How to find the possible measurements of a quantum system? A wave function of an infinite square well is given as 
$$ \psi(x) = \frac{1}{\sqrt{5a}}\sin\left(\frac{\pi x}{a}\right) +\frac{2}{\sqrt{5a}}\sin\left(\frac{3\pi x}{a}\right),\quad x\in[0,a] $$
How do I find the possible results of the measurement of the system's energy and the corresponding probabilities?
 A: In order to find the energies and corresponding probabilities, it's best to rewrite the expression in terms of eigenfunctions. The eigenfunctions of the infinite square well between $x=0$ and $x=a$ are 
$$ \phi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right). $$
Using these, your wave function is 
$$ \psi(x) = \frac{1}{\sqrt{10}}\phi_1(x) + \sqrt{\frac{2}{5}}\phi_3(x). \tag{1}$$
The coefficients in front of the eigenfunctions are the probability amplitudes. The absolute square of those yields the probability. 
So the probability of measuring $E_1$ is $p_1=1/10$ and the probability of measuring $E_3$ is $p_3=2/5$. 
...but wait, they don't add up to one. Instead, $p_1+p_3=1/2$. This indicates that our wave function was not normalized. So let's do it again, but first we normalize Eq. (1): 
$$ \psi(x) = \frac{1}{\sqrt{5}}\phi_1(x) + \frac{2}{\sqrt 5}\phi_3(x) $$
Now we can read off the amplitudes again as $a_1 = 1/\sqrt 5$ and $a_3=2/\sqrt 5$. Squaring them yields the probabilities: $p_1=1/5=20\%$ to get $E_1$ and $p_3=4/5=80\%$ to get $E_3$. 
And you probably know that the energies are 
$$ E_n = \frac{n^2\pi^2\hbar^2}{2ma^2}. $$
