Alternatively to Igor Rivin's method, here is what I think could lead to the result.

To get four independent uniformly distributed random points on the hemisphere, we first chose two (independent, uniformly distributed) diameters of its boundary in the equatorial plane, say AOB and COD, with an angle $\theta$ between them, distributed uniformly on $[0,\pi/2]$. Then we chose two planes containing AOB and COD respectively, making with the equatorial plane an angle of $\xi$ (or $\eta$, respectively) uniformly distributed on $[0,\pi]$ ; they intersect the hemisphere on half-circles. Each point M on the half-circle is defined by the angle $\alpha$ (or $\gamma$) from OA (or OC) to OM, and the half-circle is equipped with the probability measure $\sin\alpha\ d\alpha/2$ (or $\sin\gamma\ d\gamma/2$). With this choice of conditional distribution, if M is chosen randomly it will be uniformly distributed on the half-sphere, I (strongly) believe.

Clearly then, $ab\subset$ AB and $cd\subset$ CD intersect iff both contain the point M$_0$ where the two half-circles intersect (at angles $\alpha_0$ on AB, $\gamma_0$ on CD): this (given $\theta$, $\xi$ and $\eta$) has probability $\frac14 \sin^2\alpha_0\times\frac14 \sin^2\gamma_0$.

These angles $\alpha_0$ and $\gamma_0$ of M$_0$ can be determined using spherical trigonometry (considering the spherical triangle with $\theta$ as one side and $\xi,\eta$ as adjacent angles). You get$$\tan\alpha_0=\frac{2\sin\eta}{\cot(\theta/2)\sin(\xi+\eta)+\tan(\theta/2)\sin(\xi-\eta)}$$and$$\tan\gamma_0=\frac{2\sin\xi}{\cot(\theta/2)\sin(\xi+\eta)-\tan(\theta/2)\sin(\xi-\eta)}$$

Integrating in $\theta$, $\xi$ and $\eta$ should then give the desired probability that $ab$ and $cd$ intersect.