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谎言父亲

Father of lies

专题
Probability / 概率
难度
L2

题目详情

父亲声称“昨晚下雪了”。

已知：

  • 任意一晚下雪概率 P(S)=1/8P(S)=1/8；
  • 父亲有 5/65/6 的概率在说谎（即说真话概率 1/61/6）。

问：在他做出“下雪”声明后，昨晚实际下雪的概率是多少？

英文原题

A father claims about snowfall last night. First daughter tells that the probability of snowfall on a particular night is 1/8. Second daughter tells that 5 out of 6 times the father is lying! What is the probability that there actually was a snowfall?

Hint

Conditional Probability or Baye's Theorem

解析

用贝叶斯公式：

设 CC 表示“父亲声称下雪”。

P(S∣C)=P(C∣S)P(S)P(C∣S)P(S)+P(C∣Sˉ)P(Sˉ).P(S\mid C)=\frac{P(C\mid S)P(S)}{P(C\mid S)P(S)+P(C\mid \bar S)P(\bar S)}.

其中：

  • P(C∣S)=1/6P(C\mid S)=1/6（真话）；
  • P(C∣Sˉ)=5/6P(C\mid \bar S)=5/6（说谎）；
  • P(S)=1/8, P(Sˉ)=7/8P(S)=1/8,\ P(\bar S)=7/8。

代入得

P(S∣C)=(1/6)(1/8)(1/6)(1/8)+(5/6)(7/8)=136.P(S\mid C)=\frac{(1/6)(1/8)}{(1/6)(1/8)+(5/6)(7/8)}=\frac{1}{36}.

英文解析

1/36

Solution

Let S=S = Snowfall occurred, and CC be the event that the father is claiming snowfall ocurred.

Probability of (Snowfall given Claim) = P(S∣C)=P(C∣S)⋅P(S)P(C)P(S | C) = \dfrac{P(C|S) \cdot P(S)}{P(C)}

P(S)=1/8P(S) = 1/8

P(C∣S)=1−5/6=1/6P(C|S) =1 - 5/6 = 1/6 (given in the question, father lies 5/6 times)

P(C)=P(True claim)+P(False Claim)P(C) = P(\text{True claim}) + P(\text{False Claim})

=P(C∩S)+P(C∩S′)= P(C \cap S) + P(C \cap S') where S′S' means that it did not snow.

=P(C∣S)⋅P(S)+P(C∣S′)∗P(S′)= P(C|S) \cdot P(S) + P( C | S') * P(S')

=(1/6⋅1/8)+(5/6⋅7/8)= (1/6 \cdot 1/8) + (5/6 \cdot 7/8)

P(S∣C)=P(C∣S)⋅P(S)P(C)=1/6⋅1/8(1/6⋅1/8)+(5/6⋅7/8)=136P(S | C) = \dfrac{P(C|S) \cdot P(S)}{P(C)} = \dfrac{1/6 \cdot 1/8}{(1/6 \cdot 1/8) + (5/6 \cdot 7/8)} = \dfrac{1}{36}