海上遇船：已知 1 小时概率，求 10 分钟概率

Stranded at Sea

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

你在海上漂流。已知在任意 1 小时内至少看到一艘船的概率为 73.7856%。假设看到船的过程为固定速率的泊松过程，且不同时段独立。

求接下来 10 分钟内至少看到一艘船的概率。

You are boating through the Mediterranean Sea when your motor breaks down. You remember learning from your divemaster that the probability of seeing at least one boat within a given hour is $73.7856\%$ . Assume that boats are observed as a Poisson process with a fixed rate of observations per time step and that observations are independent in any mutually exclusive time period. What is the probability that you will see at least one boat within the next 10 minutes?

解析

设 10 分钟内看不到船的概率为 $p$ 。

1 小时等于 6 个 10 分钟区间，因此 1 小时内看不到船的概率为 $p^6$ 。

已知 1 小时内至少看到一艘的概率为 0.737856，所以无船概率为 $1-0.737856=0.262144=0.8^6$ 。

因此 $p=0.8$ ，所求概率为 $1-p=0.2$ 。

Original Explanation

Let $p$ be the probability that you do not see a boat in 10 minutes. Then the probability of not seeing a boat within one hour is:

p^6=1-.737856 \Rightarrow p=0.8

The probability that you will see a boat within the next 10 minutes is:

1-p=0.2