Consider the system of linear equation ax+by=0, cx+dy=0, a, b, c, d `in` {0, 1}
Statement-1 : The probability that the system of equations has a unique solution is `3/8`.
Statement-2 : The probability that the system has a solution is 1.

Options

A. Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement -4
B. Statement-1 is true, Statement-2 is true, Statement-2 is Not a correct explanation for statement -4
C. Statement-1 is true, Statement-2 is False
D. Statement-1 is False, Statement-2 is true

Answer

To solve the problem, we need to analyze the given system of linear equations:

1. **Equations**:
- \( ax + by = 0 \)
- \( cx + dy = 0 \)

where \( a, b, c, d \in \{0, 1\} \).

### Step 1: Determine Total Outcomes
Each of the coefficients \( a, b, c, d \) can take 2 values (0 or 1). Therefore, the total number of combinations for the coefficients is:

\[ \text{Total Outcomes} = 2 \times 2 \times 2 \times 2 = 16 \]

**Hint**: Count the combinations of the coefficients based on the number of choices available for each coefficient.

### Step 2: Identify Cases for Unique Solutions
For the system to have a unique solution, the two equations must not be proportional. This means:

\[ \frac{a}{b} \neq \frac{c}{d} \]

We need to consider the cases for \( a, b, c, d \) being either 0 or 1.

### Step 3: Analyze Cases
We will analyze the cases based on the values of \( a, b, c, d \):

1. **Case 1**: \( a = 0, b = 0 \) (Equation 1 becomes \( 0 = 0 \))
2. **Case 2**: \( a = 0, b = 1 \) (Equation 1 becomes \( y = 0 \))
3. **Case 3**: \( a = 1, b = 0 \) (Equation 1 becomes \( x = 0 \))
4. **Case 4**: \( a = 1, b = 1 \) (Equation 1 becomes \( x + y = 0 \))

Similarly, we can analyze the second equation \( cx + dy = 0 \) for the same cases.

### Step 4: Count Favorable Outcomes for Unique Solutions
To have a unique solution, we need to avoid the cases where both equations are equivalent or lead to infinite solutions. The valid combinations for unique solutions are:

- \( (1, 0, 0, 1) \) → \( x = 0, y = 0 \) (not unique)
- \( (1, 0, 1, 0) \) → unique solution
- \( (0, 1, 1, 0) \) → unique solution
- \( (1, 1, 0, 1) \) → unique solution
- \( (1, 1, 1, 0) \) → unique solution
- \( (0, 1, 0, 1) \) → unique solution

After analyzing all combinations, we find that there are 6 favorable outcomes that yield a unique solution.

### Step 5: Calculate Probability of Unique Solutions
The probability of having a unique solution is given by:

\[ \text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{6}{16} = \frac{3}{8} \]

### Step 6: Verify the Existence of Solutions
The system will always have at least one solution (the trivial solution \( x = 0, y = 0 \)) since both equations can be satisfied by this point. Therefore, the probability that the system has a solution is indeed 1.

### Conclusion
- **Statement 1**: The probability that the system of equations has a unique solution is \( \frac{3}{8} \) (True).
- **Statement 2**: The probability that the system has a solution is 1 (True).

Thus, both statements are correct.

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