Let two fair six-faced dice A and B be thrown simultaneously. If `E_1` is the event that die A shows up four, `E_2` is the event that die B shows up two and `E_3` is the event that the sum of numbers on both dice is odd, then which of the following statements is NOT true ? (1) `E_1` and `E_2` are independent. (2) `E_2` and `E_3` are independent. (3) `E_1` and `E_3` are independent. (4) `E_1` , `E_2` and `E_3` are independent.

Options

A. `E_(2)and E_(3)` are independent
B. `E_(1)and E_(3)` are independent
C. `E_(1)and E_(2) and E_(3)` are independent
D. `E_(1)and E_(2)` are independent

Explanation

To solve the problem, we need to analyze the events defined in the question and check the independence of the events \(E_1\), \(E_2\), and \(E_3\).

### Step 1: Define the Events
- **Event \(E_1\)**: Die A shows a 4.
- **Event \(E_2\)**: Die B shows a 2.
- **Event \(E_3\)**: The sum of the numbers on both dice is odd.

### Step 2: Calculate the Probabilities
1. **Total Outcomes**: When two dice are thrown, the total number of outcomes is \(6 \times 6 = 36\).

2. **Probability of \(E_1\)**:
- The outcomes where die A shows 4 are: (4,1), (4,2), (4,3), (4,4), (4,5), (4,6).
- There are 6 outcomes.
- \(P(E_1) = \frac{6}{36} = \frac{1}{6}\).

3. **Probability of \(E_2\)**:
- The outcomes where die B shows 2 are: (1,2), (2,2), (3,2), (4,2), (5,2), (6,2).
- There are 6 outcomes.
- \(P(E_2) = \frac{6}{36} = \frac{1}{6}\).

4. **Probability of \(E_3\)**:
- The outcomes where the sum is odd are: (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5).
- There are 18 outcomes.
- \(P(E_3) = \frac{18}{36} = \frac{1}{2}\).

### Step 3: Check Independence of Events
Two events \(A\) and \(B\) are independent if \(P(A \cap B) = P(A) \cdot P(B)\).

1. **Check Independence of \(E_1\) and \(E_2\)**:
- \(E_1 \cap E_2\): The only outcome that satisfies both \(E_1\) and \(E_2\) is (4,2).
- \(P(E_1 \cap E_2) = \frac{1}{36}\).
- Check independence:
\[ P(E_1) \cdot P(E_2) = \left(\frac{1}{6}\right) \cdot \left(\frac{1}{6}\right) = \frac{1}{36}. \]
- Since \(P(E_1 \cap E_2) = P(E_1) \cdot P(E_2)\), \(E_1\) and \(E_2\) are independent.

2. **Check Independence of \(E_2\) and \(E_3\)**:
- \(E_2 \cap E_3\): The outcomes that satisfy both \(E_2\) and \(E_3\) are (1,2), (3,2), (5,2).
- There are 3 outcomes: (1,2), (3,2), (5,2).
- \(P(E_2 \cap E_3) = \frac{3}{36} = \frac{1}{12}\).
- Check independence:
\[ P(E_2) \cdot P(E_3) = \left(\frac{1}{6}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{12}. \]
- Since \(P(E_2 \cap E_3) = P(E_2) \cdot P(E_3)\), \(E_2\) and \(E_3\) are independent.

3. **Check Independence of \(E_1\) and \(E_3\)**:
- \(E_1 \cap E_3\): The outcome that satisfies both \(E_1\) and \(E_3\) is (4,1) and (4,3), which gives us two outcomes.
- There are 2 outcomes: (4,1), (4,3).
- \(P(E_1 \cap E_3) = \frac{2}{36} = \frac{1}{18}\).
- Check independence:
\[ P(E_1) \cdot P(E_3) = \left(\frac{1}{6}\right) \cdot \left(\frac{1}{2}\right) = \frac{1}{12}. \]
- Since \(P(E_1 \cap E_3) \neq P(E_1) \cdot P(E_3)\), \(E_1\) and \(E_3\) are not independent.

### Step 4: Conclusion
From the analysis, we find that:
- \(E_1\) and \(E_2\) are independent.
- \(E_2\) and \(E_3\) are independent.
- \(E_1\) and \(E_3\) are not independent.

Thus, the statement that is NOT true is:
**(3) \(E_1\) and \(E_3\) are independent.**