To add to Tomas' answer, I will point out another reason the probability of the ground state is very high, which is that the probability depends on the square of the wavefunction.
Suppose a wavefunction is mostly $\mid 0 \rangle$ and a little bit $\mid 2 \rangle$ (these represent the ground state and second excited state respectively), so
$$\mid \Psi \rangle = \alpha \mid 0 \rangle + \epsilon \mid 2 \rangle$$
Then the probabilities to measure the energies $E_0$ and $E_2$ are
$$P(E_0) = \alpha^2$$
$$P(E_2) = \epsilon^2$$
If $\epsilon$ is small, then $\epsilon^2$ is much smaller. In your example, we have $\alpha^2 = .9985$, so $\epsilon^2 = .0015$ and $\epsilon = .04$. So the probability to be in the second excited state is very small, even though the contribution of the second excited state to the wavefunction itself is more considerable.
Note: there are actually more states involved than just these two, but I chose them alone for simplicity.