5. On the cells of a chessboard of size $8 \times 8$, 5 identical pieces are randomly placed. Find the probability that four of them will be located either on the same row, or on the same column, or on one of the two main diagonals.
Answer: $P(A)=\frac{18 \cdot\left(C_{8}^{4} \cdot C_{56}^{1}+C_{8}^{5}\right)}{C_{64}^{5}}=\frac{18 \cdot 56 \cdot 71}{31 \cdot 61 \cdot 63 \cdot 64}=\frac{71}{4 \cdot 31 \cdot 61}=\frac{71}{7564} \approx 0.0094$.