Where the smallest falls

Idea. The survival function of the minimum is easy: all three numbers must be above the level. The probability of an interval is the difference of two survival values.

Step 1. One draw is above $t$ with probability $\frac{3-t}{3}$, so $$\mathbb{P}(\min >t)={\left(\frac{3-t}{3}\right)}^3.$$

Step 2. $$\begin{array}{rl}\mathbb{P}(\min >0.5)& ={\left(\frac{5}{6}\right)}^3=\frac{125}{216},\\ \mathbb{P}(\min >1)& ={\left(\frac{2}{3}\right)}^3=\frac{64}{216}.\end{array}$$

Step 3. $$\frac{125-64}{216}=\frac{61}{216}\approx \mathrm{0.282.}$$