probability of measuring 1 in a qubit in a state

Question

Imagine we prepare a qubit in the following state : $$ |\psi\rangle = \sqrt{\frac{1}{3}}|0\rangle - \sqrt{\frac{2}{3}}|1\rangle$$ What is the probability of measuring 1?

Answer 1

The quantum state $|\psi\rangle$ of a qubit is described by a state vector $\begin{pmatrix}\alpha\\\beta\end{pmatrix}$ with $\alpha$ and $\beta$ complex numbers.
When we observe a qubit in this state, the probability of measuring 0 is $|\alpha|²$ and the probability of measuring 1 is $|\beta|²$.

We can rewrite this state as a linear combination of basis states :
$\begin{pmatrix}\alpha\\\beta\end{pmatrix} = \alpha|0\rangle + \beta|1\rangle$ with $|0\rangle = \begin{pmatrix}1\\0\end{pmatrix}$ and $|1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$.

In your state, $\alpha = \sqrt{\frac{1}{3}}$ and $\beta = -\sqrt{\frac{2}{3}}$, thus, the probability of measuring 0 is $|\sqrt{\frac{1}{3}}|² = \frac{1}{3}$ and the probability of measuring 1 is $|-\sqrt{\frac{2}{3}}|² = \frac{2}{3}$