The sum of numbers on two throws of a dice is 8. Find the probability of getting 3 in at least one throw.

Explanation

To solve the problem, we need to find the probability of getting a 3 in at least one throw of a die, given that the sum of the numbers on two throws of the die is 8. We will denote the events as follows:

- Let \( A \) be the event that the sum of the numbers on the dice is 8.
- Let \( B \) be the event that we get a 3 in at least one throw.

### Step 1: Identify the sample space
When throwing two dice, the total number of outcomes is \( 6 \times 6 = 36 \). The sample space consists of all possible pairs of outcomes from the two throws.

### Step 2: Determine the event \( A \)
We need to find all pairs of numbers that sum to 8. The pairs are:
- \( (2, 6) \)
- \( (3, 5) \)
- \( (4, 4) \)
- \( (5, 3) \)
- \( (6, 2) \)

Thus, the event \( A \) consists of the following outcomes:
\[ A = \{(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)\} \]
The number of outcomes in \( A \) is 5.

### Step 3: Determine the event \( B \)
Next, we find the pairs that contain at least one 3:
- \( (3, 1) \)
- \( (3, 2) \)
- \( (3, 3) \)
- \( (3, 4) \)
- \( (3, 5) \)
- \( (3, 6) \)
- \( (1, 3) \)
- \( (2, 3) \)
- \( (4, 3) \)
- \( (5, 3) \)
- \( (6, 3) \)

Thus, the event \( B \) consists of the following outcomes:
\[ B = \{(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1, 3), (2, 3), (4, 3), (5, 3), (6, 3)\} \]
The number of outcomes in \( B \) is 11.

### Step 4: Determine the intersection \( A \cap B \)
Now, we find the outcomes that are common to both events \( A \) and \( B \):
- From \( A \), we have \( (3, 5) \) and \( (5, 3) \).

Thus, the intersection \( A \cap B \) consists of:
\[ A \cap B = \{(3, 5), (5, 3)\} \]
The number of outcomes in \( A \cap B \) is 2.

### Step 5: Calculate the probabilities
1. **Probability of \( A \)**:
\[ P(A) = \frac{\text{Number of outcomes in } A}{\text{Total outcomes}} = \frac{5}{36} \]

2. **Probability of \( A \cap B \)**:
\[ P(A \cap B) = \frac{\text{Number of outcomes in } A \cap B}{\text{Total outcomes}} = \frac{2}{36} = \frac{1}{18} \]

3. **Conditional Probability \( P(B|A) \)**:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{\frac{1}{18}}{\frac{5}{36}} = \frac{1}{18} \times \frac{36}{5} = \frac{2}{5} \]

### Final Answer
The probability of getting a 3 in at least one throw given that the sum of the numbers on two throws of a die is 8 is:
\[ \boxed{\frac{2}{5}} \]