The problem is about calculating probabilities related to rolling two dice. (i) Find the probability that the total of the two dice is 6 or 8. (ii) Find the probability that the same number appears on both dice. (iii) Find the probability that the total is not less than 5 (meaning the total is greater than or equal to 5).

1. Problem Description

The problem is about calculating probabilities related to rolling two dice.

(i) Find the probability that the total of the two dice is 6 or

8. (ii) Find the probability that the same number appears on both dice.

(iii) Find the probability that the total is not less than 5 (meaning the total is greater than or equal to 5).

2. Solution Steps

(b)

Total number of sample space = $n(S) = 36$. This means there are 36 possible outcomes when rolling two dice.

(i)

We are looking for the probability $P$(total of 6 or 8).

The combinations that add up to 6 are: (1,5), (2,4), (3,3), (4,2), (5,1). There are 5 such combinations.

The combinations that add up to 8 are: (2,6), (3,5), (4,4), (5,3), (6,2). There are 5 such combinations.

So there are a total of $5 + 5 = 10$ combinations that sum to 6 or

8. $P$(total of 6 or 8) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{36} = \frac{5}{18}$.

(ii)

We are looking for the probability $P$(the same number on the two dice).

The combinations where the same number appears on both dice are: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).

There are 6 such combinations.

$P$(the same number on the two dice) = $\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}$.

(iii)

We are looking for the probability $P$(total of not less than 5), which means $P$(total $\ge$ 5).

It is easier to calculate the complementary probability: $P$(total $< 5$). The possible totals less than 5 are 2, 3, and

4. Combinations that add up to 2: (1,1) - 1 combination

Combinations that add up to 3: (1,2), (2,1) - 2 combinations

Combinations that add up to 4: (1,3), (2,2), (3,1) - 3 combinations

So there are a total of $1 + 2 + 3 = 6$ combinations that sum to less than

5. $P$(total $< 5$) = $\frac{6}{36}$.

$P$(total $\ge 5$) = $1 - P$(total $< 5$) = $1 - \frac{6}{36} = \frac{36 - 6}{36} = \frac{30}{36} = \frac{5}{6}$.

Another way to count the combinations is by simply looking for total number of outcomes which are not less than 5:

Total outcomes =

3

6. Outcomes that are less than 5 are = (1,1) = 2; (1,2), (2,1) = 3; (1,3),(2,2),(3,1) =

4. So, total of 6 possibilities.

Therefore outcomes that are not less than 5, are 36 - 6 =

3

0. So $P$(total $\ge 5$) = $\frac{30}{36} = \frac{5}{6}$.

3. Final Answer

(i) $P$(total of 6 or 8) = $\frac{5}{18}$

(ii) $P$(the same number on the two dice) = $\frac{1}{6}$

(iii) $P$(total of not less than 5) = $\frac{5}{6}$