20 分钟内见到车的概率已知，求 5 分钟内见到车的概率

设 5 分钟内至少看到车的概率为 $p$，则 5 分钟内没车概率为 $1-p$。

20 分钟由 4 个 5 分钟块组成，没车概率为 $(1-p)^4$，因此

$$1-(1-p)^4=\frac{609}{625}.$$

解得

$$(1-p)^4=\frac{16}{625}=\left(\frac{2}{5}\right)^4\Rightarrow 1-p=\frac{2}{5}\Rightarrow p=\frac{3}{5}.$$

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Original Explanation

Let $p$ = probability of seeing at least one car in 5 minutes. Then the probability of seeing no cars in 5 minutes is $1 - p$. Over four 5-minute blocks (20 minutes), the probability of no cars is $(1-p)^4$. Hence $$1 - (1-p)^4 = \frac{609}{625}.$$ So $$(1-p)^4 = 1 - \frac{609}{625} = \frac{16}{625}.$$ Thus $$1 - p = \bigl(\frac{16}{625}\bigr)^{\frac{1}{4}} = \frac{2}{5},$$ and $$p = \frac{3}{5}.$$