Two fair dice are rolled. Let `P(A_(i))gt0` donete the event that the sum of the faces of the dice is divisible by i.
Which one of the following events is most probable?

Options

A. `A_(3)`
B. `A_(4)`
C. `A_(5)`
D. `A_(6)`

Explanation

To solve the problem of determining which event \( P(A_i) \) is most probable when rolling two fair dice, we will calculate the probabilities for each \( i \) where \( i = 3, 4, 5, 6 \).

### Step-by-Step Solution:

1. **Calculate the Total Outcomes**:
When rolling two dice, each die has 6 faces. Therefore, the total number of outcomes when rolling two dice is:
\[ \text{Total Outcomes} = 6 \times 6 = 36 \]

2. **Determine the Event \( A_3 \)**:
We need to find the number of outcomes where the sum of the faces is divisible by 3. The possible sums when rolling two dice range from 2 to 12. The sums divisible by 3 within this range are 3, 6, 9, and 12.
- **Sums**:
- **3**: (1,2), (2,1) → 2 outcomes
- **6**: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes
- **9**: (3,6), (4,5), (5,4), (6,3) → 4 outcomes
- **12**: (6,6) → 1 outcome
- **Total for \( A_3 \)**:
\[ 2 + 5 + 4 + 1 = 12 \]
- **Probability \( P(A_3) \)**:
\[ P(A_3) = \frac{12}{36} = \frac{1}{3} \]

3. **Determine the Event \( A_4 \)**:
We find the outcomes where the sum is divisible by 4. The sums divisible by 4 are 4, 8, and 12.
- **Sums**:
- **4**: (1,3), (2,2), (3,1) → 3 outcomes
- **8**: (2,6), (3,5), (4,4), (5,3), (6,2) → 5 outcomes
- **12**: (6,6) → 1 outcome
- **Total for \( A_4 \)**:
\[ 3 + 5 + 1 = 9 \]
- **Probability \( P(A_4) \)**:
\[ P(A_4) = \frac{9}{36} = \frac{1}{4} \]

4. **Determine the Event \( A_5 \)**:
We find the outcomes where the sum is divisible by 5. The sums divisible by 5 are 5 and 10.
- **Sums**:
- **5**: (1,4), (2,3), (3,2), (4,1) → 4 outcomes
- **10**: (4,6), (5,5), (6,4) → 3 outcomes
- **Total for \( A_5 \)**:
\[ 4 + 3 = 7 \]
- **Probability \( P(A_5) \)**:
\[ P(A_5) = \frac{7}{36} \]

5. **Determine the Event \( A_6 \)**:
We find the outcomes where the sum is divisible by 6. The sums divisible by 6 are 6 and 12.
- **Sums**:
- **6**: (1,5), (2,4), (3,3), (4,2), (5,1) → 5 outcomes
- **12**: (6,6) → 1 outcome
- **Total for \( A_6 \)**:
\[ 5 + 1 = 6 \]
- **Probability \( P(A_6) \)**:
\[ P(A_6) = \frac{6}{36} = \frac{1}{6} \]

6. **Compare the Probabilities**:
- \( P(A_3) = \frac{1}{3} \)
- \( P(A_4) = \frac{1}{4} \)
- \( P(A_5) = \frac{7}{36} \)
- \( P(A_6) = \frac{1}{6} \)

To compare these probabilities, we can convert them to a common denominator (36):
- \( P(A_3) = \frac{12}{36} \)
- \( P(A_4) = \frac{9}{36} \)
- \( P(A_5) = \frac{7}{36} \)
- \( P(A_6) = \frac{6}{36} \)

The highest probability is \( P(A_3) = \frac{12}{36} \).

### Conclusion:
The most probable event is \( P(A_3) \).