车祸：1 小时概率已知，求 30 分钟概率

1 小时内无车祸概率为 $1/9$。

设 30 分钟内至少一次的概率为 $p$，则 30 分钟内无车祸概率为 $1-p$。

两个不相交 30 分钟区间独立，因此

$$(1-p)^2=\frac{1}{9}\Rightarrow 1-p=\frac{1}{3}\Rightarrow p=\frac{2}{3}.$$

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Original Explanation

Let the probability of at least $1$ car crash in a $30$ minute interval be $p$.

We know that the probability there is no car crash in the hour interval is $\dfrac{1}{9}$ by complementation. This is the same as saying that there is no car crash in each of the intervals consisting of the first and last $30$ minutes of the hour. By the question, we can say that the number of car crashes in disjoint intervals are independent. This is because they occur at a constant rate throughout time and the arrivals are independent.

The probability of no car crash in each of those two intervals individually is $1-p$, so the probability of no car crash in both intervals is $(1-p)^2$.

Thus, $(1-p)^2 = \dfrac{1}{9}$, so $p = \dfrac{2}{3}$.