# Repeated measurement of spin

I am self-studying Quantum Mechanics by Davies and Betts. Exercise 7 of Chapter 7 asks the following:

"A particle is observed initially to have spin component + along the $z$ axis. A measurement of its spin component along another axis $z'$ is then made, where the angle between $z$ and $z'$ is $\theta=\pi/3$. This measurement is not recorded. A subsequent re-measurement of the $z$ component of spin is then made. What is the probability $p$ that the spin will now point down the $z$ axis?"

This seems pretty straightforward to me, but my answer ($p=3/8$) does not coincide with the answer from the book ($p=3/4$).

I have solved the exercise as follows: We have the following relations

$\left|\uparrow\right\rangle = \cos (\theta /2) \left|\uparrow'\right\rangle - \sin(\theta/2)\left|\downarrow'\right\rangle$.

$\left|\uparrow'\right\rangle = \cos (\theta /2) \left|\uparrow\right\rangle + \sin(\theta/2)\left|\downarrow\right\rangle$.

$\left|\downarrow'\right\rangle = -\sin (\theta /2) \left|\uparrow\right\rangle + \cos(\theta/2)\left|\downarrow\right\rangle$.

By splitting the event space in the two disjoint events "the result of the first measurement was $\left|\uparrow'\right\rangle$ (resp. $\left|\downarrow'\right\rangle$)", I get for the required probability

$p = \cos^2(\theta/2)\sin^2(\theta/2) + \sin^2(\theta/2)\cos^2(\theta/2)$.

Plugging in the numbers $\cos(\theta/2) = \sqrt{3}/2$, $\sin(\theta/2) = 1/2$ gives me $p = 3/8$, which is different from the answer from the book, $p=3/4$. What am I doing wrong?