Past exam of the year 1 of the computer science course of the University of Oxford / 2019 / Discrete mathematics, continuous mathematics, and probability / 7 / c / ii / 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 $1,\dots ,6$ give respective probabilitiesTheir sum is $2$, the expected number of black beads in the white component. As a check, summing the formula from (i) and using ${\sum }_{k=1}^{n-1}\left(\genfrac{}{}{0ex}{}{k}{3}\right)=\left(\genfrac{}{}{0ex}{}{n}{4}\right)$ gives the general value $2\left(\genfrac{}{}{0ex}{}{n}{4}\right)/\left(\genfrac{}{}{0ex}{}{n}{3}\right)=(n-3)/2$, again equal to $2$ when $n=7$.