10 → 3 even sums
- Row 4 (Red die = 4): 6, 8, 10, 12 → 4 even sums
- Row 5 (Red die = 5): 6, 8, 10 → 3 even sums
- Row 6 (Red die = 6): 8, 10, 12 → 3 even sums

So, the total number of even sums is \(3 + 4 + 3 + 4 + 3 + 3 = 20\).

Therefore, the probability of getting an even number is:
\[
P(\text{Even number}) = \frac{20}{36} = \frac{5}{9}
\]

2. **(ii) What is the probability that the score on the two dice will add up to a number divisible by 5?**

The numbers divisible by 5 are 5 and 10. Let's count how many times these sums appear in the table:

- Row 1 (Red die = 1): no 5 or 10
- Row 2 (Red die = 2): 10 → 1 sum
- Row 3 (Red die = 3): 5, 10 → 2 sums
- Row 4 (Red die = 4): 10 → 1 sum
- Row 5 (Red die = 5): 5, 10 → 2 sums
- Row 6 (Red die = 6): 10 → 1 sum

So, the total number of sums divisible by 5 is \(1 + 2 + 1 + 2 + 1 = 7\).

Therefore, the probability of getting a number divisible by 5 is:
\[
P(\text{Divisible by 5}) = \frac{7}{36}
\]

3. **(iii) What is the probability that the score on the two dice will add up to a number divisible by 5 or an even number?**

This is the union of two events: sums divisible by 5 and sums that are even. We already know the number of even sums (20) and sums divisible by 5 (7), but we need to subtract any sums that are both divisible by 5 and even to avoid double-counting.

The only even number divisible by 5 in this case is 10. From the table, 10 appears **5 times**.

So, the number of sums that are either divisible by 5 or even is:
\[
20 + 7 - 5 = 22
\]

Therefore, the probability is:
\[
P(\text{Divisible by 5 or even}) = \frac{22}{36} = \frac{11}{18}
\]

4. **(iv) What is the probability that the score on the two dice will add up to a number divisible by 5 and even?**

From part (iii), we already found that the only even number divisible by 5 is 10, which appears 5 times in the table.

Therefore, the probability is:
\[
P(\text{Divisible by 5 and even}) = \frac{5}{36}
\]

5. **(v) What is the probability that the score on the two dice will add up to a number that is greater than 10 and odd?**

The numbers greater than 10 and odd are 11. Let's count how many times 11 appears in the table:

- Row 5 (Red die = 5): 11 → 1 sum
- Row 6 (Red die = 6): 11 → 1 sum

So, the number of sums greater than 10 and odd is \(1 + 1 = 2\).

Therefore, the probability is:
\[
P(\text{Greater than 10 and odd}) = \frac{2}{36} = \frac{1}{18}
\]

---

Let me know if you would like more details or have any further questions!

### Related questions:
1. What is the probability of getting a sum of exactly 7?
2. How many sums are greater than 6?
3. What is the probability of getting a sum of less than 5?
4. How many odd sums are there?
5. What is the probability of getting a sum divisible by 3?

### Tip:
When calculating probabilities, ensure to count each event carefully and distinguish between mutually exclusive events (like "or" and "and" situations).

What is the probability that the score on the two dice will add up to: (i) an even number, (ii) a number divisible by 5, (iii) a number divisible by 5 or even, (iv) a number divisible by 5 and even, (v) a number that is greater than 10 and odd?

Explore the probability of different sums when rolling two dice: even numbers, sums divisible by 5, and other scenarios. This guide covers core probability concepts, including the addition rule of probability and mutually exclusive events. Ideal for middle school students (Grades 6-8).

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