= Solution
The <Lorentz factor> is $\gamma=E/(m_\mu c^2)=20000/106=188.6792$ and $\beta=\sqrt{1-\gamma^{-2}}$. Traversal of the height $H$ takes <proper time> $H/(\beta\gamma c)$. Using the <exponential decay> law with the given <muon lifetime>,
$$
\boxed{P=\exp\left[-\frac{H}{c\tau_0\sqrt{\gamma^2-1}}\right]\simeq0.8180.}
$$
The laboratory flight lasts about $83.39\,\mu\mathrm s$, while the elapsed <proper time> is about $0.4420\,\mu\mathrm s$. This is the probability of surviving decay to the surface, with atmospheric <energy> loss and detector efficiency neglected.
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