Consider all six-digit numbers that can be created from the digits 0-9
 where the first and last digits must be odd and no digit can repeat. What is the probability of choosing a random number that starts with 1
 from this group? Enter a fraction or round your answer to 4
 decimal places, if necessary.

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.

Consider all six-digit numbers that can be created from the digits 0-9 where the first and last digits must be odd and no digit can repeat. What is the probability of choosing a random number that starts with 1 from this group? Enter a fraction or round your answer to 4 decimal places, if necessary.

Explore how to calculate the probability of choosing a six-digit number, where the first and last digits are odd, and no digits repeat. Learn the step-by-step process to find the probability that the first digit is 1. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation