3×3 灯泡网格

共有 $2^9=512$ 种等可能状态。计数满足约束的状态数为 63，因此概率为

$$\frac{63}{512}.$$

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Original Explanation

Consider the light bulbs arranged in a matrix:

$$\begin{bmatrix} 1 & 2 & 3\\ 4 & 5 & 6\\ 7 & 8 & 9 \end{bmatrix}$$

There are $2^9 = 512$ equally likely outcomes for the light bulbs being on or off. We need to count the outcomes that meet our criteria.

Focusing on light bulb $5$ (the middle light bulb): If it is on, then $2, 4, 6,$ and $8$ cannot be on. However, $1, 3, 7,$ and $9$ can be on or off, giving $2^4 = 16$ outcomes.

The case where light bulb $5$ is off is more complex.

If bulb $5$ is off, we consider 3 sub-cases based on the states of bulbs $4$ and $6$:

Case 1 - Both On:

If $4$ and $6$ are on, bulbs $1, 3, 7,$ and $9$ must be off, while $2$ and $8$ can be on or off. Thus, there are $2^2 = 4$ outcomes.

Case 2 - One On:

Assume $4$ is on and $6$ is off; we multiply by 2 for the reverse case. With $4$ on, $1$ and $7$ must be off, and $2$ and $3$ can be on or off, yielding 3 combinations. The same logic applies for bulbs $8$ and $9$, resulting in $2 \cdot 3 \cdot 3 = 18$ outcomes.

Case 3 - Both Off:

If both $4$ and $6$ are off, bulbs $1-3$ can be all off (1 case), exactly one on (3 cases), or 1 and 3 on with 2 off (1 case), totaling 5 combinations. The same holds for bulbs $7-9$, resulting in $5^2 = 25$ combinations.

Summing these, we find $16 + 4 + 18 + 25 = 63$ favorable outcomes, leading to a final answer of $\dfrac{63}{512}$.