Using the Poisson distribution to approximate the binomial distribution | 用泊松分布近似二项分布

📚 Using the Poisson distribution to approximate the binomial distribution | 用泊松分布近似二项分布

In A-Level Mathematics, the binomial distribution models the number of successes in a fixed number of independent trials, each with the same probability of success. However, when the number of trials is large and the success probability is small, calculating binomial probabilities directly becomes cumbersome. In such cases, the Poisson distribution provides a convenient and accurate approximation. This article explains the conditions, method, and practical use of this approximation, with worked examples and exam insights.

在A-Level数学中，二项分布用于描述固定次数的独立试验中成功次数的概率，每次试验成功概率相同。然而，当试验次数很大且成功概率很小时，直接计算二项概率会非常繁琐。此时，泊松分布可作为一种方便且精确的近似方法。本文介绍该近似的条件、方法及实际应用，并配有例题和考试要点。

1. Understanding the Binomial Distribution | 理解二项分布

The binomial distribution, denoted X ~ B(n, p), describes the number of successes X in n independent trials, where each trial has probability p of success. Its probability mass function is:

二项分布记作 X ~ B(n, p)，描述在 n 次独立试验中成功的次数 X，每次试验成功的概率为 p。其概率质量函数为：

P(X = r) = ⁿC_r p^r (1−p)^n−r, for r = 0, 1, 2, …, n.

P(X = r) = ⁿC_r p^r (1−p)^n−r，其中 r = 0, 1, 2, …, n。

Here, ⁿC_r (or 'n choose r') is the binomial coefficient, equal to n! / [r! (n−r)!].

其中 ⁿC_r（或“n 选 r”）是二项式系数，等于 n! / [r! (n−r)!]。

This model is useful for counts of 'successes' in fixed-size samples, such as the number of defective items in a batch or the number of heads in coin tosses.

该模型适用于固定样本量中的“成功”计数，例如一批产品中的次品数或抛硬币中正面的次数。

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2. Understanding the Poisson Distribution | 理解泊松分布

The Poisson distribution, written X ~ Po(λ), models the number of events occurring in a fixed interval of time or space when events happen independently at a constant average rate λ. Its probability function is:

泊松分布记作 X ~ Po(λ)，用于模拟在固定时间或空间区间内事件发生的次数，当事件独立发生且平均发生率为常数 λ 时。其概率函数为：

P(X = r) = (e^−λ λ^r) / r!, for r = 0, 1, 2, …

P(X = r) = (e^−λ λ^r) / r!，其中 r = 0, 1, 2, …

The parameter λ is both the mean and the variance of the distribution. This makes it particularly simple to use when λ is small or moderate.

参数 λ 既是分布的均值也是方差。这使得它在 λ 较小或中等时使用特别简便。

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3. The Need for Approximation | 近似法的必要性

direct binomial calculations involve enormous powers and factorials. For example, finding ²⁰⁰C₃ × (0.01)³ × (0.99)¹⁹⁷ manually or even with a standard calculator can be time‑consuming and error‑prone. Moreover, statistical tables rarely cover large n for every p.

当 n 很大（例如 n = 200）而 p 很小（例如 p = 0.01）时，直接进行二项计算会涉及巨大的幂和阶乘。例如，手工或用普通计算器求 ²⁰⁰C₃ × (0.01)³ × (0.99)¹⁹⁷ 既费时又容易出错。此外，统计用表很少覆盖每种 p 值下的大 n。

In such situations, the Poisson distribution offers an elegant shortcut, provided certain conditions are met. It turns a complicated binomial expression into a simpler exponential one, often requiring only a few terms to be added for cumulative probabilities.

在这种情况下，只要满足一定条件，泊松分布就能提供一种简洁的捷径。它将复杂的二项式表达式转化为较简单的指数式，求累积概率时通常只需累加少数几项即可。

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4. Conditions for a Valid Approximation | 有效近似的条件

For the Poisson approximation to be valid, two conditions must hold:

要使用泊松近似有效，必须满足两个条件：

(i) The number of trials n must be large.

(i) 试验次数 n 必须很大。

(ii) The probability of success p on each trial must be small.

(ii) 每次试验的成功概率 p 必须很小。

These ensure that the binomial distribution’s shape closely matches that of a Poisson distribution with the same mean. Essentially, when n is large and p is tiny, the binomial probability of rare events depends almost entirely on the expected number of successes, not on the exact values of n and p separately.

这些条件保证了二项分布的形状与具有相同均值的泊松分布非常接近。本质上看，当 n 很大且 p 很小时，罕见事件的二项概率几乎完全取决于期望成功次数，而非 n 和 p 的具体数值。

The approximation is often used when dealing with rare events such as industrial defects, road accidents, or call centre calls arriving per minute.

该近似常用于处理罕见事件，例如工业缺陷、交通事故或每分钟打入呼叫中心的电话。

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5. The Rule of Thumb: n > 50 and p < 0.1 | 经验法则：n > 50 且 p < 0.1

should be less than 0.1. More precisely, many textbooks state n ≥ 50 and p ≤ 0.1, or n ≥ 100 and p ≤ 0.05 for a highly accurate approximation.

在 Edexcel A-Level 考试大纲中，使用泊松近似的标准经验法则是 n 应大于 50，p 应小于 0.1。更精确地说，许多教材指出 n ≥ 50 且 p ≤ 0.1，或者若要求高精度近似则 n ≥ 100 且 p ≤ 0.05。

ied. Check that np < 5 is sometimes given as an additional indicator that λ is not too large, because Poisson tables often go up to λ = 10 or 20, but for very large λ a normal approximation might be better.

即使 n = 30 且 p = 0.05，近似结果也可能可以接受，但为应对考试，你应该只在两个条件都明显满足时才使用该近似。注意 np < 5 有时作为 λ 不太大的补充指示，因为泊松表格通常达到 λ = 10 或 20，但对很大的 λ，正态近似可能更合适。

Typical exam questions will state the condition clearly, or ask you to justify why the approximation is suitable.

典型的考试题目会明确说明条件，或要求你解释为何适合使用该近似。

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6. Setting λ = np | 设 λ = np

n to the parameter of the Poisson distribution. Since E(X) = np for a binomial, we define:

关键一步是将二项分布的均值等同于泊松分布的参数。由于二项分布的 E(X) = np，我们定义：

λ = np

This ensures that the approximating Poisson distribution has the same expected number of successes.

这样可以确保近似所用的泊松分布具有相同的期望成功次数。

Then, for any r, P(X = r) ≈ (e^−np (np)^r) / r! . All binomial probabilities are replaced by Poisson probabilities using this λ.

然后，对于任何 r，有 P(X = r) ≈ (e^−np (np)^r) / r! 。所有二项概率都用这个 λ 对应的泊松概率来替代。

Remember that the Poisson distribution has no upper bound (r can be any non‑negative integer), whereas the binomial stops at n. The approximation will assign very small probabilities to r > n, which is acceptable as long as p is small.

请记住，泊松分布没有上界（r 可以是任何非负整数），而二项分布在 n 处停止。该近似会给 r > n 分配极小的概率，只要 p 很小，这就可以接受。

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7. Mathematical Justification (Limit Theorem) | 数学依据（极限定理）

t n → ∞ and p → 0 in such a way that np = λ remains constant, the binomial expression converges to the Poisson probability:

泊松近似不仅仅是一个经验法则，它有着严格的数学基础。如果我们将二项概率质量函数中的 n → ∞、p → 0 且保持 np = λ 为常数，则二项表达式收敛为泊松概率：

lim_{n→∞, p→0, np=λ} ⁿC_r p^r (1−p)^n−r = (e^−λ λ^r) / r!

This result is known as the “law of rare events”. The derivation uses the fact that ⁿC_r ≈ n^r/r! for large n and small r, and the exponential limit (1 − λ/n)ⁿ → e^−λ.

这一结果被称为“稀有事件定律”。推导中用到了当 n 很大且 r 很小时 ⁿC_r ≈ n^r/r!，以及指数极限 (1 − λ/n)ⁿ → e^−λ。

In A‑Level exams you are not required to reproduce the proof, but knowing it strengthens your understanding of why the approximation works and when it might break down.

在A‑Level考试中，你不需要重现该证明，但了解它能加深你对近似为何有效以及何时会失效的理解。

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8. Worked Example 1: Direct Calculation | 例题1：直接计算

A machine produces components, and the probability that any one component is defective is 0.02. If a sample of 100 components is taken, find the probability that exactly 2 are defective, using a suitable approximation.

一台机器生产零件，每个零件是次品的概率为 0.02。若抽取 100 个零件作为样本，用适当的近似方法求恰好有 2 个次品的概率。

t, check conditions: n = 100 > 50, p = 0.02 < 0.1, so a Poisson approximation is suitable.

首先检查条件：n = 100 > 50，p = 0.02 < 0.1，故适合使用泊松近似。

Calculate λ = np = 100 × 0.02 = 2. Let X ~ Po(2) approximate the binomial number of defectives.

计算 λ = np = 100 × 0.02 = 2。令 X ~ Po(2) 近似表示二项分布的次品数。

Then P(X = 2) ≈ (e⁻² × 2²) / 2! = (e⁻² × 4) / 2 = 2 e⁻².

则 P(X = 2) ≈ (e⁻² × 2²) / 2! = (e⁻² × 4) / 2 = 2 e⁻²。

Using e⁻² 135335, we get 2 × 0.135335 = 0.27067. So the probability is approximately 0.271 (3 d.p.).

利用 e⁻² ≈ 0.135335，得到 2 × 0.135335 = 0.27067。因此概率约为 0.271（保留三位小数）。

For comparison, the exact binomial probability is ¹⁰⁰C₂ 0.02² × 0.98⁹⁸, which evaluates to about 0.2734. The approximation is already very close.

作为比较，精确的二项概率为 ¹⁰⁰C₂ × 0.02² × 0.98⁹⁸，计算值约为 0.2734。近似值已经非常接近。

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9. Worked Example 2: Multiple Probabilities | 例题2：多个概率

In the same scenario, find the probability that there are at most 3 defective components.

在上面的场景中，求次品零件数不超过 3 的概率。

Using X ~ Po(2), we need P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3).

使用 X ~ Po(2)，需要计算 P(X ≤ 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)。

P(X=0) = e⁻² × 2⁰/0! = e⁻² ≈ 0.1353

P(X=0) = e⁻² × 2⁰/0! = e⁻² ≈ 0.1353

P(X=1) = e⁻² × 2¹/1! = 2 e⁻² ≈ 0.2707

P(X=1) = e⁻² × 2¹/1! = 2 e⁻² ≈ 0.2707

P(X=2) = e⁻² × 2²/2! = 2 e⁻² ≈ 0.2707

P(X=2) = e⁻² × 2²/2! = 2 e⁻² ≈ 0.2707

P(X=3) = e⁻² × 2³/3! = (8 e⁻²)/6 ≈ 1.3333 × 0.135335 ≈ 0.1804

P(X=3) = e⁻² × 2³/3! = (8 e⁻²)/6 ≈ 1.3333 × 0.135335 ≈ 0.1804

Summing these: 0.1353 + 0.2707 + 0.2707 + 0.1804 = 0.8571.

求和得：0.1353 + 0.2707 + 0.2707 + 0.1804 = 0.8571。

ial cumulative probability for this case is about 0.858, again showing excellent agreement.

因此 P(X ≤ 3) ≈ 0.857。该情况下的精确二项累积概率约为 0.858，再次显示了高度一致性。

In an exam, you may be asked to use the Poisson cumulative distribution tables to find these values directly, especially for larger λ.

在考试中，你可能会被要求直接使用泊松累积分布表来查找这些值，特别当 λ 较大时。

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10. Comparison of Binomial and Poisson Probabilities | 二项分布与泊松分布概率对比

le lists r, the exact binomial probability (B), the Poisson probability (P), and the absolute error.

我们构建一个小的对比表，n = 100，p = 0.02 (λ = 2)。表列出 r、精确二项概率 (B)、泊松概率 (P) 和绝对误差。

r	Binomial B(100,0.02)	Poisson Po(2)	|B−P|
0	0.1326	0.1353	0.0027
1	0.2707	0.2707	~0
2	0.2734	0.2707	0.0027
3	0.1823	0.1804	0.0019
4	0.0902	0.0902	~0

As the table shows, the largest error here is only about 0.0027, which is perfectly acceptable in most practical and exam settings. The approximation works remarkably well because n is large and p is small.

如表中所示，这里最大的误差只有约 0.0027，在大多数实际和考试场合完全可接受。由于 n 大、p 小，该近似效果非常出色。

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11. Accuracy, Errors, and When Not to Use | 精度、误差及不适用情况

the binomial distribution is somewhat symmetric, and the Poisson (λ = 10) would assign non‑ hich is impossible in the binomial. The approximation would then be noticeably inaccurate.

如果 p 不是很小，泊松近似效果会变差。例如 n = 50, p = 0.2（因此 np = 10），二项分布具有一定对称性，而泊松分布（λ = 10）会给 r > 50 分配不可忽略的概率，这在二项分布中是不可能的。此时近似误差会很明显。

Also, when np is large (say > 20), the binomial distribution itself starts

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