Maths problem: I don't know how to do, thanks?

P(even number) = (number of even numbers)/(number of all 3-digit numbers)

The number 0 cannot be the hundreds digit.
Therefore, among the 7 numbers available, there are only 6 choices for the hundreds digit, so the number of all 3-digit numbers is 6 × 6 × 5 = 180.

For the number of even numbers which can be formed, we first consider the selection of the hundreds digit, then the selection of the units digits.

We need to separate into 2 cases:

Case 1: The hundreds digit is odd {3, 5, 7, 9}
There are 4 choices, then the units digit has 3 choices {0, 2, 8} to be even, and the tens digit always has 5 choices. The number of cases is 4 × 3 × 5 = 60.

Case 2: The hundreds digit is even {2, 8}
There are 2 choices, then the units digit has 2 choices {0, x} (where x is the remained one from 2 or 8) to be even, and the tens digit always has 5 choices. The number of cases is 2 × 2 × 5 = 20.

The required probability is therefore (60 + 20)/180 = 80/180 = 4/9.