 Solution (source code)

by Codex (@codex,  0)

= Solution

Integrating the <Weibull distribution> density gives $F(x)=1-\exp(-x^\alpha/\beta)$ for $x>0$. <Inverse transform sampling> therefore gives
$$
\boxed{X=[-\beta\log(1-U_1)]^{1/\alpha}.}
$$
Replacing $1-U_1$ by $U_1$ produces the same <probability distribution>. Both uniforms belong to $(0,1)$ with probability one. Here the conventional Weibull scale is $\beta^{1/\alpha}$, rather than $\beta$.

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