A fair die is thrown twice. Let (a, b) denote the outcome in which the first throw shows 'a' and the second throw shows 'b' . Let A and B be the following events :
A={(a,b): a is even}, B={(a,b): b is even}
Statement-1: If C={(a,b): a+b is odd}, then
`P(A cap B cap C)=(1)/(8)`
Statement-2: If D={(a,b): a+b is even}, then
`P(A cap B cap D// A cup B)=(1)/(3)`

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 step by step, we will analyze the two statements provided regarding the outcomes of throwing a fair die twice.

### Step 1: Define the Sample Space
When a die is thrown twice, the sample space consists of all possible outcomes, which can be represented as pairs (a, b) where a is the result of the first throw and b is the result of the second throw. Since each die has 6 faces, the total number of outcomes is:
\[ S = \{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)\} \]
The total number of outcomes is \(6 \times 6 = 36\).

### Step 2: Define Events A, B, C, and D
- **Event A**: The first throw (a) is even. Possible values for a are {2, 4, 6}.
- **Event B**: The second throw (b) is even. Possible values for b are {2, 4, 6}.
- **Event C**: The sum \(a + b\) is odd. This occurs when one of a or b is even and the other is odd.
- **Event D**: The sum \(a + b\) is even. This occurs when both a and b are either even or both are odd.

### Step 3: Calculate Probabilities for Statement 1
**Statement 1**: \(P(A \cap B \cap C) = \frac{1}{8}\)

- **Finding \(A \cap B\)**:
- A consists of outcomes where a is even: {2, 4, 6}.
- B consists of outcomes where b is even: {2, 4, 6}.
- The intersection \(A \cap B\) consists of outcomes where both a and b are even:
\[ A \cap B = \{(2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), (6,6)\} \]
This gives us 9 outcomes.

- **Finding \(C\)**: Since \(C\) requires \(a + b\) to be odd, and both a and b are even in \(A \cap B\), it is impossible for \(a + b\) to be odd. Therefore:
\[ A \cap B \cap C = \emptyset \]
Hence,
\[ P(A \cap B \cap C) = 0 \]

### Conclusion for Statement 1
Since \(P(A \cap B \cap C) = 0\), Statement 1 is false.

### Step 4: Calculate Probabilities for Statement 2
**Statement 2**: \(P(A \cap B \cap D) / P(A \cup B) = \frac{1}{3}\)

- **Finding \(D\)**:
- \(D\) consists of outcomes where \(a + b\) is even. This includes outcomes where both a and b are even or both are odd.

- **Finding \(A \cap B \cap D\)**: Since \(A \cap B\) consists of outcomes where both a and b are even, all of these outcomes will also satisfy \(D\). Thus:
\[ A \cap B \cap D = A \cap B = 9 \text{ outcomes} \]

- **Finding \(A \cup B\)**:
- Outcomes where a is even or b is even can be calculated as:
\[ A \cup B = \{(2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1,2), (3,2), (5,2), (1,4), (3,4), (5,4), (1,6), (3,6), (5,6)\} \]
This gives us 27 outcomes.

- **Calculating the Probability**:
\[ P(A \cap B \cap D) = \frac{9}{36} = \frac{1}{4} \]
\[ P(A \cup B) = \frac{27}{36} = \frac{3}{4} \]
Thus,
\[ \frac{P(A \cap B \cap D)}{P(A \cup B)} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} \]

### Conclusion for Statement 2
Since \(P(A \cap B \cap D) / P(A \cup B) = \frac{1}{3}\), Statement 2 is true.

### Final Answer
- Statement 1 is false.
- Statement 2 is true.