To solve the problem, we will break it down into three parts as given in the question.
### Step-by-Step Solution
**Step 1: Determine the Sample Space**
When two unbiased dice are rolled, each die has 6 faces. Therefore, the total number of outcomes (sample space) when rolling two dice is:
\[
\text{Total outcomes} = 6 \times 6 = 36
\]
**Step 2: Part (a) - Probability of obtaining a total of at least 10**
To find the probability of obtaining a total of at least 10, we need to identify the combinations that yield a sum of 10, 11, or 12.
- **Sum = 10**: (4,6), (5,5), (6,4) → 3 outcomes
- **Sum = 11**: (5,6), (6,5) → 2 outcomes
- **Sum = 12**: (6,6) → 1 outcome
Adding these outcomes together gives:
\[
\text{Total favorable outcomes} = 3 + 2 + 1 = 6
\]
Thus, the probability is:
\[
P(\text{total} \geq 10) = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}
\]
**Step 3: Part (b) - Probability of getting a multiple of 2 on one die and a multiple of 3 on the other die**
Multiples of 2 on a die are: 2, 4, 6.
Multiples of 3 on a die are: 3, 6.
Now, we can list the combinations:
- (2,3), (2,6)
- (4,3), (4,6)
- (6,3), (6,6)
- (3,2), (3,4), (3,6)
- (6,2), (6,4)
Counting these, we find:
- From die 1 (multiple of 2): (2,3), (2,6), (4,3), (4,6), (6,3), (6,6) → 6 outcomes
- From die 2 (multiple of 3): (3,2), (3,4), (3,6), (6,2), (6,4) → 5 outcomes
However, we need to avoid double counting combinations like (6,6) and (2,6).
Total unique combinations = 11.
Thus, the probability is:
\[
P(\text{multiple of 2 on one die and multiple of 3 on the other}) = \frac{11}{36}
\]
**Step 4: Part (c) - Probability of getting a multiple of 3 as the sum**
The possible sums that are multiples of 3 are: 3, 6, 9, and 12.
- **Sum = 3**: (1,2), (2,1) → 2 outcomes
- **Sum = 6**: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes
- **Sum = 9**: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- **Sum = 12**: (6,6) → 1 outcome
Adding these outcomes gives:
\[
\text{Total favorable outcomes} = 2 + 5 + 4 + 1 = 12
\]
Thus, the probability is:
\[
P(\text{sum is a multiple of 3}) = \frac{12}{36} = \frac{1}{3}
\]
### Final Answers:
(a) \( \frac{1}{6} \)
(b) \( \frac{11}{36} \)
(c) \( \frac{1}{3} \)
