= Solution
Only $n=1,2$ have energies below $6\epsilon$. Squaring the corresponding expansion coefficients gives $0.2^2+0.3^2=0.13$.
The printed rounded coefficients have norm
$$
S=0.2^2+0.3^2+0.4^2+0.843^2=1.000649.
$$
Thus the expected rounded answer is \b[$P(E<6\epsilon)\simeq0.13$]. If the decimal coefficients are treated as literal exact inputs and the state normalized, the exact <probability> for that input is \b[$0.13/S\simeq0.129916$]. This distinction avoids treating rounding error as a physical departure from <wavefunction normalization>.
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