I managed to solve it by plotting a graph and then calculating the needed area.
To find the area I solved the following integral:
$\int_{0}^{15} \int_{0}^{x} \frac{1}{225}dy \space dx - \int_{5}^{15} \int_{0}^{x-5} \frac{1}{225}dy \space dx $
Which gives an answer of $\frac{5}{18}$