To find the probability that the numbers drawn from the three boxes are in arithmetic progression, we can follow these steps:
### Step 1: Identify the total number of outcomes
- Box 1 has 3 cards: {1, 2, 3}
- Box 2 has 5 cards: {1, 2, 3, 4, 5}
- Box 3 has 7 cards: {1, 2, 3, 4, 5, 6, 7}
The total number of outcomes when drawing one card from each box is calculated by multiplying the number of cards in each box:
\[
\text{Total Outcomes} = 3 \times 5 \times 7 = 105
\]
### Step 2: Determine the conditions for arithmetic progression
For three numbers \(x_1\), \(x_2\), and \(x_3\) to be in arithmetic progression, they must satisfy the condition:
\[
2x_2 = x_1 + x_3
\]
This means that \(x_2\) must be the average of \(x_1\) and \(x_3\).
### Step 3: Find the favorable outcomes
We will check all possible combinations of \(x_1\), \(x_2\), and \(x_3\) to see which combinations satisfy the arithmetic progression condition.
1. **When \(x_1 = 1\)**:
- \(x_2 = 1\): \(x_3 = 1\) (Valid)
- \(x_2 = 2\): \(x_3 = 3\) (Valid)
- \(x_2 = 3\): \(x_3 = 5\) (Valid)
- \(x_2 = 4\): \(x_3 = 7\) (Valid)
- Total valid combinations: 4
2. **When \(x_1 = 2\)**:
- \(x_2 = 1\): \(x_3 = 1\) (Valid)
- \(x_2 = 2\): \(x_3 = 2\) (Valid)
- \(x_2 = 3\): \(x_3 = 4\) (Valid)
- Total valid combinations: 3
3. **When \(x_1 = 3\)**:
- \(x_2 = 1\): \(x_3 = 5\) (Valid)
- \(x_2 = 2\): \(x_3 = 7\) (Valid)
- \(x_2 = 3\): \(x_3 = 9\) (Invalid)
- Total valid combinations: 2
Adding all valid combinations gives us:
\[
\text{Total Favorable Outcomes} = 4 + 3 + 2 = 9
\]
### Step 4: Calculate the probability
The probability that the numbers drawn from the boxes are in arithmetic progression is given by the ratio of favorable outcomes to total outcomes:
\[
\text{Probability} = \frac{\text{Favorable Outcomes}}{\text{Total Outcomes}} = \frac{9}{105} = \frac{3}{35}
\]
### Final Answer
The probability that \(x_1, x_2, x_3\) are in an arithmetic progression is:
\[
\frac{3}{35}
\]
