Is it possible to determine a quantum state given the probabilities of +/- in the x and y components.
For example, a quantum state $|\psi\rangle$ has the following probabailites: \begin{equation} \begin{cases} |\langle +_x|\psi\rangle|^2=p(ℏ/2) = 0.7 \\ |\langle -_x|\psi\rangle|^2=p(-ℏ/2) = 0.3 \end{cases} \ \text{and} \ \begin{cases} |\langle +_y|\psi\rangle|^2=p(ℏ/2) = 0.2 \\ |\langle -_y|\psi\rangle|^2=p(-ℏ/2) = 0.8 \end{cases} \end{equation}
Assuming you are talking about pure state, and assuming you mean, that you have lots of identical states which you can try against x and y Pauli matrices as much times as you want, and those are the probabilities from the limit of doing infinite experiments with different copies of the same original state, then:
From those probabilities, you know that the original state has to be proportional to:
\begin{equation} |\psi\rangle = \frac{1}{\sqrt{5}}\left(\sqrt{3}|+_x\rangle + e^{i\theta_1}\sqrt{2}|-_x\rangle\right) = \frac{1}{\sqrt{10}}\left(\sqrt{3}(|+_z\rangle + |-_z\rangle) + e^{i\theta_1}\sqrt{2}(|+_z\rangle - |-_z\rangle)\right) \end{equation}
\begin{equation} |\psi\rangle = \frac{1}{\sqrt{5}}\left(|+_y\rangle + e^{i\theta_2}\sqrt{4}|-_y\rangle\right) = \frac{1}{\sqrt{10}}\left((|+_z\rangle + i |-_z\rangle) + e^{i\theta_2}\sqrt{4}(|+_z\rangle -i |-_z\rangle)\right) \end{equation}
where we have substituted the eigenvectors of the x and y Pauli matrices. Now we just need to write the + and - components of the z together and compare:
\begin{equation} |\psi\rangle = \frac{1}{\sqrt{10}}\left( (\sqrt{3}+e^{i\theta_1}\sqrt{2} )|+_z\rangle + (\sqrt{3}-e^{i\theta_1}\sqrt{2} )|-_z\rangle\right) \end{equation}
\begin{equation} |\psi\rangle = \frac{1}{\sqrt{10}}\left( (1+e^{i\theta_2}\sqrt{4} )|+_z\rangle + i(1-e^{i\theta_2}\sqrt{4} )|-_z\rangle\right) \end{equation}
where now we can do a system of equations and see if it is solvable deterministic or not:
\begin{equation} \begin{cases} \sqrt{3}+e^{i\theta_1}\sqrt{2} = 1+e^{i\theta_2}\sqrt{4} \\ \sqrt{3}-e^{i\theta_1}\sqrt{2} = i(1-e^{i\theta_2}\sqrt{4}) \end{cases} \ \xrightarrow[]{} \ \begin{cases} 2\sqrt{3} = (1+i) + (1-i) e^{i\theta_2}\sqrt{4} \\ 2e^{i\theta_1}\sqrt{2} = (1-i) + (1+i) e^{i\theta_2}\sqrt{4} \end{cases} \end{equation}
from where it is pretty easy to find that for this case $\theta_2$ has to fulfill: \begin{equation} \frac{2\sqrt{3} - (1+i)}{\sqrt{4}(1-i)} = e^{i\theta_2} \end{equation} which is imposible, because the left part doesn't have modulus 1.
This does not mean, that you can't find a state from given probabilities, it just simply states that the probabilities you gave, can't really be happening.
In general you if you have a state with probability a of $+_x$, your state then can be found in a circle in the x axis in the bloch sphere, so considering both $+_x$ and $+_y$ probablities your state should be in both of those circles at the same time, which for your concrete example can't happen because the circles are too small:
But for another case, you would have one or two solutions (yellow dots), such as in this example:
This can be easily seen by, picking the x circle of 0 radius, which means our state is exactly $|+_x\rangle$ or $|-_x\rangle$, then the only possibility for the y circle would be a maximum radius one, which mean 1/2 probability of $+_y$ or $-_y$:
such as we know the eigen states of x have.
