Let A and B be two events such that `P(A cup B)=P(A cap B)`. Then,
Statement-1: `P(A cap overline(B))=P(overline(A) cap B)=0`
Statement-2: `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.

Explanation

To solve the problem, we need to analyze the given conditions and statements step by step.

### Given:
We have two events A and B such that:
\[ P(A \cup B) = P(A \cap B) \]

### Step 1: Use the formula for the probability of union of two events
Using the formula for the probability of the union of two events, we can express \( P(A \cup B) \):
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

### Step 2: Set up the equation
Since we know that \( P(A \cup B) = P(A \cap B) \), we can set up the following equation:
\[ P(A) + P(B) - P(A \cap B) = P(A \cap B) \]

### Step 3: Rearranging the equation
Rearranging the equation gives us:
\[ P(A) + P(B) = 2P(A \cap B) \]

### Step 4: Analyze the implications
From the equation \( P(A) + P(B) = 2P(A \cap B) \), we can infer that the probabilities of A and B can be expressed in terms of their intersection.

### Step 5: Consider the complements
Now, we need to analyze the first statement:
\[ P(A \cap \overline{B}) = P(\overline{A} \cap B) = 0 \]

This means that the probabilities of A occurring without B and B occurring without A are both zero.

### Step 6: Conclusion from the complements
If \( P(A \cap \overline{B}) = 0 \), it implies that whenever A occurs, B must also occur. Similarly, if \( P(\overline{A} \cap B) = 0 \), it implies that whenever B occurs, A must also occur. Therefore, A and B must occur together or not at all.

### Step 7: Verify the first statement
Since both \( P(A \cap \overline{B}) \) and \( P(\overline{A} \cap B) \) are zero, we conclude that:
\[ P(A \cap \overline{B}) = 0 \quad \text{and} \quad P(\overline{A} \cap B) = 0 \]

Thus, Statement 1 is correct.

### Step 8: Verify the second statement
Now, we check Statement 2:
\[ P(A) + P(B) = 1 \]

From our earlier equation \( P(A) + P(B) = 2P(A \cap B) \), we cannot conclude that \( P(A) + P(B) = 1 \) unless \( P(A \cap B) = \frac{1}{2} \). Therefore, Statement 2 is not necessarily true.

### Final Conclusion:
- Statement 1 is true: \( P(A \cap \overline{B}) = P(\overline{A} \cap B) = 0 \)
- Statement 2 is false: \( P(A) + P(B) \) does not necessarily equal 1.

### Correct Option:
Thus, the correct option is that Statement 1 is true and Statement 2 is false.

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