📚 Using the Poisson Distribution to Approximate the Binomial Distribution | 使用泊松分布近似二项分布
ecomes unmanageable by hand. When n is large and p is small, the Poisson distribution provides a much simpler approximation because it depends only on the single parameter λ = np. This article explains when the approximation is allowed, why it works, how to apply it accurately in an exam, and where candidates typically lose marks.
在 A-Level 统计中，二项分布有时会涉及非常大的幂和非常大的组合数，纯手算很快就变得难以处理。当 n 很大且 p 很小时，泊松分布能提供一种简单得多的近似，因为它只依赖单一参数 λ = np。本文将说明该近似何时允许使用、为什么成立、考试中如何准确应用，以及考生通常在哪里失分。
1. Why the Binomial Can Become Unmanageable | 为什么二项分布会变得难以计算
r any value of n, but correctness is not the same as practicality. To find P(X = r) when X ~ on C(200, 3) equals 1 313 400, and you still have to multiply by (0.01)³ and by (0.99)¹⁹⁷.
二项概率公式对任何 n 值都是完全正确的，但"正确"并不等于"实用"。当 X ~ B(n, p) 时，为求 P(X = r)，你必须计算一个组合数和两个幂，而这些数字可能极其巨大。例如 n = 200、r = 3 时，组合数 C(200, 3) 等于 1 313 400，而且你还要再乘以 (0.01)³ 与 (0.99)¹⁹⁷。
se tables only list the cumulative distribution. A calculator will handle the computation, but it is slow and offers no insight into the structure of the problem. An approximation that captures the essential behaviour of the distribution is therefore valuable, both in examinations and in applied work.
纸质二项分布表很少超过 n = 20 或 n = 30，而且许多表只列出累积分布。计算器虽然能算，但速度慢，也无法让你看清问题的结构。因此，一种能抓住分布本质特征的近似，在考试和实际应用中都很有价值。
d its probabilities are easy to compute recursively because each term is obtained from the previous one by a simple multiplication.
泊松近似正是这样的工具。它不需要 n 和 p 两个参数，只需要 λ = np，而且各项概率可以通过简单的递推快速计算——每一项都由前一项乘以一个简单因子得到。
2. When the Poisson Approximation Is Valid | 泊松近似何时适用
Suppose X ~ en n is large, p is small, and the product np is moderate, so that the binomial probabilities become very close to the corresponding Poisson probabilities.
假设 X ~ B(n, p)。泊松近似把 X 替换为一个均值为 λ = np 的泊松变量，记作 Po(λ)。当 n 很大、p 很小且乘积 np 适中时可以使用此近似，此时二项概率会非常接近对应的泊松概率。
- n is large (commonly n ≥ 50) | n 很大（通常 n ≥ 50）
- p is small (commonly p ≤ 0.1) | p 很小（通常 p ≤ 0.1）
- λ = np is moderate (often λ < 10) | λ = np 适中（通常 λ < 10）
hnique is to state the conditions explicitly in your answer, quote n and p, compute λ = np, and then write "so a Poisson approximation is appropriate". That sentence is often worth a mark on its own.
这些阈值是经验法则而非数学定律，不同考试局的说法略有差异。最稳妥的应试做法是：在答案中明确写出条件，列出 n 与 p，算出 λ = np，然后写"因此泊松近似适用"。这句话本身往往就能得一分。
| Condition 条件 | Rule of thumb 经验法则 | Reason 原因 |
| n large n 很大 | n ≥ 50 | The limit that produces e⁻λ needs a large number of trials |
| p small p 很小 | p ≤ 0.1 | Keeps the variance np(1 − p) close to np |
| λ moderate λ 适中 | λ < 10 | Large λ calls for the normal approximation instead |
3. Comparing the Binomial and Poisson Distributions | 二项分布与泊松分布的比较
Both distributions are discrete, both take values r = 0, 1, 2, 3, … trials, each with the same success probability. A Poisson variable counts the number of rare events occurring in a fixed interval of time, length, area or volume, at a constant average rate.
两种分布都是离散的，都取值 r = 0, 1, 2, 3, …，都描述计数。区别在于计数背后的机制：二项变量统计的是"固定次数独立试验中成功的次数"，每次试验成功概率相同；泊松变量统计的是"在固定的时间、长度、面积或体积区间内稀有事件发生的次数"，事件以恒定平均速率出现。
The binomial probability mass function is
二项分布的概率质量函数为
P(X = r) = ⁿCᵣ pʳ (1 − p)ⁿ⁻ʳ
while the Poisson probability mass function is
而泊松分布的概率质量函数为
P(X = r) = e⁻λ λʳ / r!
where λ = np, e ≈ 2.71828, and r! = r × (r − 1) × … × 2 × 1 with 0! = 1. Notice that as p → 0 with λ = np fixed, the factor (1 − p)ⁿ⁻ʳ tends to e⁻λ, and the combination term tends to λʳ / r!.
其中 λ = np，e ≈ 2.71828，r! = r × (r − 1) × … × 2 × 1，且 0! = 1。可以看到，当 p → 0 而 λ = np 固定时，因子 (1 − p)ⁿ⁻ʳ 趋于 e⁻λ，组合数项趋于 λʳ / r!。
| Feature 特征 | Binomial B(n, p) 二项分布 | Poisson Po(λ) 泊松分布 |
| Parameters 参数 | two: n, p 两个：n, p | one: λ 一个：λ |
| Mean 均值 | np | λ |
| Variance 方差 | np(1 − p) | λ |
| Maximum value 最大值 | n | unbounded 无上界 |
4. The Single Parameter λ = np | 单一参数 λ = np
The parameter λ is the mean number of occurrences in the interval considered, and it is also the variance. This double role is the key to the approximation. A binomial variable has mean np and variance np(1 − p); when p is small, (1 − p) is close to 1, so np(1 − p) is close to np, and the two distributions have almost the same mean and almost the same variance.
参数 λ 是所考虑区间内事件发生次数的均值，同时也是方差。这种"一个参数兼两职"正是近似的关键。二项变量的均值为 np、方差为 np(1 − p)；当 p 很小时，(1 − p) 接近 1，于是 np(1 − p) 接近 np，两种分布便具有几乎相同的均值和几乎相同的方差。
The Poisson distribution Po(λ) therefore has both mean and variance equal to λ, which is why it can match the binomial very closely under these conditions. The smaller p is, the closer the match becomes.
因此泊松分布 Po(λ) 的均值和方差都等于 λ，这正是它在此条件下能与二项分布高度吻合的原因。p 越小，吻合程度越高。
In modelling terms, λ represents a rate. If a factory produces 500 items and 0.4% efective items per batch on average, and the count of defectives in a batch is approximately Po(2).
从建模角度看，λ 代表一种"速率"。如果工厂生产 500 件产品，其中 0.4% 是次品，则 λ = 500 × 0.004 = 2，即每批次平均 2 件次品，该批次的次品数近似服从 Po(2)。
5. Deriving the Approximation from the Binomial Formula | 从二项公式推导近似
h the binomial probability and write p = λ / n, where λ is fixed and n is allowed to grow.
这个近似并非巧合，而是一个真正的极限。从二项概率出发，令 p = λ / n，其中 λ 固定，n 趋于无穷。
P(X = r) = [n(n − 1) … (n − r + 1) / r!] × (λ / n)ʳ × (1 − λ / n)ⁿ⁻ʳ
Rearrange the factors so that the behaviour of each part is visible as n becomes large:
重新整理各因子，使 n 变大时每一部分的行为都清晰可见：
P(X = r) = [n(n − 1) … (n − r + 1) / nʳ] × (λʳ / r!) × (1 − λ / n)ⁿ × (1 − λ / n)⁻ʳ
Now examine the three limits separately, holding r and λ fixed.
现在分别考察三个极限，保持 r 与 λ 不变。
The first factor has r terms in the numerator and nʳ in the denominator, so it tends to 1. The second factor, λʳ / r! ve n at all and stays unchanged. The third factor uses the standard limit (1 + x / n)ⁿ → eˣ with x = −λ, so it tends to e⁻λ. The fourth factor tends to 1.
第一个因子分子有 r 项、分母为 nʳ，因此趋于 1。第二个因子 λʳ / r! 不含 n，保持不变。第三个因子利用标准极限 (1 + x / n)ⁿ → eˣ（取 x = −λ），趋于 e⁻λ。第四个因子趋于 1。
Multiplying the limits together gives the Poisson probability mass function.
把各极限相乘，即得泊松概率质量函数。
P(X = r) → e⁻λ λʳ / r!
The derivation shows exactly where each condition comes from: "n large" makes the first limit valid, and "p small" is equivalent to λ / n being small, which makes the exponential limit accurate.
该推导清楚地说明了每个条件的来源："n 很大"保证第一个极限成立；"p 很小"等价于 λ / n 很小，从而保证指数极限足够精确。
6. Worked Example 1 — A Single Probability | 例题一：单个概率
A manufacturer produces electronic components in batches of 200. Historically 1% of components are defective, independently of one another. A batch is selected at random. Use a Poisson approximation to find the probability that exactly 3 components are defective.
某厂商以每批 200 件生产电子元件。历史数据显示 1% 的元件是次品，且各元件是否次品相互独立。随机抽取一批，用泊松近似求恰好 3 件为次品的概率。
ble and check the conditions. Let X be the number of defective components in a batch, so X ~ Here n = 200 is large and p = 0.01 is small, so the Poisson approximation is appropriate.
第一步：定义变量并检验条件。设 X 为该批中的次品数，则 X ~ B(200, 0.01)。这里 n = 200 很大，p = 0.01 很小，因此泊松近似适用。
Step 2: compute the parameter. λ = np = 200 × 0.01 = 2, so X is approximately Po(2).
第二步：计算参数。λ = np = 200 × 0.01 = 2，故 X 近似服从 Po(2)。
Step 3: substitute into the Poisson formula with r = 3.
第三步：代入泊松公式，取 r = 3。
P(X = 3) = e⁻² × 2³ / 3! = e⁻² × 8 / 6 = 1.3333 × 0.13534 ≈ 0.1804
so the approximation is out by only about 0.001. Notice how much easier the Poisson calculation is.
因此近似概率为 0.180（保留三位小数）。作为对比，精确的二项值为 C(200, 3)(0.01)³(0.99)¹⁹⁷ ≈ 0.1814，误差仅约 0.001。可见泊松计算要简便得多。
7. Worked Example 2 — Cumulative and Tail Probabilities | 例题二：累积与尾部概率
Using the same batch, X ~ B(200, 0.01) approximated by Po(2), find P(X ≤ 2), P(X ≥ 4) and P(X ≥ 5).
沿用同一批次，X ~ B(200, 0.01) 用 Po(2) 近似，求 P(X ≤ 2)、P(X ≥ 4) 与 P(X ≥ 5)。
Compute the individual terms first, using
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