The probablities of events , `AnnB`, `A`, `B` and `AuuB` are respectively in `A.P.` with second term equal to the common difference. Therefore `A` and `B` are

Options

A. mutually exclusive
B. independent
C. such that one of them must occur
D. such that one is twice as likely as the order

Explanation

To solve the problem, we need to analyze the given probabilities of events A ∩ B, A, B, and A ∪ B, which are in Arithmetic Progression (A.P.) with the second term equal to the common difference.

### Step-by-Step Solution:

1. **Define the Probabilities**:
Let:
- \( P(A \cap B) = a \) (first term)
- \( P(A) = a + d \) (second term)
- \( P(B) = a + 2d \) (third term)
- \( P(A \cup B) = a + 3d \) (fourth term)

2. **Set Up the Condition**:
According to the problem, the second term is equal to the common difference:
\[ a + d = d \]

3. **Solve for \( a \)**:
Rearranging the equation from step 2 gives:
\[ a + d - d = 0 \implies a = 0 \]

4. **Substitute \( a \) Back**:
Now substituting \( a = 0 \) into the expressions for the probabilities:
- \( P(A \cap B) = 0 \)
- \( P(A) = 0 + d = d \)
- \( P(B) = 0 + 2d = 2d \)
- \( P(A \cup B) = 0 + 3d = 3d \)

5. **Interpret the Results**:
Since \( P(A \cap B) = 0 \), it indicates that events A and B do not have any outcomes in common. Thus, A and B are mutually exclusive events.

6. **Check the Probabilities**:
- The probabilities must satisfy the condition \( P(A) + P(B) - P(A \cap B) = P(A \cup B) \):
\[ d + 2d - 0 = 3d \]
This condition holds true.

7. **Conclusion**:
Therefore, we conclude that:
- Events A and B are mutually exclusive.
- The probability of event B is twice that of event A.

### Final Answer:
- Events A and B are mutually exclusive.
- The probability of event B is twice that of event A.