= Solution
View the necklace as a <cycle graph>. Between the white bead and $B$ there are two <edge-disjoint paths>, containing $k$ and $n-k$ links. The beads remain connected if and only if at least one of these paths is uncut.
There are $\binom n3$ equally likely cut <sets>. Keeping the $k$-link path uncut requires all three cuts to be among its $n-k$ complementary links, giving $\binom{n-k}{3}$ choices. Keeping the other path uncut gives $\binom k3$ choices. These two events are disjoint because three links are cut and they cannot all lie in both complementary paths. Therefore
$$
\boxed{\mathbb P(B\text{ shares the white component})
=\frac{\binom{n-k}{3}+\binom k3}{\binom n3}.}
$$
Use $\binom a3=0$ for $0\le a<3$. The answer is unchanged by exchanging $k$ and $n-k$, so either orientation around the necklace gives the same value. This is the connectivity calculation for <random cuts of a cycle>.
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