= Solution
Let $I_B$ be the <indicator> that black bead $B$ belongs to the white bead's <connected component>. The number of such black beads is $\sum_B I_B$. By <linearity of expectation>, no <independence> between these <indicators> is needed.
For $n=7$, the six oriented distances $1,\ldots,6$ give respective <probabilities>
$$
\frac47,\ \frac27,\ \frac17,\ \frac17,\ \frac27,\ \frac47.
$$
Their sum is \b[$2$], the expected number of black beads in the white component. As a check, summing the formula from (i) and using $\sum_{k=1}^{n-1}\binom k3=\binom n4$ gives the general value $2\binom n4/\binom n3=(n-3)/2$, again equal to $2$ when $n=7$.
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