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.
The number of all possible ordered pair (I,j) for which the events `A_(i) and a_(j)` are independent is

Options

A. 6
B. 12
C. 13
D. 25

Explanation

To solve the problem, we need to determine the number of ordered pairs \((i, j)\) for which the events \(A_i\) and \(A_j\) are independent when two fair dice are rolled.

### Step 1: Understanding the Events
The event \(A_i\) represents the situation where the sum of the faces of the two dice is divisible by \(i\). The possible sums when rolling two dice range from 2 (1+1) to 12 (6+6).

### Step 2: Total Possible Outcomes
When rolling two dice, there are a total of \(6 \times 6 = 36\) possible outcomes.

### Step 3: Finding the Probability of Each Event
To find the probability \(P(A_i)\), we need to find how many sums are divisible by \(i\) and divide that by the total outcomes (36).

- For \(i = 1\): All sums (2 to 12) are divisible by 1. So, \(P(A_1) = 36/36 = 1\).
- For \(i = 2\): The sums divisible by 2 are 2, 4, 6, 8, 10, 12. There are 18 outcomes. So, \(P(A_2) = 18/36 = 1/2\).
- For \(i = 3\): The sums divisible by 3 are 3, 6, 9, 12. There are 12 outcomes. So, \(P(A_3) = 12/36 = 1/3\).
- For \(i = 4\): The sums divisible by 4 are 4, 8, 12. There are 9 outcomes. So, \(P(A_4) = 9/36 = 1/4\).
- For \(i = 5\): The sums divisible by 5 are 5, 10. There are 6 outcomes. So, \(P(A_5) = 6/36 = 1/6\).
- For \(i = 6\): The sums divisible by 6 are 6, 12. There are 3 outcomes. So, \(P(A_6) = 3/36 = 1/12\).
- For \(i = 7\): The only sum divisible by 7 is 7. There are 6 outcomes. So, \(P(A_7) = 6/36 = 1/6\).
- For \(i = 8\): The only sum divisible by 8 is 8. There are 3 outcomes. So, \(P(A_8) = 3/36 = 1/12\).
- For \(i = 9\): The only sum divisible by 9 is 9. There are 4 outcomes. So, \(P(A_9) = 4/36 = 1/9\).
- For \(i = 10\): The only sum divisible by 10 is 10. There are 3 outcomes. So, \(P(A_{10}) = 3/36 = 1/12\).
- For \(i = 11\): The only sum divisible by 11 is 11. There are 2 outcomes. So, \(P(A_{11}) = 2/36 = 1/18\).
- For \(i = 12\): The only sum divisible by 12 is 12. There is 1 outcome. So, \(P(A_{12}) = 1/36\).

### Step 4: Independence of Events
Two events \(A_i\) and \(A_j\) are independent if:
\[ P(A_i \cap A_j) = P(A_i) \cdot P(A_j) \]

### Step 5: Finding Ordered Pairs \((i, j)\)
We need to find pairs \((i, j)\) such that the events are independent.

1. **For \(i = 1\)**: \(A_1\) is independent of all events \(A_j\) for \(j = 1, 2, \ldots, 12\). This gives us 12 pairs: \((1,1), (1,2), \ldots, (1,12)\).

2. **For \(i = 2\)**: Check independence with \(A_j\) for \(j = 2, 3, \ldots, 12\). We find that \(A_2\) is independent with \(A_4\) and \(A_6\) (2 pairs).

3. **For \(i = 3\)**: Check independence with \(A_j\) for \(j = 3, 4, \ldots, 12\). We find that \(A_3\) is independent with \(A_6\) (1 pair).

4. **For \(i = 4\)**: Check independence with \(A_j\) for \(j = 4, 5, \ldots, 12\). We find that \(A_4\) is independent with \(A_8\) (1 pair).

5. **For \(i = 5\)**: Check independence with \(A_j\) for \(j = 5, 6, \ldots, 12\). No independent pairs.

6. **For \(i = 6\)**: Check independence with \(A_j\) for \(j = 6, 7, \ldots, 12\). No independent pairs.

7. **For \(i = 7\)**: Check independence with \(A_j\) for \(j = 7, 8, \ldots, 12\). No independent pairs.

8. **For \(i = 8\)**: Check independence with \(A_j\) for \(j = 8, 9, \ldots, 12\). No independent pairs.

9. **For \(i = 9\)**: Check independence with \(A_j\) for \(j = 9, 10, \ldots, 12\). No independent pairs.

10. **For \(i = 10\)**: Check independence with \(A_j\) for \(j = 10, 11, 12\). No independent pairs.

11. **For \(i = 11\)**: Check independence with \(A_j\) for \(j = 11, 12\). No independent pairs.

12. **For \(i = 12\)**: No pairs since it only pairs with itself.

### Step 6: Count the Total Ordered Pairs
From our analysis, we have:
- 12 pairs from \(A_1\)
- 2 pairs from \(A_2\)
- 1 pair from \(A_3\)
- 1 pair from \(A_4\)

Thus, the total number of ordered pairs \((i, j)\) for which the events \(A_i\) and \(A_j\) are independent is:
\[ 12 + 2 + 1 + 1 = 16 \]

### Final Answer
The number of all possible ordered pairs \((i, j)\) for which the events \(A_i\) and \(A_j\) are independent is **16**.