For any two independent events `E_1` and `E_2` `P{(E_1uuE_2)nn(bar(E_1)nnbar(E_2)}` is equal to

Options

A. `le 1 //4`
B. `ge 1//4`
C. `gt 1//2`
D. none of these

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. **Understanding the Events**:
- We have two independent events \( E_1 \) and \( E_2 \).
- The notation \( \overline{E_1} \) represents the complement of event \( E_1 \), and \( \overline{E_2} \) represents the complement of event \( E_2 \).

2. **Using the Formula for Probability**:
- We want to find \( P(E_1 \cup E_2 \cap \overline{E_1} \cap \overline{E_2}) \).
- By the properties of set operations, we can rewrite this as:
\[ P(E_1 \cup (E_2 \cap \overline{E_1} \cap \overline{E_2})) \]

3. **Simplifying the Intersection**:
- The term \( E_2 \cap \overline{E_1} \cap \overline{E_2} \) simplifies to just \( E_2 \cap \overline{E_1} \) because \( E_2 \cap \overline{E_2} \) is empty.
- Therefore, we can rewrite our expression as:
\[ P(E_1 \cup (E_2 \cap \overline{E_1})) \]

4. **Using the Independence of Events**:
- Since \( E_1 \) and \( E_2 \) are independent, we can find \( P(E_2 \cap \overline{E_1}) \) as:
\[ P(E_2 \cap \overline{E_1}) = P(E_2) \cdot P(\overline{E_1}) = P(E_2) \cdot (1 - P(E_1)) \]

5. **Final Calculation**:
- Now we can calculate \( P(E_1 \cup (E_2 \cap \overline{E_1})) \) using the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Here, let \( A = E_1 \) and \( B = E_2 \cap \overline{E_1} \).
- Thus, we have:
\[ P(E_1 \cup (E_2 \cap \overline{E_1})) = P(E_1) + P(E_2 \cap \overline{E_1}) - P(E_1 \cap (E_2 \cap \overline{E_1})) \]
- The term \( P(E_1 \cap (E_2 \cap \overline{E_1})) \) is zero because \( E_1 \) and \( \overline{E_1} \) cannot occur simultaneously.

6. **Final Expression**:
- Therefore, we can conclude:
\[ P(E_1 \cup (E_2 \cap \overline{E_1})) = P(E_1) + P(E_2) \cdot (1 - P(E_1)) \]

### Conclusion:
The final result is:
\[ P(E_1 \cup E_2 \cap \overline{E_1} \cap \overline{E_2}) = P(E_1) + P(E_2)(1 - P(E_1)) \]