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The Poisson distribution may be useful to model events such as
The Poisson distribution may be useful to model events such as
泊松分布模型可以用来模拟事件，比如
* The number of meteorites greater than 1 meter diameter that strike Earth in a year
* The number of meteorites greater than 1 meter diameter that strike Earth in a year
* The number of laser photons hitting a detector in a particular time interval
* The number of laser photons hitting a detector in a particular time interval
*一年内撞击地球的直径大于1米的陨石数量
* 一年内撞击地球的直径大于1米的陨石数量
*晚上10点到11点到达急诊室的病人人数
* 晚上10点到11点到达急诊室的病人人数
*在特定时间间隔内撞击探测器的激光光子数
* 在特定时间间隔内撞击探测器的激光光子数
===Assumptions and validity假设与有效条件===
===Assumptions and validity假设与有效条件===
The Poisson distribution is an appropriate model if the following assumptions are true:{{r|Koehrsen2019}}
The Poisson distribution is an appropriate model if the following assumptions are true:{{r|Koehrsen2019}}
The Poisson distribution is an appropriate model if the following assumptions are true:
The Poisson distribution is an appropriate model if the following assumptions are true:
以下假设成立时，泊松分布模型适用:
* {{mvar|k}} is the number of times an event occurs in an interval and {{mvar|k}} can take values 0, 1, 2, ....
* {{mvar|k}} is the number of times an event occurs in an interval and {{mvar|k}} can take values 0, 1, 2, ....
* Two events cannot occur at exactly the same instant; instead, at each very small sub-interval exactly one event either occurs or does not occur.
* Two events cannot occur at exactly the same instant; instead, at each very small sub-interval exactly one event either occurs or does not occur.
*事件在一个时间间隔内发生且{mvar | k}可以取值0，1，2，...
* 事件在一个时间间隔内发生且{mvar | k}可以取值0，1，2，...；
*一个事件的发生不影响第二个事件发生的概率
* 一个事件的发生不影响第二个事件发生的概率，也就是时间发生相互独立；
*事件发生的平均速率与任何事件无关。为简单起见，通常假定其为常数，但实际上可能随时间而变化
* 事件发生的平均速率与任何事件无关。为简单起见，通常假定其为常数，但实际上可能随时间而变化；
*两个事件不可能在完全相同的时刻发生，即在每一小段的时间内正好有一个事件发生或不发生
* 两个事件不可能在完全相同的时刻发生，即在每一小段的时间内正好有一个事件发生或不发生。
If these conditions are true, then {{mvar|k}} is a Poisson random variable, and the distribution of {{mvar|k}} is a Poisson distribution.
If these conditions are true, then {{mvar|k}} is a Poisson random variable, and the distribution of {{mvar|k}} is a Poisson distribution.
If these conditions are true, then is a Poisson random variable, and the distribution of is a Poisson distribution.
If these conditions are true, then is a Poisson random variable, and the distribution of is a Poisson distribution.
如果这些条件成立，那么它是一个泊松随机变量，其分布是一个'''<font color="#ff8000"> {泊松分佈 Poisson distribution</font>'''。
如果这些条件成立，那么它是一个泊松随机变量，其分布是一个'''<font color="#ff8000"> {泊松分布 Poisson distribution</font>'''。
The Poisson distribution is also the limit of a binomial distribution, for which the probability of success for each trial equals divided by the number of trials, as the number of trials approaches infinity (see Related distributions).
The Poisson distribution is also the limit of a binomial distribution, for which the probability of success for each trial equals divided by the number of trials, as the number of trials approaches infinity (see Related distributions).
每次试验的成功概率除以总试验次数，随着试验的数量趋于无穷大，泊松分布也是'''<font color="#ff8000"> 二项式分布Binomial distribution</font>'''的极限。（可参考相关分布）
This equation is the probability mass function (PMF) for a Poisson distribution.
This equation is the probability mass function (PMF) for a Poisson distribution.
这个方程就是概率质量函数的泊松分布。
: <math>P(k \text{ events in interval } t) = \frac{(r t)^k e^{-r t}}{k!}</math>
: <math>P(k \text{ events in interval } t) = \frac{(r t)^k e^{-r t}}{k!}</math>
< math > p (k text { events in interval } t) = frac {(r t) ^ k e ^ {-r t }{ k！{/math >
< math > p (k text { events in interval } t) = frac {(r t) ^ k e ^ {-r t }{ k！{/math >
Ugarte and colleagues report that the average number of goals in a World Cup soccer match is approximately 2.5 and the Poisson model is appropriate.
Ugarte and colleagues report that the average number of goals in a World Cup soccer match is approximately 2.5 and the Poisson model is appropriate.
乌加特和他的同事们在一篇报道中提到，世界杯足球赛的平均进球数约为2.5个，这也适用泊松模型。
Because the average event rate is 2.5 goals per match, ''λ'' = 2.5.
Because the average event rate is 2.5 goals per match, ''λ'' = 2.5.
Suppose that astronomers estimate that large meteorites (above a certain size) hit the earth on average once every 100 years (λ = 1 event per 100 years), and that the number of meteorite hits follows a Poisson distribution. What is the probability of = 0 meteorite hits in the next 100 years?
Suppose that astronomers estimate that large meteorites (above a certain size) hit the earth on average once every 100 years (λ = 1 event per 100 years), and that the number of meteorite hits follows a Poisson distribution. What is the probability of = 0 meteorite hits in the next 100 years?
假设天文学家估计，大型陨石(超过一定大小)平均每100年撞击地球一次(= 每100年撞击一次) ，而且陨石撞击的次数紧随'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''之后。在接下来的100年里，被陨石击中k=0的概率是多少？
假设天文学家估计，大型陨石(超过一定大小)平均每100年撞击地球一次(= 每100年撞击一次) ，而且陨石撞击的次数紧随泊松分布之后。在接下来的100年里，被陨石击中k=0的概率是多少？
In an example above, an overflow flood occurred once every 100 years (λ = 1). The probability of no overflow floods in 100 years was roughly 0.37, by the same calculation.
In an example above, an overflow flood occurred once every 100 years (λ = 1). The probability of no overflow floods in 100 years was roughly 0.37, by the same calculation.
在上面的一个例子中，洪水每100年发生泛滥一次(''λ'' = 1)。根据同样的计算，100年内不会有洪水泛滥的概率大约是0.37。
在上面的一个例子中，洪水每100年发生泛滥一次(''λ''= 1)。根据同样的计算，100年内不会有洪水泛滥的概率大约是0.37。
In general, if an event occurs on average once per interval (λ = 1), and the events follow a Poisson distribution, then . In addition, P(exactly one event in next interval) = 0.37, as shown in the table for overflow floods.
In general, if an event occurs on average once per interval (λ = 1), and the events follow a Poisson distribution, then . In addition, P(exactly one event in next interval) = 0.37, as shown in the table for overflow floods.
一般来说，如果一个事件平均每个时间间隔发生一次(''λ'' = 1) ，并且事件遵循'''<font color="#ff8000"> 泊松分佈Poisson distribution.</font>'''，那么p (下一个间隔中正好有一个事件) = 0.37，如洪水泛滥的表所示。
一般来说，如果一个事件平均每个时间间隔发生一次(''λ''= 1) ，并且事件遵循泊松分布，那么p (下一个间隔中正好有一个事件) = 0.37，如洪水泛滥的表所示。