an integer between 500 and 999, inclusive, is chosen at random. what is the probability that it is an even integer whose digits are all different?

Answer 1

To solve the problem, we need to calculate the probability of selecting an even integer with all different digits from the range of integers between 500 and 999.

1. Identify the total number of integers between 500 and 999: The range is inclusive of both endpoints. The total count of integers from 500 to 999 can be calculated as follows: \[ 999 - 500 + 1 = 500 \]

2. Determine the criteria for even integers between 500 and 999: An integer is even if its last digit (the ones place) is one of 0, 2, 4, 6, or 8.

3. Count valid even integers with all different digits: The even integers between 500 and 999 will have their first digit between 5 and 9 (i.e., 5, 6, 7, 8, 9).

   Let's analyze each case based on the first digit:

   • First digit is 5:

     • Remaining choices for second digit: 0, 1, 2, 3, 4, 6, 7, 8, 9 (9 choices)
     • Remaining choices for last digit (even): choose among remaining even digits, making sure it's different from the first and second: 0, 2, 4, 6, 8 (4 choices depending on selected second digit).
     • Total combinations: \( 9 \text{ (second digit)} \times 4 \text{ (last digit)} = 36 \).

   • First digit is 6:

     • Remaining choices for second digit: 0, 1, 2, 3, 4, 5, 7, 8, 9 (9 choices)
     • Remaining choices for last digit (even): same selection rules as above apply.
     • Total combinations: \( 9 \times 4 = 36 \).

   • First digit is 7:

     • Remaining choices for second digit: 0, 1, 2, 3, 4, 5, 6, 8, 9 (9 choices)
     • Remaining choices for last digit (even): must pick even number and avoid duplicates.
     • Total combinations: \( 9 \times 4 = 36 \).

   • First digit is 8:

     • Remaining choices for second digit: 0, 1, 2, 3, 4, 5, 6, 7, 9 (9 choices)
     • Total combinations: \( 9 \times 4 = 36 \).

   • First digit is 9:

     • Remaining choices for second digit: 0, 1, 2, 3, 4, 5, 6, 7, 8 (8 choices since 9 is used)
     • Last digit can only be 0, 2, 4, 6, 8.
     • Total combinations: \( 8 \times 5 = 40 \).

4. Total Count of Valid Integers: \[ \text{Total valid integers} = 36 + 36 + 36 + 36 + 40 = 184 \]

5. Compute the Probability: We finally compute the probability of selecting a valid integer: \[ P(\text{even integer with distinct digits}) = \frac{\text{Number of valid outcomes}}{\text{Total outcomes}} = \frac{184}{500} = \frac{92}{250} = \frac{46}{125} \]

Thus, the probability that a randomly chosen integer between 500 and 999 (inclusive) is an even integer with all different digits is \( \frac{46}{125} \).