+4 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?
+281 Let's break down the problem:
Total Possible Outcomes:
We have numbers from 500 to 999, inclusive.
That's a total of 999 - 500 + 1 = 500 numbers.
Favorable Outcomes (Even Numbers with Distinct Digits):
Even Numbers: The last digit must be 0, 2, 4, 6, or 8. That's 5 choices.
Distinct Digits:
For the first digit (hundreds place), we have 5 choices (5, 6, 7, 8, or 9, as we can't use 0).
For the second digit (tens place), we have 8 choices left (since we've used one digit).
So, for each of the 5 choices for the last digit, we have 5 * 8 = 40 choices for the first two digits.
Therefore, the total number of favorable outcomes is 5 (last digit choices) * 40 (first two digit choices) = 200.
Probability:
Probability = Favorable Outcomes / Total Outcomes
Probability = 200 / 500
Probability = 2/5
So, the probability of choosing an even integer between 500 and 999 with distinct digits is 2/5.
