= Solution
An idealized underlying <probability space> is
$$
\Omega=(0,60)\times(1,20)\times\mathbb R^{25},\qquad
\mathbb P=\operatorname{Unif}(0,60)\otimes\operatorname{Unif}(1,20)\otimes N(0,1)^{\otimes25},
$$
with the product <Borel sigma-algebra>. Its 27 independent random coordinates are $\mu$, $\sigma$, twenty past standard-normal noises, and five future standard-normal noises. Define $T_i(\omega)=\mu+\sigma\varepsilon_i$. Sampling an unknown <variance> does not mean drawing a new <variance> on each day; a single draw of $\sigma^2$ applies to all durations in a trace.
Let $A$ be the <event> that the first twenty durations satisfy their corresponding observed inequalities. The observations are the twenty late/early labels, conditional on the fixed departure thresholds. They are not exact values of the latent $T_i$. The posterior <probability measure> on traces is
$$
\mathbb P_A(B)=\mathbb P(B\mid A)=\frac{\mathbb P(B\cap A)}{\mathbb P(A)}.
$$
Here $\mathbb P(A)>0$: at every interior parameter pair each strict normal-tail <probability> lies in $(0,1)$, so their finite product has positive prior integral.
The returned outcome is
$$
Y(\omega)=\mathbf1\{T_{21}>30,\ldots,T_{25}>30\}\in\{0,1\}.
$$
Thus \b[the underlying <sample space> has dimension 27, the successful output space is $\{0,1\}$, and the observations define the conditioning event $A$]. The desired output <probability distribution> is the <pushforward measure> of $\mathbb P_A$ under $Y$. Rejection is an implementation device, not a third scientific answer: `None` is discarded before forming the conditional output law.
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