If `A` and `B` are two events such that `P(A)=0.3`, `P(B)=0.25`, `P(AnnB)=0.2`, then `P(((A^(C ))/(B^(C )))^(C ))` is equal to

Options

A. `(2)/(15)`
B. `(11)/(15)`
C. `(13)/(15)`
D. `(14)/(15)`

Explanation

To solve the problem, we need to find \( P\left(\left(\frac{A^C}{B^C}\right)^C\right) \).

### Step-by-step Solution:

1. **Understanding the Expression**:
We start with the expression \( P\left(\left(\frac{A^C}{B^C}\right)^C\right) \). This can be rewritten using the complement rule:
\[ P\left(\left(\frac{A^C}{B^C}\right)^C\right) = 1 - P\left(\frac{A^C}{B^C}\right) \]

2. **Finding \( P\left(\frac{A^C}{B^C}\right) \)**:
The probability \( P\left(\frac{A^C}{B^C}\right) \) can be expressed as:
\[ P\left(\frac{A^C}{B^C}\right) = \frac{P(A^C \cap B^C)}{P(B^C)} \]

3. **Calculating \( P(A^C) \) and \( P(B^C) \)**:
We know:
\[ P(A^C) = 1 - P(A) = 1 - 0.3 = 0.7 \]
\[ P(B^C) = 1 - P(B) = 1 - 0.25 = 0.75 \]

4. **Finding \( P(A^C \cap B^C) \)**:
Using the formula for the union of two events:
\[ P(A^C \cap B^C) = 1 - P(A \cup B) \]
We can find \( P(A \cup B) \) using:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting the known values:
\[ P(A \cup B) = 0.3 + 0.25 - 0.2 = 0.35 \]
Therefore:
\[ P(A^C \cap B^C) = 1 - 0.35 = 0.65 \]

5. **Calculating \( P\left(\frac{A^C}{B^C}\right) \)**:
Now we can substitute back into our equation:
\[ P\left(\frac{A^C}{B^C}\right) = \frac{P(A^C \cap B^C)}{P(B^C)} = \frac{0.65}{0.75} \]

6. **Simplifying the Fraction**:
Simplifying \( \frac{0.65}{0.75} \):
\[ \frac{0.65}{0.75} = \frac{65}{75} = \frac{13}{15} \]

7. **Final Calculation**:
Now substituting back into our first equation:
\[ P\left(\left(\frac{A^C}{B^C}\right)^C\right) = 1 - P\left(\frac{A^C}{B^C}\right) = 1 - \frac{13}{15} = \frac{2}{15} \]

### Conclusion:
Thus, the final answer is:
\[ \boxed{\frac{2}{15}} \]