Start with numbers ending in 5.
The prime restriction is already satisfied, so we don't worry about that.
Then, there are 8 choices for the hundreds digit (anything but 0 or 5) and 8 for the tens digit (anything but the hundreds digit and 5), which makes for \(8 \times 8 = 64\)
Now, if the final digit is 0, there are 9 choices for the hundreds (anything but 0), and 8 for the tens (anything but the hundreds and 0), which makes for \(9 \times 8 = 72\)
But, we need to account for numbers that have no primes. The digits that aren't prime are (0, 1, 4, 6, 8, and 9).
There are 5 choices for the hundreds and 4 for the tens, which makes for \(5 \times 4 = 20\) numbers that don't work.
So, there are \(64 + 72 - 20 = \color{brown}\boxed{116}\) numbers that work.