If Aand B are two events than the value of the determinant choosen at random from all the determinants of order 2 with entries 0 or 1 only is positive or negative respectively. Then (a) `P(A) ge P(B)` (b) `P(A) le P(B)` (c) `P(A)= P(B)` (d)None of these

Options

A. `P(A) gt P(B)`
B. `P(A) lt P(B)`
C. `P(A) = P(B)=1//2`
D. `P(A)=P(B)`

Answer

To solve the problem, we need to analyze the determinants of order 2 with entries consisting only of 0 or 1. Let's go through the solution step by step.

### Step 1: Determine the total number of 2x2 determinants with entries 0 or 1.

A 2x2 determinant can be represented as:
\[ \begin{vmatrix} a & b \\ c & d \end{vmatrix} \]
where \(a\), \(b\), \(c\), and \(d\) can each be either 0 or 1.

Since there are 4 entries and each entry has 2 possible values (0 or 1), the total number of different determinants is:
\[ 2^4 = 16 \]

### Step 2: Identify the conditions for positive and negative determinants.

The value of a 2x2 determinant is calculated as:
\[ \text{determinant} = ad - bc \]

- **Positive Determinants:** We need \(ad - bc > 0\).
- **Negative Determinants:** We need \(ad - bc < 0\).

### Step 3: Count the number of positive determinants.

We will analyze the combinations of \(a\), \(b\), \(c\), and \(d\) to find the positive determinants:

1. **Case 1:** \(a = 1\), \(d = 1\) (then \(ad = 1\)):
- \(b = 0\), \(c = 0\) → \(1 - 0 = 1\) (positive)
- \(b = 0\), \(c = 1\) → \(1 - 0 = 1\) (positive)
- \(b = 1\), \(c = 0\) → \(1 - 0 = 1\) (positive)
- \(b = 1\), \(c = 1\) → \(1 - 1 = 0\) (not positive)

Total positive determinants from this case: 3.

2. **Case 2:** \(a = 1\), \(d = 0\) (then \(ad = 0\)):
- All combinations yield a negative determinant or zero.

3. **Case 3:** \(a = 0\), \(d = 1\) (then \(ad = 0\)):
- All combinations yield a negative determinant or zero.

4. **Case 4:** \(a = 0\), \(d = 0\) (then \(ad = 0\)):
- All combinations yield a negative determinant or zero.

Thus, the total number of positive determinants is 3.

### Step 4: Count the number of negative determinants.

Following a similar analysis for negative determinants:

1. **Case 1:** \(a = 0\), \(d = 1\) (then \(ad = 0\)):
- \(b = 1\), \(c = 1\) → \(0 - 1 = -1\) (negative)
- \(b = 0\), \(c = 1\) → \(0 - 0 = 0\) (not negative)
- \(b = 1\), \(c = 0\) → \(0 - 0 = 0\) (not negative)

Total negative determinants from this case: 1.

2. **Case 2:** \(a = 1\), \(d = 0\) (then \(ad = 0\)):
- \(b = 1\), \(c = 1\) → \(0 - 1 = -1\) (negative)
- \(b = 0\), \(c = 1\) → \(0 - 0 = 0\) (not negative)
- \(b = 1\), \(c = 0\) → \(0 - 0 = 0\) (not negative)

Total negative determinants from this case: 1.

3. **Case 3:** \(a = 0\), \(d = 1\) (then \(ad = 0\)):
- \(b = 1\), \(c = 1\) → \(0 - 1 = -1\) (negative)
- \(b = 0\), \(c = 1\) → \(0 - 0 = 0\) (not negative)
- \(b = 1\), \(c = 0\) → \(0 - 0 = 0\) (not negative)

Total negative determinants from this case: 1.

4. **Case 4:** \(a = 0\), \(d = 0\) (then \(ad = 0\)):
- All combinations yield a negative determinant or zero.

Thus, the total number of negative determinants is also 3.

### Step 5: Calculate the probabilities.

Now we can calculate the probabilities:
- \(P(A)\) (positive determinants) = \(\frac{3}{16}\)
- \(P(B)\) (negative determinants) = \(\frac{3}{16}\)

### Conclusion

Since both probabilities are equal, we conclude:
\[ P(A) = P(B) \]

Thus, the correct answer is:
(c) \(P(A) = P(B)\)