In order for $\chi$ to be normalized, $A$ must equal
$\begin{align*}A = \frac{1}{\sqrt{6}}\end{align*}$
So, the probability to find the particle in the spin down state is
$\begin{align*}\left|\frac{2}{\sqrt{6}}\right|^2 = \boxed{\frac{2}{3}}\end{align*}$
Therefore, answer (D) is correct.

Solution to 2008 Problem 88