We are tasked with determining the probability of selecting a six-digit number, where the first and last digits must be odd, and no digit can repeat. We also want to find the probability that the first digit is 1.

Step 1: Total possible numbers with restrictions

The first and last digits must be odd, and the digits must not repeat. The odd digits are $\{1, 3, 5, 7, 9\}$, giving us 5 possible choices for both the first and last digits.

Total number of valid numbers:

1. First digit: The first digit can be any odd number (1, 3, 5, 7, 9), so there are 5 choices.
2. Last digit: The last digit must also be odd, and it cannot be the same as the first digit, so there are 4 remaining choices for the last digit.
3. Middle digits: The remaining four digits can be chosen from the remaining 8 digits (since two digits are already used, the first and last digits).
   • For the second digit, we can choose any of the remaining 8 digits.
   • For the third digit, we can choose any of the remaining 7 digits.
   • For the fourth digit, we can choose any of the remaining 6 digits.
   • For the fifth digit, we can choose any of the remaining 5 digits.

Thus, the total number of valid six-digit numbers is: $5 \times 4 \times 8 \times 7 \times 6 \times 5 = 33600$

Step 2: Total numbers where the first digit is 1

Now, we find how many of these numbers start with the digit 1.

1. First digit: This is fixed as 1, so there is only 1 choice.
2. Last digit: The last digit must be odd and cannot be 1, so there are 4 choices (3, 5, 7, 9).
3. Middle digits: The remaining four digits are chosen from the remaining 8 digits (since 1 and the last digit are used).
   • For the second digit, we can choose any of the remaining 8 digits.
   • For the third digit, we can choose any of the remaining 7 digits.
   • For the fourth digit, we can choose any of the remaining 6 digits.
   • For the fifth digit, we can choose any of the remaining 5 digits.

Thus, the total number of valid six-digit numbers where the first digit is 1 is: $1 \times 4 \times 8 \times 7 \times 6 \times 5 = 6720$

Step 3: Probability calculation

The probability of selecting a number where the first digit is 1 is the ratio of favorable outcomes to total outcomes: $\text{Probability} = \frac{6720}{33600} = \frac{1}{5} = 0.2$

Thus, the probability is $\boxed{0.2}$.

Do you want more details or have any questions?

Here are 5 related questions for further exploration:

1. What if the first digit had to be even instead of odd?
2. How would the probability change if the number had to be 7 digits long?
3. What is the probability that the last digit is 9 instead of any odd digit?
4. How would you approach this problem if digits could repeat?
5. What is the probability of choosing a number that starts with 1 and ends with 9?

Tip: When calculating probabilities for counting problems, focus on calculating the total number of outcomes first and then focus on the favorable outcomes.