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By rotational invariance, fix the first landing point at the north pole of a sphere of radius $R$. A band of central angle $\theta$ and width $d\theta$ has area $2\pi R^2\sin \theta d\theta$. Dividing by total area gives the central angle between independent uniform sphere points densityand zero outside this range. The radius cancels.

Given the first pair's central angle $\gamma$, the third point can communicate with each of them exactly when it lies in that point's positive hemisphere. Choose their intersection line as a polar axis. Each hemisphere permits an azimuth interval of length $\pi$, and the intervals' centres are separated by $\gamma$. Their overlap is a spherical lune of opening $\pi -\gamma$, with area $2R^2(\pi -\gamma )$. ThereforeThe union probability follows from inclusion-exclusion for two hemispheres:Thus the requested conditional answers are“Either” here is inclusive: a third ship linked to both is also linked to at least one.

For network connectivity, if $\gamma <\pi /2$ the first pair is already linked, and the third must link to at least one of them. If $\gamma >\pi /2$, the first pair is not linked, and both must link through the third. Consequently the connectivity of three acute-angle sphere points isUsing ${\int }_0^{\pi /2}\gamma \sin \gamma d\gamma =1$ and ${\int }_{\pi /2}^{\pi }\gamma \sin \gamma d\gamma =\pi -1$ givesAngles exactly equal to the threshold have probability zero. Connectivity allows relaying and is weaker than requiring direct links between every pair.