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The digits 1, 2, 3, 4, and 5 are randomly aranged to form 3-digit number. (Digits are not repeated). Find the probability that the number is even and greater than 500.

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Answer to a math question The digits 1, 2, 3, 4, and 5 are randomly aranged to form 3-digit number. (Digits are not repeated). Find the probability that the number is even and greater than 500.

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Sigrid
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To find the total number of ways to arrange the digits 1, 2, 3, 4, 5 to form a 3-digit number without repetition, we use the permutation formula:

Total number of ways = P(n,r) = \frac{n!}{(n-r)!}
Total number of ways = P(5,3) = \frac{5!}{(5-3)!} = \frac{5 \cdot 4 \cdot 3}{2 \cdot 1} = 60

Now, let's find the number of ways to form a 3-digit number that is both even and greater than 500.

For the number to be even, the units digit must be 2 or 4.
For the number to be greater than 500, the hundreds digit must be 5. Therefore, the hundreds digit is fixed and can only be 5.

So, the hundreds digit is 5, there are 2 choices for the units digit (2 or 4), and the tens digit can be any of the remaining 3 digits.

Number of even 3-digit numbers greater than 500 = 1 (for hundreds digit 5) * 2 (for units digit 2 or 4) * 3 (for remaining tens digit) = 6

Finally, the probability of forming an even 3-digit number greater than 500 is given by:
Probability = \frac{\text{Number of successful outcomes}}{\text{Total number of outcomes}}
Probability = \frac{6}{60} = \frac{1}{10}

\boxed{\frac{1}{10}}

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