Suppose I have an electron that is in the spin state
$$\chi =A\begin{bmatrix}3i \\4\end{bmatrix} $$
If I calculate the expectation values of its spin components $S_x$ $S_y$ $S_z$, I get
$$\langle S_x \rangle= 0$$
$$\langle S_y \rangle = -\hbar \dfrac{12}{25}$$
$$\langle S_z \rangle= -\hbar \dfrac{7}{50}$$
I can do the math part easily, however I am having a bit of difficulty trying to apply a physical interpretation to my answers.
My attempt at a physical interpretation
Due to the uncertainty principle, we can not having a well defined $S_x$ $S_y$ $S_z$ component simultaneously. Hence, we can only calculate the expectation value of its components. For $S_x$, the probability that the spin will lie in the $S_x$ component will be zero. For $S_y$ and $S_z$ we the spin has either $-\hbar \dfrac{12}{25}$ or $
-\hbar \dfrac{7}{50}$.
Best Answer
Average values do not tell you about probabilities (which must be non-negative numbers). In particular, if your particle is describe by a state $\vert\xi\rangle=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1 \end{array}\right)$ the average $\langle S_z\rangle=0$ but the probability of having the system in the spin-up state is $1/2$, as is the probability of having the system in the spin-down state.
Average values are weighted sums of probabilities: in the case of $\vert\xi\rangle$, the average value is $$ \langle S_z\rangle = P(\uparrow)\frac{\hbar}{2}+P(\downarrow)\left(-\frac{\hbar}{2}\right) =\frac{1}{2}\frac{\hbar}{2}+ \frac{1}{2}\left(-\frac{\hbar}{2}\right)=0 $$ where $P(\uparrow)$ is the probability of having your system in the spin-up state, and $\hbar/2$ is the value of $S_z$ when the system is in the spin-up state.
Clearly here, the fact that the average value is $0$ does not in any way mean the probability of the spin being along $\hat z$ is $0$, since the system is half the time along $+\hat z$ and half the time along $-\hat z$. In fact, the only two possible values of spin are $\pm\frac{\hbar }{2}$, which excludes $\langle S_z\rangle=0$ for this example. More generally, the only possible outcomes of $S_y$ or $S_x$ are $\pm \frac{\hbar}{2}$, and never numbers such as $-\frac{12\hbar}{25}$.
In general, the probability of a system described by $\vert \chi\rangle$ to be found in a state having definite spin along - say - $\boldsymbol{\hat y}$ is given by $$ P(\uparrow)_{\hat y}= \vert\langle\chi\vert \uparrow\rangle_{\hat y}\vert^2\, ,\qquad P(\downarrow)_{\hat y}=\vert\langle\chi\vert \downarrow\rangle_{\hat y}\vert^2 $$ where $\vert\uparrow\rangle_{\hat y}$ is the eigenstate of $\sigma_y$ with eigenvalue $+\hbar/2$. In your case, we have $$ P(\uparrow)_{\hat y}=\frac{1}{50}\, ,\qquad P(\downarrow)_{\hat y}=\frac{49}{50} $$ so $$ \langle S_y\rangle = \frac{\hbar}{2}\frac{1}{50}-\frac{\hbar}{2}\frac{49}{50}=-\frac{12\hbar}{50}\, . $$
What the average values tell you is the "direction" of the spin. The unit vector $\boldsymbol{\hat n}=(\sigma_x,\sigma_y,\sigma_z)=(0,-24/25,-7/25)$ points in a definite direction, and if you were to choose your quantization axis along this direction, the system would be in a spin-up state, i.e it would be an eigenstate of $$ \boldsymbol{\hat n}\cdot \vec S= \frac{1}{25}\left( \begin{array}{cc} -7 & 24 i \\ -24 i & 7 \\ \end{array} \right)\, . \tag{1} $$ You can indeed verify that your vector $\chi$ is an eigenvector of (1) with eigenvalue $\frac{\hbar}{2}$