Statement-1 Let A and B be the two events, such that `P(AcupB)=P(AcapB)`, then `P(AcapB')=P(A'capB)=0`
Statement-2 Let A and B be the two events, such that `P(AcupB)=P(AcapB),` then `P(A)+P(B)=1`

Options

A. Statement-1 is true, Statement-2 is true: Statement-2 is a correct explanation for Statement-1
B. Statement-1 is true, Statement-2 is true: Statement-2 is not a correct explanation for Statement-1
C. Statement-1 is true, Statement-2 is false
D. Statement-1 is false, Statement-2 is true

Answer

To solve the problem, we will analyze both statements step by step.

### Step 1: Analyze Statement 1
**Statement 1:** Let A and B be two events such that \( P(A \cup B) = P(A \cap B) \). Then \( P(A \cap B') = P(A' \cap B) = 0 \).

1. **Understanding the equation \( P(A \cup B) = P(A \cap B) \)**:
- From set theory, we know that:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Setting this equal to \( P(A \cap B) \):
\[ P(A) + P(B) - P(A \cap B) = P(A \cap B) \]
- Rearranging gives:
\[ P(A) + P(B) = 2P(A \cap B) \]

2. **Finding \( P(A \cap B') \) and \( P(A' \cap B) \)**:
- \( P(A \cap B') \) represents the probability of event A occurring while event B does not occur.
- \( P(A' \cap B) \) represents the probability of event B occurring while event A does not occur.
- Since \( P(A \cup B) = P(A \cap B) \), it implies that the only area where A and B overlap is the area of their intersection. Therefore, the areas outside this intersection (i.e., \( A \cap B' \) and \( A' \cap B \)) must have zero probability.
- Thus, we conclude:
\[ P(A \cap B') = 0 \quad \text{and} \quad P(A' \cap B) = 0 \]

**Conclusion for Statement 1**: True.

### Step 2: Analyze Statement 2
**Statement 2:** Let A and B be two events such that \( P(A \cup B) = P(A \cap B) \). Then \( P(A) + P(B) = 1 \).

1. **Using the previous result**:
- We already established that \( P(A \cup B) = P(A \cap B) \) leads to:
\[ P(A) + P(B) = 2P(A \cap B) \]
- This does not imply that \( P(A) + P(B) = 1 \). In fact, it suggests that the sum of the probabilities of A and B is twice the probability of their intersection.

2. **Counterexample**:
- If we take \( P(A) = 0.5 \) and \( P(B) = 0.5 \), then:
\[ P(A \cap B) = 0.5 \quad \text{(if they overlap completely)} \]
- Therefore, \( P(A \cup B) = 0.5 \) which is not equal to \( 1 \).

**Conclusion for Statement 2**: False.

### Final Conclusion
- **Statement 1 is true**.
- **Statement 2 is false**.