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电动车检测

Electric car detection

专题
Probability / 概率
难度
L4

题目详情

某城市估计 20% 的汽车是电动车。高速收费站安装的新传感器能检测到 95% 的电动车；但有 5% 的概率会把非电动车误判为电动车。每次车辆经过传感器的结果相互独立。

  1. 若一辆车被传感器检测为电动车，那么它实际上是非电动车的概率是多少？
  2. 已知同一辆车第一次通过时被传感器判为电动车，若这辆车再次通过传感器，那么它再次被判为电动车的概率是多少？

In a certain city, it’s estimated that 20% of cars are electric. A new sensor installed at a highway toll can detect 95% of electric cars. However, 5% of the time, it mistakenly identifies a non-electric car as electric. Every time a car passes the sensor, the result is independent.

  1. If a car is detected as electric by the sensor, what is the probability that it is in fact a non-electric car?
  2. Knowing that the sensor flagged a car as electric the first time it passed, if the same car passes the sensor again, what is the probability that the sensor will flag the car as electric again?
解析

记事件 EE 表示“车是电动车”，++ 表示“传感器判为电动车”。

已知

P(E)=0.2,P(+∣E)=0.95,P(+∣Ec)=0.05.P(E)=0.2,\quad P(+\mid E)=0.95,\quad P(+\mid E^c)=0.05.

(1) 由贝叶斯公式

P(Ec∣+)=P(+∣Ec)P(Ec)P(+∣E)P(E)+P(+∣Ec)P(Ec)=0.05⋅0.80.95⋅0.2+0.05⋅0.8=423.P(E^c\mid +)=\frac{P(+\mid E^c)P(E^c)}{P(+\mid E)P(E)+P(+\mid E^c)P(E^c)} =\frac{0.05\cdot 0.8}{0.95\cdot 0.2+0.05\cdot 0.8} =\frac{4}{23}.

(2) 第一次判 ++ 后，后验为 P(E∣+)=1−423=1923P(E\mid +)=1-\frac{4}{23}=\frac{19}{23}。

第二次再次判 ++ 的概率为

P(+(2nd)∣+(1st))=P(+∣E)P(E∣+)+P(+∣Ec)P(Ec∣+)=0.95⋅1923+0.05⋅423=7392.P(+\text{(2nd)}\mid +\text{(1st)}) =P(+\mid E)P(E\mid +)+P(+\mid E^c)P(E^c\mid +) =0.95\cdot\frac{19}{23}+0.05\cdot\frac{4}{23} =\frac{73}{92}.

因此答案为 423\boxed{\frac{4}{23}} 与 7392\boxed{\frac{73}{92}}。