Typo in Bell's original paper on the EPR paradox? I am trying to understand Bell's famous paper from 1964 on the EPR paradox.
I stumbled over equation $(9)$ where it says
$$\left.\begin{align}A(\vec a,\vec\lambda)&=\operatorname{sign}\vec a\cdot\vec\lambda\\B(a,b)&=-\operatorname{sign}\vec b\cdot\vec\lambda \end{align}\right\}\tag9$$
Apart from the missing vector arrows I don't understand why it's $B(a, b)$ there. Could Bell actually have meant $B(b, \lambda)$ instead?
I wondered about this because I also don't understand how he concludes $(10)$ from this, but this might be another question.
 A: Yes, this is a typo on the left-hand side of the second equation; it is correct that $$B(\vec{b},\vec{\lambda})=-\mathrm{sign}\,\vec{b}\cdot\vec{\lambda}.$$
To see how he concludes (10), let's rewrite and put the normalization in explicitly for this uniform distribution:
$$P(\vec{a},\vec{b})=\frac{\int d\lambda A(\vec{a},\vec{\lambda})B(\vec{b},\vec{\lambda})}{\int d\lambda}.$$ The integral is over the surface of a sphere with unit radius, so the denominator will be $4\pi$. The product $A(\vec{a},\vec{\lambda})B(\vec{b},\vec{\lambda})=-\mathrm{sign}\,(\vec{a}\cdot\vec{\lambda})\mathrm{sign}\,(\vec{b}\cdot\vec{\lambda})$ will either be $+1$ or $-1$, and we need to determine for what fraction of the values of $\vec{\lambda}$ each of the $\pm1$ values occur.
The rest is geometry. Set the coordinate system, without loss of generality, to have $\vec{a}$ point to the top of the sphere and $\vec{b}$ be at polar angle $\theta$ and azimuthal angle $0$. This is always allowed because the integral over $\lambda$ is spherically symmetric. The region of values of $\vec{\lambda}$ for which $\mathrm{sign}\vec{a}\cdot\vec{\lambda}$ is $+1$ is the top half of the sphere; the region for which $\mathrm{sign}\vec{b}\cdot\vec{\lambda}$ is $+1$ is the hemisphere whose pole points in the direction of $\vec{b}$. The overlap of the two $+1$ regions forms a spherical wedge with angular width $\pi-\theta$. By symmetry, the spherical wedge on the opposite side of the sphere also has angular width $\pi-\theta$, and there both values are $-1$. (You can check the limits: when $\theta=0$, they fully overlap, and when $\theta=\pi$ they do not.)
From this picture, we see that $\mathrm{sign}\,(\vec{a}\cdot\vec{\lambda})\mathrm{sign}\,(\vec{b}\cdot\vec{\lambda})=1$ for $2\pi-2\theta$ out of the possible $2\pi$ angular width and $\mathrm{sign}\,(\vec{a}\cdot\vec{\lambda})\mathrm{sign}\,(\vec{b}\cdot\vec{\lambda})=-1$ for the other $2\theta$. The total probability is thus (remembering the extra $-$ sign)
$$P(\vec{a},\vec{b})=-\frac{2\pi-2\theta}{2\pi}+\frac{2\theta}{2\pi}=-1+\frac{2\theta}{\pi},
$$ as per Bell.
