= Solution
For any chosen proper <prior density> $\rho(\mu,\sigma)$ on $\mathbb R\times(0,\infty)$ with positive evidence, <Bayes' theorem> gives
$$
\boxed{\mathbb P(Y=1\mid A)=\frac{\displaystyle\int_0^\infty\int_{-\infty}^{\infty}\rho(\mu,\sigma)L(\mu,\sigma)\Phi\left(\frac{\mu-30}{\sigma}\right)^5\,d\mu\,d\sigma}{\displaystyle\int_0^\infty\int_{-\infty}^{\infty}\rho(\mu,\sigma)L(\mu,\sigma)\,d\mu\,d\sigma}.}
$$
The <prior density> comes from the first two random draws. The <likelihood> $L$ is the <probability> of satisfying all observed-label constraints after integrating the past duration noises. The factor $\Phi((\mu-30)/\sigma)^5$ is the <probability> of the five-day outcome conditional on the shared parameters. The denominator is the <probability> of acceptance, which converts the numerator's joint <probability> into the required <conditional probability>.
For the independent uniform priors in part (a), the constant $\rho=1/(60\cdot19)$ cancels and both integrals reduce to $0<\mu<60$, $1<\sigma<20$. A prior on the <variance> $v=\sigma^2$ would require transforming the density: this chosen prior has density $1/(38\sqrt v)$ for $1<v<400$ conditional on $\mu$. One must not silently replace it by a uniform prior on $v$.
There is no unique numerical <probability> until the <prior distribution> is chosen. As a check for the stated uniform priors, independent <Gaussian quadrature> calculations with 256 and 512 nodes per coordinate agree at the shown precision:
$$
\mathbb P(Y=1\mid A)\approx0.15038909,\qquad \mathbb E[\pi\mid A]\approx0.61265126.
$$
The prior-predictive <probability> of all observed labels is approximately $0.00030422536$, explaining the rejection cost. These numbers are illustrations of the explicit model, not additional observed data.
A subtle point is that the five future days are independent only conditional on $(\mu,\sigma)$. Integrating shared parameter uncertainty gives $\mathbb E[\pi^5\mid A]$, \b[not] $(\mathbb E[\pi\mid A])^5$. The latter is about $0.08631110$ for these priors and understates the five-day <probability>. <Jensen's inequality> gives $(\mathbb E[\pi\mid A])^5\leq\mathbb E[\pi^5\mid A]$, while $0\leq\pi\leq1$ gives $\mathbb E[\pi^5\mid A]\leq\mathbb E[\pi\mid A]$. For two future late-arrival <indicator functions>, their posterior <covariance> is $\operatorname{Var}(\pi\mid A)\geq0$. This is the <posterior predictive repeated-success identity>.
\Image[/past-exam-of-the-msc-in-advanced-computer-science-course-of-the-university-of-oxford/2021/bayesian-statistical-probabilistic-programming-censored-journeys.png]
{title=Posterior uncertainty and the five-day predictive probability for the explicitly chosen uniform priors. The contour is cropped to the parameter region carrying the visible mass; integration uses the full prior rectangle.}
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