For any two independent events `E_(1) and E_(2)`, `P{E_(1) cup (E_(2)) cap (overline(E_(1)))cap(overline(E_(2)))}` is

Options

A. `le1//4`
B. `gt1//4`
C. `ge1//2`
D. none

Answer

To solve the problem, we need to find the probability of the event \( P(E_1 \cup E_2 \cap \overline{E_1} \cap \overline{E_2}) \) for two independent events \( E_1 \) and \( E_2 \).

### Step-by-Step Solution:

1. **Understand the Events**:
- \( E_1 \) and \( E_2 \) are independent events.
- \( \overline{E_1} \) and \( \overline{E_2} \) are the complements of \( E_1 \) and \( E_2 \), respectively.

2. **Apply the Definitions**:
- The expression \( E_1 \cup E_2 \) represents the event that either \( E_1 \) or \( E_2 \) occurs.
- The intersection \( \overline{E_1} \cap \overline{E_2} \) represents the event that neither \( E_1 \) nor \( E_2 \) occurs.

3. **Rewrite the Expression**:
- We can rewrite \( P(E_1 \cup E_2 \cap \overline{E_1} \cap \overline{E_2}) \) as:
\[ P((E_1 \cup E_2) \cap (\overline{E_1} \cap \overline{E_2})) \]

4. **Use De Morgan's Law**:
- According to De Morgan's laws, \( \overline{E_1} \cap \overline{E_2} = \overline{(E_1 \cup E_2)} \).
- Thus, we can rewrite the expression as:
\[ P((E_1 \cup E_2) \cap \overline{(E_1 \cup E_2)}) \]

5. **Analyze the Intersection**:
- The intersection of an event with its complement is always the empty set:
\[ (E_1 \cup E_2) \cap \overline{(E_1 \cup E_2)} = \emptyset \]

6. **Calculate the Probability**:
- The probability of the empty set is zero:
\[ P(\emptyset) = 0 \]

### Final Answer:
Thus, the final result is:
\[ P(E_1 \cup E_2 \cap \overline{E_1} \cap \overline{E_2}) = 0 \]