 Solution

Expand the original two-spinor as $\chi =c_{+}{\chi }_{+}+c_{-}{\chi }_{-}$. Orthogonality givesLet $r=|\mathbf{p}|/(E+m)$ and $w_{\pm }=({\chi }_{\pm },-{\chi }_{\pm }{)}^{\mathsf{T}}/\sqrt{2}$. These $w_{\pm }$ are normalized, left-chiral helicity eigenvectors. ThenThe two contributions have squared normsThese are weights relative to the original normalized state; divide by their sum if conditioning on a normalized left-chiral projection. In the ultrarelativistic limit $N\to 1/\sqrt{2}$, $r\to 1$, the positive-helicity amplitude is suppressed as $m/(2E)$, while the negative-helicity amplitude tends to $c_{-}$. HenceFor momentum specifically along $+z$, $c_{+}=\sqrt{5}/3$ and $c_{-}=2/3$: the original helicity probabilities are $5/9,4/9$, the left-chiral Norm is $1/2-\beta /18$, and its limiting weight is $4/9$. The source does not specify that momentum direction, so these numerical probabilities are a specialization, not the general answer.