A and B each throw a die. Then it is 7:5 that A's throw is not greater than B's.

Answer

To solve the problem, we need to analyze the situation where A and B each throw a die. We are given that the ratio of the number of ways A's throw is not greater than B's throw is 7:5.

### Step-by-Step Solution:

1. **Understanding the Total Outcomes**:
Each die has 6 faces, so when A and B each throw a die, the total number of outcomes is:
\[ 6 \times 6 = 36 \]

2. **Identifying the Events**:
We need to find the number of outcomes where A's throw is not greater than B's. This includes the cases where A's throw is equal to B's throw and where A's throw is less than B's throw.

3. **Counting the Outcomes**:
- **A's throw equals B's throw**: The pairs are (1,1), (2,2), (3,3), (4,4), (5,5), (6,6). There are 6 outcomes where A's throw equals B's throw.
- **A's throw is less than B's throw**: We can list the pairs:
- If A rolls 1: B can roll 2, 3, 4, 5, 6 (5 outcomes)
- If A rolls 2: B can roll 3, 4, 5, 6 (4 outcomes)
- If A rolls 3: B can roll 4, 5, 6 (3 outcomes)
- If A rolls 4: B can roll 5, 6 (2 outcomes)
- If A rolls 5: B can roll 6 (1 outcome)
- If A rolls 6: B cannot roll anything greater (0 outcomes)

Adding these gives:
\[ 5 + 4 + 3 + 2 + 1 + 0 = 15 \]

4. **Total Outcomes Where A's Throw is Not Greater Than B's**:
The total number of outcomes where A's throw is not greater than B's is:
\[ 15 \text{ (A < B)} + 6 \text{ (A = B)} = 21 \]

5. **Calculating the Outcomes Where A's Throw is Greater Than B's**:
The outcomes where A's throw is greater than B's can be calculated as:
\[ 36 \text{ (total outcomes)} - 21 \text{ (A not greater than B)} = 15 \]

6. **Setting Up the Ratio**:
We have:
- Outcomes where A's throw is not greater than B's = 21
- Outcomes where A's throw is greater than B's = 15

The ratio of A's throw not greater than B's to A's throw greater than B's is:
\[ \frac{21}{15} = \frac{7}{5} \]

### Conclusion:
The given condition of the problem is satisfied, confirming that the calculations are correct.