Given a $3 \times 3$ table (like for playing tic-tac-toe). Four chips are placed randomly in four randomly chosen cells.

Find the probability that among these four chips, there will be three that are in a row either vertically, horizontally, or diagonally.
The total number of possible arrangements is $C_{9}^{4}=126$. Arrangements where a vertical, horizontal, or diagonal line is occupied are easily counted: there are $8 \cdot 6=48$. Due to randomness and equal probability of arrangements, the desired probability is $48 / 126=8 / 21$.

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