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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?

 Oct 27, 2024
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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.

 Oct 27, 2024

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