Find the probability that A,B,C are connected I was given the following problem as a homework assignment:

Denote with $S$ the ball with center $O$. Three points $A, B$ and $C$ are chosen at random on its surface, their positions being independent and each
  uniformly distributed on the surface. Points A and B can be connected
  together if the angle $AOB<\pi/2$. What is the probability that they can be connected (with, for example, $A$ connecting with $B$ via $C$ if necessary)?

I was given the answer by my professor as a hint: $(\pi+2)/(4\pi)$.
I thought about it in the following way. Let $r$ denote the line that passes through the center $O$, which will intersect the ball at a point, call it $A$. Now, take a plane perpendicular to $r$ and make it pass through $O$. The plane divides the sphere into $2$ hemispheres, one that contains the point $A$ and one that does not. If $B$ is placed in the hemisphere that contains $A$, then  $AOB<\pi/2$ and so they can connect. Otherwise, they cannot. So the probability of $A$ connecting with $B$ is equivalent to the prbability that $B$ falls in one of the two hemispheres that contains $A$ which is $1/2$. (The idea of this reasonsing is to keep $A$ fixed and set it at the "center" of the hemisphere's surface). Now I am left with calculating the probability that $B$ is placed on the other hemisphere so that $A$ must connect with $B$ through $C$. Which is again $1/2$? What is wrong with this reasoning? And how would one solve this problem?
 A: We arbitrarily assume $S$ is the unit sphere, $O$ is the origin, $A=(1,0,0)$ and $B$ lies in the upper $xy$-plane; this can always be done by a suitable rotation and $x=\angle AOB$ has pdf $\frac12\sin x$ for $x\in[0,\pi]$. Looking at the sphere on the $z$-axis then gives two cases, depending on the comparison between $x$ and $\pi/2$:



*

*If $x<\pi/2$ then $A$ and $B$ are connected. $C$ then only needs to lie in either of the hemispheres defined by $A$ and $B$; the probability of this happening is the proportion of the disc covered by the two hemispheres in the top-down view. The angle covered by either hemisphere is $\pi+x$.

*If $x>\pi/2$ then $A$ and $B$ are not connected, so $C$ must lie in both their hemispheres. In the top-down view, the angle covered by both hemispheres is $\pi-x$.
Thus we have the final probability that $A,B,C$ are connected as
$$\int_0^{\pi/2}\frac12\left(\frac{\pi+x}{2\pi}\right)\sin x\,dx+\int_{\pi/2}^\pi\frac12\left(\frac{\pi-x}{2\pi}\right)\sin x\,dx$$
$$=\frac1{4\pi}\left(\int_0^{\pi/2}(\pi+x)\sin x\,dx+\int_{\pi/2}^\pi(\pi-x)\sin x\,dx\right)$$
$$=\frac1{4\pi}\left(\int_0^{\pi/2}(\pi+x)\sin x\,dx+\int_0^{\pi/2}x\sin x\,dx\right)$$
$$=\frac1{4\pi}\left(\pi\int_0^{\pi/2}\sin x\,dx+2\int_0^{\pi/2}x\sin x\,dx\right)$$
$$=\frac1{4\pi}\left(\pi+2\left([-x\cos x]_0^{\pi/2}-\int_0^{\pi/2}-\cos x\,dx\right)\right)$$
$$=\frac1{4\pi}\left(\pi+2[\sin x]_0^{\pi/2}\right)=\frac{\pi+2}{4\pi}$$
