If `A` and `B` are exhaustive events in a sample space such that probabilites of the events `AnnB`, `A`, `B` and `AuuB` are in `A.P.` If `P(A)=K`, where `0 lt K le 1`, then

Options

A. `P(B)=(K+1)/(2)`
B. `P(AnnB)=(3K-1)/(2)`
C. `P(AuuB)=1`
D. `P(A'uuB')=(3(1-K))/(2)`

Explanation

To solve the problem, we need to use the information given about the probabilities of the events A, B, A ∩ B (intersection of A and B), and A ∪ B (union of A and B) being in arithmetic progression (AP).

### Step-by-Step Solution:

1. **Understanding the Events**:
- Let \( P(A) = k \)
- Since A and B are exhaustive events, we have \( P(A \cup B) = 1 \).
- Let \( P(B) = p \), \( P(A \cap B) = q \), and \( P(A \cup B) = 1 \).

2. **Using the Arithmetic Progression Condition**:
- The probabilities \( P(A \cap B) \), \( P(A) \), \( P(B) \), and \( P(A \cup B) \) are in AP.
- In AP, the middle term is the average of the other two terms. Thus, we can write the equation:
\[ 2P(A) = P(A \cap B) + P(B) \]
- Substituting the known values:
\[ 2k = q + p \quad \text{(1)} \]

3. **Using the Exhaustive Events Condition**:
- Since A and B are exhaustive, we also know:
\[ P(A) + P(B) = 1 \quad \Rightarrow \quad k + p = 1 \quad \Rightarrow \quad p = 1 - k \quad \text{(2)} \]

4. **Substituting Equation (2) into Equation (1)**:
- Replace \( p \) in Equation (1):
\[ 2k = q + (1 - k) \]
- Simplifying gives:
\[ 2k = q + 1 - k \quad \Rightarrow \quad q = 3k - 1 \quad \text{(3)} \]

5. **Finding \( P(A \cup B) \)**:
- We know \( P(A \cup B) = 1 \), which is consistent with our earlier findings.

6. **Finding \( P(A \cap B) \)**:
- From Equation (3), we have:
\[ P(A \cap B) = 3k - 1 \]

7. **Finding \( P(A^c \cup B^c) \)**:
- We can express \( P(A^c \cup B^c) \) using the complement rule:
\[ P(A^c \cup B^c) = 1 - P(A \cap B) = 1 - (3k - 1) = 2 - 3k \]

### Summary of Results:
- \( P(A) = k \)
- \( P(B) = 1 - k \)
- \( P(A \cap B) = 3k - 1 \)
- \( P(A^c \cup B^c) = 2 - 3k \)