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半小时事故概率

Half Time

专题
Probability / 概率
难度
L2

题目详情

某条路在 1 小时内发生事故的概率是 34\frac{3}{4}

问:在半小时内发生事故的概率是多少?

提示:假设两个半小时区间的“无事故”联合概率等于 1 小时无事故的概率。

The probability of having accidents on a road in one hour is 3/43/4. What is the probability of accidents in half an hour?

Hint

The joint probability of no events in two intervals of half an hour each is the same as the probability of no event in one hour.

解析

答案是 12\frac{1}{2}

1 小时无事故概率为 134=141-\frac{3}{4}=\frac{1}{4}

设半小时无事故概率为 pp,则两个半小时都无事故的概率为 p2p^2,于是

p2=14p=12.p^2=\frac{1}{4}\Rightarrow p=\frac{1}{2}.

因此半小时发生事故概率为 1p=121-p=\frac{1}{2}


Original Explanation

1/2

Solution

Probability of no accident in 11 hour = P(no accidents in the first half an hour ) * P(no accidents in the next half an hour)

    13/4=p2    p=1/2\implies 1 - 3/4 =p^2 \implies p=1/2

Hence the probability of an accident in half an hour =11/2=1/2=1 - 1/2 = 1/2


For those who are more inclined towards the inner wirings

Assume that the disjoint time intervals are independent, the probability of no accidents in [0,2t)[0, 2t) = the probability of no accidents in [0,t)[0,t) AND [t,2t)[t,2t)

Define p(t)p(t) = probability of an accident in an interval of tt hours

    1p(2t)=(1p(t))(1p(2tt))=(1p(t))2\implies 1-p(2t) = (1-p(t)) \cdot (1-p(2t - t)) = (1 - p(t))^2

Given that p(1h)=3/4    p(0.5h)=1/2p(1h)=3/4 \implies p(0.5h)=1/2 and p(2h)=15/16p(2h)=15/16.

The way we have defined p(t)p(t) above, it is the probability that at least 11 accident happens in time interval tt. Thus, as tt increases, p(t)p(t) also increases.


Trivia

  1. The assumption for the independence of disjoint intervals is called Memorylessness

  2. The probability of having at least one accident = 3/4. But there can be any number of accidents. Typically, it follows a Poisson Distribution, with parameter λ=\lambda = average number of accidents in that period.

  3. The waiting time for the next accident is a continuous random variable. Typically it follows an exponential distribution, with the parameter β=\beta = average waiting time for the next event.

  4. The occurrence rate (λ\lambda) and the average waiting time (β\beta) are related as β=1/λ\beta = 1/\lambda. This means that if on average, there are 3 meteorites per hour, then on average we wait for (1/3) hours to see the next meteorite.