= Solution
Because the <events> are pairwise <disjoint events>, their count can only be zero or one. The <event> that the count is zero is the <complement of an event> of their union. Finite additivity therefore gives
$$
\boxed{\mathbb P(N=0)=1-\mathbb P\left(\bigcup_{i=1}^r A_i\right)
=1-\sum_{i=1}^r\mathbb P(A_i).}
$$
Now use $N\geq1$ for the fixed number of independently located ships. Normalize the radius to one and write their positions as unit vectors $a_1,\ldots,a_N$. A point $x$ on the <sphere> is reached by ship $i$ precisely when $a_i\cdot x>0$, so each ship reaches an open <hemisphere>. Coverage fails exactly when there is a unit vector $x$ with $a_i\cdot x\leq0$ for every $i$. Equivalently, all ship positions lie in some closed <hemisphere>.
With <probability> one no pair of ship vectors is antipodal and no three are linearly dependent: the exceptional choices lie on sets of zero surface area. For such a configuration, containment in a closed <hemisphere> implies containment in an open <hemisphere>. To see this, choose its inward normal $u$, so $u\cdot a_i\geq0$. At most two vectors lie on its boundary. If there are two, their sum has positive dot product with each, since they are not antipodal; if there is one, use that vector itself. Perturb $u$ slightly in this direction. The boundary dot products become positive while all already positive dot products remain positive. Thus the strict and non-strict containment conditions have the same <probability> here.
Condition on the unoriented axes $\{a_i,-a_i\}$. Choose representatives $b_i$ by any fixed sign convention. By uniform sampling and <independence>, the actual positions are $a_i=\varepsilon_i b_i$ with $N$ <independent> fair signs $\varepsilon_i\in\{-1,1\}$. Conditional on these axes, each of the $2^N$ sign choices has <probability> $2^{-N}$.
The <great circles> $b_i\cdot u=0$ partition the <sphere> into $R_N=N(N-1)+2$ open regions. In each region the sign vector $\sigma(u)=(\operatorname{sgn}(b_i\cdot u))_i$ is constant. No sign vector labels two different regions: its strict inequalities define a convex cone, and normalized line segments within that cone connect any two points of its spherical section. The <regions of a general-position great-circle arrangement> therefore yield exactly $R_N$ realized sign vectors.
The chosen positions lie in an open <hemisphere> if and only if some $u$ has $u\cdot a_i>0$ for every $i$, that is, $\varepsilon_i=\operatorname{sgn}(b_i\cdot u)$ for every $i$. Failure of coverage thus corresponds exactly to one of the $R_N$ realized sign choices. These choices are mutually exclusive <events>, each of <conditional probability> $2^{-N}$, so the first part applies:
$$
\mathbb P(\text{failure}\mid\text{axes})=\frac{R_N}{2^N}.
$$
The result is independent of the axes. Averaging gives the <coverage of a sphere by random open hemispheres>:
$$
\boxed{\mathbb P(\text{every point is reached})=1-\frac{N(N-1)+2}{2^N},\qquad N\geq1.}
$$
For $N=1,2,3$ this is zero; for $N=4$ it is $1/8$. If no ship lands, coverage has <probability> zero separately. The counting uses equiprobable orientation signs, not an assumption that the spherical regions have equal area.
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