Short answer:
Assuming you are measuring in the computational basis (Z basis), $\{|0\rangle, |1\rangle \}$, there is no randomness upon measurement in the following quantum circuit (you will always get back the state $|1\rangle$):
Thus, measurement needs not be random. However, if you try this following circuit, you will have 50% to see a $|0\rangle$ and 50% to see $|1\rangle$:
So your answer is not so deterministic here... when designing a quantum algorithm, we want somehow create interference within the system and so upon measurement, we have something like the first circuit. Where the result is deterministic.
Long answer:
A state vector is a vector describing the state of the system. In quantum computing, your system is a qubit, hence a two-level quantum system. Thus, it can be described by a complex Euclidian space $\mathbb{C}^2$. Thus, the state of a qubit, $|\psi \rangle$ can be written as
$$ |\psi \rangle = \alpha|0\rangle + \beta|1\rangle \hspace{1 cm} \alpha, \beta \in \mathbb{C}, \hspace{0.5 cm} |0\rangle = \begin{pmatrix} 1 \\ 0\end{pmatrix}, \hspace{0.25 cm} |1\rangle = \begin{pmatrix} 0 \\ 1\end{pmatrix} \hspace{1cm} |\alpha|^2 +|\beta|^2 = 1$$
And if you have an $n$ qubit state then it can be written as a normalized vector in $\mathbb{C}^{2 ^{\otimes n}}$.
So suppose you have the state $|\psi \rangle = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix} $ then you can see that $|\psi\rangle$ can be written as a superposition (in a linear combination) of the states $|0\rangle$ and $|1\rangle$ as
$$ |\psi \rangle = \dfrac{1}{\sqrt{2}}|0\rangle + \dfrac{1}{\sqrt{2}}|1\rangle $$
Now, it is a postulate of quantum mechanics that any device that measures a two-state quantum system (a qubit) must have two preferred states $\{|e_1\rangle, |e_2\rangle \}$ that form an orthonormal basis for the associated vector space (here would be $\mathbb{C}^2$). A measurement on the state $|\psi\rangle$ transforms $|\psi\rangle$ into one of the these basis vectors $|e_1 \rangle$ or $|e_2\rangle$. The probability that the state $|\psi\rangle$ is measured as $|e_1\rangle$ or $|e_2\rangle$ is the square of the magnitude of the amplitude of the component of the state in the direction of the basis vector $|e_1\rangle$ or $|e_2\rangle$.
So if we picked $|e_1\rangle = 0 $ and $|e_2 \rangle = |1\rangle$ then upon measuring the state $|\psi \rangle = \dfrac{1}{\sqrt{2}}|0\rangle + \dfrac{1}{\sqrt{2}}|1\rangle$, you will have 50% observing the state $|0\rangle$ and 50% observing the state $|1\rangle$. In this case, a single measurement doesn't tell you anything... you need many, many measurements to build up a statistical distribution.
But the uncertainty described above is only because we have picked $|e_1\rangle$ and $e_2\rangle$ the way we did. If we have picked, $|e_1\rangle = |+\rangle = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}$ and $|e_2\rangle = |-\rangle = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{pmatrix}$ then upon measuring, we will observe the state $|+\rangle$ with a 100% probability (assuming no noise). There is no randomness here. This is because $|\psi \rangle$ is NOT in a superposition in the $\{|+\rangle, |-\rangle \}$ basis.
Therefore, the notion of superposition is basis-dependent. All states are superposition with respect to some bases but not with respect to others. That is, a state $|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$ is only a superposition with respect to the computational basis $\{|0\rangle, |1\rangle \}$ but not a superposition with respect to the bases $\{ \alpha |0\rangle + \beta |1\rangle, \beta^* |0\rangle - \alpha^* |1\rangle \}$.
Since measuring a superposition state $|\psi \rangle = \alpha |0\rangle + \beta |1\rangle$ is probabilisitc, it is tempted to say that the state $|\psi \rangle$ is a probabilistic mixture of $|0\rangle$ and $|1\rangle$ and we just don't know which, when in fact, $|\psi \rangle$ is actually a definite state. That is, if we measure $|\psi\rangle$ in certain bases, we will get a deterministic result.
Thus, when designing a quantum algorithm, we want the final state which contains the answer we are looking for be in a single eigenstate and not in superposition with respect to the computational basis (Z basis). For instance, the following circuit will create a state that is NOT in superposition in the computational basis...
so if you run the above circuit on a quantum computer, you will only need to measure it ONCE to obtain your result... This circuit is constructed to solve the problem of "secret bitstring". In fact, the state upon measurement is the state $|111\rangle$, which tells you that the secret bitstring is 11... This is known as the Bernstein–Vazirani algorithm. I would recommend you to read up on it as it will help you to understand where the advantage of quantum computation come about.
