= Solution
{c}
The laboratory is the parent's <rest frame>. In this <relativistic two-body decay>, <conservation of momentum> gives equal and opposite daughter <momenta>, of magnitude $p$. Their total <relativistic energies> satisfy
$$
E_1+E_2=4mc^2,\qquad
E_1^2=p^2c^2+m^2c^4,\qquad
E_2^2=p^2c^2+4m^2c^4.
$$
Subtracting the <relativistic energy-momentum relations> eliminates $p$:
$$
(E_2-E_1)(E_2+E_1)=3m^2c^4,
\qquad E_2-E_1=\frac34mc^2.
$$
Solving the two <linear equations> gives
$$
E_1=\frac{13}{8}mc^2,\qquad E_2=\frac{19}{8}mc^2,\qquad
p=\frac{\sqrt{105}}8mc.
$$
Since <three-momentum> obeys $pc^2=Ev$, the daughter <speeds> are
$$
\boxed{v_1=\frac{\sqrt{105}}{13}c=0.7882c,\qquad
v_2=\frac{\sqrt{105}}{19}c=0.5393c.}
$$
Their directions are opposite; the initial state does not specify their common axis. The <kinetic energies> are $K_1=5mc^2/8$ and $K_2=3mc^2/8$. Their sum $mc^2$ is the loss of <rest energy> when the total daughter <rest mass> is $3m$ instead of $4m$.
The <proper mean lifetime> $\tau_0$ is the <mean lifetime> measured in the unstable particle's own <rest frame>, or equivalently the mean elapsed <proper time> before decay. In the <exponential decay> model, its survival <probability> to <proper time> $\tau$ is $e^{-\tau/\tau_0}$. For the daughter of <rest mass> $m$, the <Lorentz factor> is $\gamma_1=E_1/(mc^2)=13/8$, so <time dilation> gives a laboratory <mean lifetime> of $\gamma_1\tau_0=3.25\,\mu\mathrm s$.
At constant <speed> $v_1$, travelling a <distance> $L$ requires laboratory time $L/v_1$ and <proper time> $L/(\gamma_1v_1)$. Its <mean decay length> is
$$
\ell_d=\gamma_1v_1\tau_0
=\frac{\sqrt{105}}8c\tau_0
=768.0\,\mathrm m
$$
for $\tau_0=2\times10^{-6}\,\mathrm s$. Hence
$$
\boxed{\mathbb P(\text{distance before decay}>1000\,\mathrm m)
=\exp\!\left[-\frac{1000\,\mathrm m}{\ell_d}\right]
=0.272\quad\text{approximately}.}
$$
The <mean lifetime> sets the scale of an <exponential distribution>; it is not a fixed deadline, so travelling beyond one <mean decay length> has a nonzero <probability>.
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