An urn contains four balls bearing numbers 1, 2, 3 and 123 respectively. A ball is drawn at random from the urn. Let `E_i`, i = 1, 2, 3 donote the event that digit i appears on the ball drawn.
Statement -1: `P(E_1 nn E_2) =P(E_1 nn E_3) = P (E_2 nn E_3)=1/4`
Statement-2: `P(E_1)=P(E_2)=P(E_3)=1/2`

Answer

To solve the problem, we need to analyze the events and their probabilities based on the balls in the urn.

### Step-by-Step Solution:

1. **Identify the Balls**: The urn contains four balls with the following numbers: 1, 2, 3, and 123.

2. **Define Events**:
- Let \( E_1 \) be the event that the digit 1 appears on the ball drawn.
- Let \( E_2 \) be the event that the digit 2 appears on the ball drawn.
- Let \( E_3 \) be the event that the digit 3 appears on the ball drawn.

3. **Total Outcomes**: There are 4 balls in total, so the total number of outcomes when drawing a ball is 4.

4. **Calculate \( P(E_1 \cap E_2) \)**:
- \( E_1 \cap E_2 \) means that both digits 1 and 2 appear on the drawn ball. The only ball that has both digits is the ball numbered 123.
- Therefore, the number of favorable outcomes for \( E_1 \cap E_2 \) is 1 (the ball numbered 123).
- Thus, \( P(E_1 \cap E_2) = \frac{1}{4} \).

5. **Calculate \( P(E_1 \cap E_3) \)**:
- \( E_1 \cap E_3 \) means that both digits 1 and 3 appear on the drawn ball. Again, the ball numbered 123 contains both digits.
- Thus, the number of favorable outcomes for \( E_1 \cap E_3 \) is also 1.
- Therefore, \( P(E_1 \cap E_3) = \frac{1}{4} \).

6. **Calculate \( P(E_2 \cap E_3) \)**:
- \( E_2 \cap E_3 \) means that both digits 2 and 3 appear on the drawn ball. The ball numbered 123 contains both digits.
- Thus, the number of favorable outcomes for \( E_2 \cap E_3 \) is also 1.
- Therefore, \( P(E_2 \cap E_3) = \frac{1}{4} \).

7. **Conclusion for Statement 1**:
- Since \( P(E_1 \cap E_2) = P(E_1 \cap E_3) = P(E_2 \cap E_3) = \frac{1}{4} \), Statement 1 is correct.

8. **Calculate \( P(E_1) \)**:
- The event \( E_1 \) occurs if the ball drawn is either 1 or 123 (which contains the digit 1).
- Therefore, the favorable outcomes for \( E_1 \) are 2 (the balls numbered 1 and 123).
- Thus, \( P(E_1) = \frac{2}{4} = \frac{1}{2} \).

9. **Calculate \( P(E_2) \)**:
- The event \( E_2 \) occurs if the ball drawn is either 2 or 123 (which contains the digit 2).
- Thus, \( P(E_2) = \frac{2}{4} = \frac{1}{2} \).

10. **Calculate \( P(E_3) \)**:
- The event \( E_3 \) occurs if the ball drawn is either 3 or 123 (which contains the digit 3).
- Thus, \( P(E_3) = \frac{2}{4} = \frac{1}{2} \).

11. **Conclusion for Statement 2**:
- Since \( P(E_1) = P(E_2) = P(E_3) = \frac{1}{2} \), Statement 2 is also correct.

### Final Answer:
Both statements are correct.