$\begin{aligned} \mathrm{P}\left(\frac{1}{4} < x < \frac{1}{3}\right)= & \int_{\frac{1}{4}}^{\frac{1}{3}} \mathrm{f}(x) \mathrm{d} x=\int_{\frac{1}{4}}^{\frac{1}{3}} 3\left(1-2 x^2\right) \mathrm{d} x \\ & =\left[3 x-2 x^3\right]_{\frac{1}{4}}^{\frac{1}{3}} \\ & =\left(1-\frac{2}{27}\right)-\left(\frac{3}{4}-\frac{1}{32}\right) \\ & =\frac{1}{4}+\frac{1}{32}-\frac{2}{27} \\ & =\frac{179}{864}\end{aligned}$