If A and B are two independent events such that `P(A) gt (1)/(2), P(A nn B^(C ))=(3)/(25) and P(A^(C )nnB)=(8)/(25)`, then `P(A)` is equal to `("where, "A^("c") and B^("c")` represent the complement of events A and B respectively)

Options

A. `(1)/(5)`
B. `(3)/(5)`
C. `(3)/(4)`
D. `(4)/(5)`

Answer

To solve the problem, we need to find the probability \( P(A) \) given the conditions about events A and B.

### Step-by-Step Solution:

1. **Define the Variables**:
Let \( P(A) = x \) and \( P(B) = y \). We know that \( P(A) > \frac{1}{2} \).

2. **Use the Given Information**:
- From the problem, we have:
\[ P(A \cap B^C) = P(A) \cdot P(B^C) = x(1 - y) = \frac{3}{25} \]
- This can be rewritten as:
\[ x - xy = \frac{3}{25} \quad \text{(Equation 1)} \]

3. **Use the Other Given Information**:
- The second piece of information gives us:
\[ P(A^C \cap B) = P(A^C) \cdot P(B) = (1 - x)y = \frac{8}{25} \]
- This can be rewritten as:
\[ y - xy = \frac{8}{25} \quad \text{(Equation 2)} \]

4. **Set Up the Equations**:
- From Equation 1:
\[ x - xy = \frac{3}{25} \implies xy = x - \frac{3}{25} \quad \text{(Equation 3)} \]
- From Equation 2:
\[ y - xy = \frac{8}{25} \implies xy = y - \frac{8}{25} \quad \text{(Equation 4)} \]

5. **Equate the Two Expressions for \( xy \)**:
- From Equation 3 and Equation 4, we can set them equal:
\[ x - \frac{3}{25} = y - \frac{8}{25} \]
- Rearranging gives:
\[ x - y = -\frac{5}{25} \implies x - y = -\frac{1}{5} \implies y = x + \frac{1}{5} \quad \text{(Equation 5)} \]

6. **Substitute Equation 5 into Equation 1**:
- Substitute \( y \) from Equation 5 into Equation 1:
\[ x(1 - (x + \frac{1}{5})) = \frac{3}{25} \]
- This simplifies to:
\[ x(1 - x - \frac{1}{5}) = \frac{3}{25} \implies x(\frac{4}{5} - x) = \frac{3}{25} \]
- Multiplying through by 25 to eliminate the fraction:
\[ 25x(\frac{4}{5} - x) = 3 \implies 20x - 25x^2 = 3 \]
- Rearranging gives:
\[ 25x^2 - 20x + 3 = 0 \]

7. **Solve the Quadratic Equation**:
- Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 25 \cdot 3}}{2 \cdot 25} \]
\[ = \frac{20 \pm \sqrt{400 - 300}}{50} = \frac{20 \pm \sqrt{100}}{50} = \frac{20 \pm 10}{50} \]
- This gives us two potential solutions:
\[ x = \frac{30}{50} = \frac{3}{5} \quad \text{or} \quad x = \frac{10}{50} = \frac{1}{5} \]

8. **Select the Valid Solution**:
- Since \( P(A) > \frac{1}{2} \), we discard \( \frac{1}{5} \) and keep:
\[ P(A) = \frac{3}{5} \]

### Final Answer:
\[ P(A) = \frac{3}{5} \]