= Solution
The nuclear <beta decay> is fast compared with the electronic orbital timescale. In the <sudden approximation>, the electronic <wavefunction> does not have time to change during the increase from $Z_i=1$ to $Z_f=2$. The subsequent <probability> of a particular daughter <energy eigenstate> is therefore its squared overlap with the initial state.
For the final $1s$ state, write $\beta_i=1/a_0$ and $\beta_f=2/a_0$. The <sudden hydrogenic charge change> gives the <amplitude>
$$
A_{1s}=\int\psi_f^*\psi_i\,d^3r=4(\beta_i\beta_f)^{3/2}\int_0^\infty r^2e^{-(\beta_i+\beta_f)r}\,dr=\frac{8(\beta_i\beta_f)^{3/2}}{(\beta_i+\beta_f)^3}=\frac{16\sqrt2}{27}.
$$
Thus
$$
\boxed{P_{1s}=|A_{1s}|^2=\frac{512}{729}\simeq0.702.}
$$
The final $2p$ states have <spherical harmonics> $Y_{1m}$, whereas the initial state is proportional to $Y_{00}$. Their angular overlap $\int Y_{1m}^*Y_{00}\,d\Omega$ vanishes by <orthogonality>. Accordingly $\boxed{P_{2p}=0}$ in the ideal spherically symmetric sudden-change model. Nuclear recoil or interactions with the emitted particles can produce small corrections. This is an overlap <selection rule> for a sudden change, not a photon-induced <electric dipole transition>.
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