Given that the two number appearing on throwing two dice are different. Find the probability of the event the sum of numbers on the dice is 4.

Explanation

To solve the problem, we need to find the probability of the event that the sum of the numbers on two dice is 4, given that the two numbers appearing on the dice are different.

### Step-by-Step Solution:

1. **Define the Events**:
- Let \( E \) be the event that the sum of the numbers on the dice is 4.
- Let \( F \) be the event that the numbers on the two dice are different.

2. **Identify the Sample Space**:
- The total number of outcomes when throwing two dice is \( 6 \times 6 = 36 \).

3. **Find the Outcomes for Event \( E \)**:
- The pairs of numbers that give a sum of 4 are:
- (1, 3)
- (2, 2)
- (3, 1)
- Thus, the outcomes for event \( E \) are \( \{(1, 3), (2, 2), (3, 1)\} \).
- However, since we are given that the numbers are different, we exclude (2, 2).
- Therefore, the valid outcomes for \( E \) given \( F \) are \( \{(1, 3), (3, 1)\} \).

4. **Count the Outcomes for Event \( E \cap F \)**:
- The outcomes that satisfy both \( E \) and \( F \) are \( (1, 3) \) and \( (3, 1) \).
- Thus, \( E \cap F = \{(1, 3), (3, 1)\} \) which has 2 outcomes.

5. **Count the Outcomes for Event \( F \)**:
- The outcomes where the numbers on the two dice are different can be calculated by excluding the outcomes where the numbers are the same (i.e., (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)).
- There are 6 outcomes where the numbers are the same, so the number of outcomes where the numbers are different is \( 36 - 6 = 30 \).

6. **Calculate the Probabilities**:
- The probability of \( E \cap F \) is:
\[ P(E \cap F) = \frac{\text{Number of outcomes in } E \cap F}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18} \]
- The probability of \( F \) is:
\[ P(F) = \frac{\text{Number of outcomes in } F}{\text{Total outcomes}} = \frac{30}{36} = \frac{5}{6} \]

7. **Use Conditional Probability Formula**:
- The conditional probability \( P(E|F) \) is given by:
\[ P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{\frac{1}{18}}{\frac{5}{6}} = \frac{1}{18} \times \frac{6}{5} = \frac{1 \times 6}{18 \times 5} = \frac{6}{90} = \frac{1}{15} \]

### Final Answer:
Thus, the probability that the sum of the numbers on the dice is 4, given that the two numbers are different, is \( \frac{1}{15} \).