= Solution
Integrate over the region $0\leq s<t$:
$$
P(T>S)=\int_0^\infty\frac{e^{-s/\bar s}}{\bar s}
\left[\int_s^\infty\frac{e^{-t/\bar t}}{\bar t}dt\right]ds
=\frac1{\bar s}\int_0^\infty e^{-s(1/\bar s+1/\bar t)}ds.
$$
Therefore
$$
\boxed{P(\text{no interruption})=\frac{\bar t}{\bar s+\bar t}.}
$$
This is also the two-clock case of <competing exponential clocks>. A larger <mean> time to noise increases this <probability>, while a larger <mean> service <length> decreases it.
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