Let A, B and C be three events such that P(C )=0
Statement-1: `P(A cap B cap C)=0`
Statement-2: `P(A cup B cup C)=P(A cup B)`

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 two statements given the condition that \( P(C) = 0 \).

### Step 1: Analyze Statement 1
**Statement 1**: \( P(A \cap B \cap C) = 0 \)

Since \( P(C) = 0 \), it means that the event \( C \) does not occur at all. The intersection of any event with an event that has zero probability will also have zero probability. Therefore, regardless of what events \( A \) and \( B \) are, the intersection \( A \cap B \cap C \) must also be zero.

**Conclusion for Statement 1**:
\[ P(A \cap B \cap C) = 0 \text{ is true.} \]

### Step 2: Analyze Statement 2
**Statement 2**: \( P(A \cup B \cup C) = P(A \cup B) \)

Using the property of probabilities, we can express the union of events as follows:
\[ P(A \cup B \cup C) = P((A \cup B) \cup C) \]
Since \( P(C) = 0 \), the event \( C \) does not contribute to the probability of the union. Thus, we can simplify:
\[ P((A \cup B) \cup C) = P(A \cup B) + P(C) - P((A \cup B) \cap C) \]
Given that \( P(C) = 0 \) and \( P((A \cup B) \cap C) = P(A \cap B \cap C) = 0 \) (from Statement 1), we have:
\[ P(A \cup B \cup C) = P(A \cup B) + 0 - 0 = P(A \cup B) \]

**Conclusion for Statement 2**:
\[ P(A \cup B \cup C) = P(A \cup B) \text{ is also true.} \]

### Final Conclusion
Both statements are true:
- Statement 1 is true because \( P(A \cap B \cap C) = 0 \).
- Statement 2 is true because \( P(A \cup B \cup C) = P(A \cup B) \).

Thus, both statements are valid under the condition that \( P(C) = 0 \).