Question

the digits 5, 6, 7, 8, and 9 are randomly arranged to form a three - digit number. (digits are not repeated.) find the probability that the number is even and greater than 900. (type an integer or a simplified fraction.)

Explanation:

Step1: Calculate total number of three - digit numbers

The number of ways to arrange 5 non - repeated digits (5, 6, 7, 8, 9) to form a three - digit number is given by the permutation formula $P(n,r)=\frac{n!}{(n - r)!}$, where $n = 5$ and $r=3$. So $P(5,3)=\frac{5!}{(5 - 3)!}=\frac{5!}{2!}=5\times4\times3 = 60$.

Step2: Determine the first - digit condition for numbers greater than 900

For a three - digit number to be greater than 900, the first digit must be 9.

Step3: Determine the last - digit condition for even numbers

For a number to be even, the last digit must be 6 or 8.
When the first digit is 9 (1 way to choose the first digit), and the last digit is 6 or 8 (2 ways to choose the last digit), then for the middle digit, there are 3 remaining digits to choose from.
The number of three - digit numbers that are even and greater than 900 is $1\times3\times2=6$.

Step4: Calculate the probability

The probability $P$ is the number of favorable outcomes divided by the number of total outcomes. So $P=\frac{6}{60}=\frac{1}{10}$.

Answer:

$\frac{1}{10}$