To solve the problem, we need to calculate the probability that a randomly chosen five-digit number has all distinct digits, with odd digits in the odd places and even digits in the even places.
### Step-by-Step Solution:
1. **Determine Total Cases**:
- A five-digit number cannot start with 0. Therefore, the first digit can be any digit from 1 to 9 (9 options).
- The remaining four digits can be any digit from 0 to 9, allowing repetition. Thus, the total number of five-digit numbers is:
\[
\text{Total Cases} = 9 \times 10^4
\]
2. **Determine Favorable Cases**:
- **Odd and Even Digits**: There are 5 odd digits (1, 3, 5, 7, 9) and 5 even digits (0, 2, 4, 6, 8).
- **Positions**: In a five-digit number, the odd positions are 1st, 3rd, and 5th, while the even positions are 2nd and 4th.
- **Choosing Odd Digits**: We need to choose 3 odd digits from the 5 available. The number of ways to choose and arrange 3 odd digits is given by:
\[
\text{Ways to choose odd digits} = P(5, 3) = 5 \times 4 \times 3 = 60
\]
- **Choosing Even Digits**: We need to choose 2 even digits from the 5 available. The number of ways to choose and arrange 2 even digits is given by:
\[
\text{Ways to choose even digits} = P(5, 2) = 5 \times 4 = 20
\]
- **Total Favorable Cases**: The total number of favorable cases is the product of the arrangements of odd and even digits:
\[
\text{Favorable Cases} = P(5, 3) \times P(5, 2) = 60 \times 20 = 1200
\]
3. **Calculate Probability**:
- The probability is given by the ratio of favorable cases to total cases:
\[
\text{Probability} = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{1200}{9 \times 10^4}
\]
- Simplifying this gives:
\[
\text{Probability} = \frac{1200}{90000} = \frac{12}{900} = \frac{1}{75}
\]
### Final Answer:
The probability that all the digits are distinct and that the digits at odd places are odd and the digits at even places are even is \(\frac{1}{75}\).
