In a $k \times k$ chessboard, a set $S$ of 25 cells that are in a $5 \times 5$ square is chosen uniformly at random. The probability that there are more black squares than white squares in $S$ is $48 \%$. Find $k$.
Solution: We know that there must be fewer black squares than white squares, and $k$ must be odd. Additionally, we know that there are $k-4$ ways to pick the left column of the $5 \times 5$ square so that the right column can fit within the $k \times k$ grid, and $k-4$ ways to pick the top row by similar logic. Therefore, there are $(k-4)^{2}$ of these $5 \times 5$ squares on this chessboard, and because there will be more black squares than white squares whenever there exists a black square in the top left corner, there are $\frac{(k-4)^{2}-1}{2}$ of them have more black squares than white squares, corresponding to the number of black squares in the upper $(k-4) \times(k-4)$ grid. Thus, we have \[ \frac{\frac{(k-4)^{2}-1}{2}}{(k-4)^{2}}=0.48 \Longrightarrow k=9 \] $\fbox{9}$.