观察事故：1 小时概率已知，求 30 分钟概率

1 小时内无事故概率为 $1/4$。

设 30 分钟内至少一次概率为 $p$，则无事故为 $1-p$。两个半小时区间独立：

$$(1-p)^2=\frac{1}{4}\Rightarrow 1-p=\frac{1}{2}\Rightarrow p=\frac{1}{2}.$$

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Original Explanation

The hour interval can be broken down into two disjoint thirty-minute intervals such that the probability $p$ of observing any accident is uniform. The probability that we do not observe an accident within a thirty-minute interval is $1-p$. Furthermore, the probability that we do not observe any accidents within two independent thirty-minute intervals (or an hour) can be written as:

$$(1-p)^2 = 1 - \frac{3}{4} \Rightarrow p=1/2$$