Other Common Discrete Distributions | 其他常见离散分布概览

📚 Other Common Discrete Distributions | 其他常见离散分布概览

In IB Mathematics, the binomial distribution is often the first discrete model students meet. However, real-world data frequently come from situations that do not fit a fixed number of independent trials. This article gives an overview of the other common discrete distributions: the geometric, negative binomial, hypergeometric and Poisson distributions. We will examine their probability mass functions, means and variances, as well as the conditions under which each should be used.

在 IB 数学中，二项分布往往是同学们最先接触的离散模型。然而，现实世界中的数据常常并不符合“固定次数的独立试验”这一情境。本文概述其他常见的离散分布：几何分布、负二项分布、超几何分布和泊松分布。我们将讨论它们的概率质量函数、均值与方差，以及各自适用的条件。

1. Why Go Beyond the Binomial? | 为何超越二项分布？

The binomial distribution describes the number of successes in a fixed number of independent trials, each with the same probability of success. But many random processes are not structured that way. For instance, we may count the number of trials until the first success, or the number of defective items in a sample drawn without replacement, or the number of emails arriving in one hour. Each of these situations requires a different discrete distribution.

二项分布描述的是在固定次数的独立试验中，每次成功概率相同的前提下，成功次数的分布。但许多随机过程并不是这种结构。例如，我们可能统计“直到第一次成功所需的试验次数”，或者“无放回抽样中抽到的次品数量”，又或者“一小时内到达的电子邮件数”。这些情况都需要不同的离散分布。

The key to success in probability questions is to match the distribution to the real-world mechanism: fixed trials, waiting times, sampling with or without replacement, and rare events over time or space.

在概率问题中取得成功的关键，是将分布与实际机制相匹配：固定试验次数、等待时间、有放回或无放回抽样，以及时间或空间中的稀有事件。

2. Geometric Distribution: Modelling the First Success | 几何分布：首次成功的建模

The geometric distribution models the number of trials needed to obtain the first success in a sequence of independent Bernoulli trials, each with success probability \(p\). The only possible values are 1, 2, 3, … If \(X\) is the number of trials until the first success, then the probability that the first success occurs on trial \(x\) is that we must first have \(x-1\) failures, then one success.

几何分布用于描述在一系列独立伯努利试验中，获得第一次成功所需的试验次数，每次试验的成功概率为 \(p\)。其可能取值为 1, 2, 3, …。若 \(X\) 表示直到第一次成功所需的试验次数，那么第一次成功发生在第 \(x\) 次试验的概率要求：前 \(x-1\) 次都是失败，最后一次是成功。

P(X = x) = (1 – p)^x-1 p, x = 1, 2, 3, …

For example, rolling a fair die until we get a 6. The probability that the first 6 appears on the 4th roll is \(P(X=4) = (5/6)^3 \times (1/6)\).

例如，掷一颗均匀骰子直到出现 6。第一次出现 6 在第 4 次掷出的概率为 \(P(X=4) = (5/6)^3 \times (1/6)\)。

3. Key Properties of the Geometric Distribution | 几何分布的关键性质

For the geometric distribution defined as counting trials to the first success, the mean and variance have simple forms:

对于将几何分布定义为“直到首次成功的试验次数”时，其均值与方差具有简洁的形式：

E(X) = 1/p, Var(X) = (1 – p)/p²

These results make sense intuitively: if the probability of success is large, we expect to wait only a few trials. If \(p\) is small, the waiting time is longer.

这些结果在直观上是合理的：如果成功概率很大，我们预期只需等待很少的试验；如果 \(p\) 很小，等待时间就会更长。

A famous property of the geometric distribution is its lack of memory. The conditional probability that we need at least \(x+y\) trials given that we have already waited \(x\) trials is the same as the probability we need at least \(y\) trials from the start. This is because the trials are independent.

几何分布的一个著名性质是“无记忆性”。在已经等待了 \(x\) 次试验的条件下，还需要至少 \(y\) 次试验的概率，与一开始就需要至少 \(y\) 次试验的概率相同。这是因为各次试验相互独立。

Some textbooks define the geometric distribution as the number of failures before the first success, with values 0,1,2,…. In that case, the PMF is \(P(X=x) = (1-p)^x p\) and the mean is \((1-p)/p\). Always check which definition your exam uses.

有些教材将几何分布定义为“第一次成功之前的失败次数”，其取值为 0, 1, 2, …。此时概率质量函数为 \(P(X=x) = (1-p)^x p\)，均值为 \((1-p)/p\)。请务必看清考试采用哪种定义。

4. Negative Binomial Distribution: Waiting for the r-th Success | 负二项分布：等待第 r 次成功

The negative binomial distribution generalises the geometric distribution. It models the number of trials needed to obtain \(r\) successes in independent Bernoulli trials with success probability \(p\). If \(X\) is the total number of trials required, then the \(r\)-th success occurs on the \(x\)-th trial. In the previous \(x-1\) trials, there must be \(r-1\) successes and \(x-r\) failures, arranged in any order.

负二项分布是几何分布的推广。它描述在成功概率为 \(p\) 的独立伯努利试验中，获得第 \(r\) 次成功所需的试验次数。若 \(X\) 为所需试验总数，则第 \(r\) 次成功发生在第 \(x\) 次试验。在前 \(x-1\) 次试验中，必须有 \(r-1\) 次成功和 \(x-r\) 次失败，且排列顺序任意。

P(X = x) = C(x-1, r-1) (1-p)^x-r p^r, x = r, r+1, …

Here \(C(x-1, r-1)\) is the number of ways to place the \(r-1\) successes among the first \(x-1\) trials. The mean and variance are:

其中 \(C(x-1, r-1)\) 是前 \(x-1\) 次试验中放置 \(r-1\) 次成功的方法数。均值和方差为：

E(X) = r/p, Var(X) = r(1-p)/p²

When \(r=1\), the negative binomial reduces to the geometric distribution.

当 \(r=1\) 时，负二项分布就退化为几何分布。

5. Hypergeometric Distribution: Sampling Without Replacement | 超几何分布：无放回抽样

The hypergeometric distribution arises when we sample \(n\) items without replacement from a finite population of size \(N\), in which \(K\) items are labelled “successes”. Let \(X\) be the number of successes in the sample. Because sampling is without replacement, the trials are not independent, so the binomial distribution is not appropriate.

当我们从大小为 \(N\) 的有限总体中不放回地抽取 \(n\) 个物品，而总体中有 \(K\) 个物品被标为“成功”时，就得到超几何分布。设 \(X\) 为样本中的成功次数。由于抽样不放回，试验并不独立，因此二项分布并不适用。

P(X = x) = C(K, x) × C(N-K, n-x) / C(N, n)

The possible values of \(x\) are restricted by the sample size and the number of successes and failures in the population: \(\max(0, n-(N-K)) \leq x \leq \min(n, K)\).

\(x\) 的可能取值受到样本量以及总体中成功与失败数量的限制：\(\max(0, n-(N-K)) \leq x \leq \min(n, K)\)。

The mean is \(E(X) = nK/N\). The variance contains a correction factor for sampling without replacement:

其均值为 \(E(X) = nK/N\)。方差包含一个针对无放回抽样的修正因子：

Var(X) = n (K/N) (1 – K/N) × (N-n)/(N-1)

If the population size \(N\) is very large compared with the sample size \(n\), the hypergeometric distribution is well approximated by the binomial distribution with \(p = K/N\).

当总体大小 \(N\) 相对于样本量 \(n\) 很大时，超几何分布可以用参数 \(p = K/N\) 的二项分布很好地近似。

6. Poisson Distribution: Modelling Rare Events | 泊松分布：稀有事件建模

The Poisson distribution models the number of events occurring in a fixed interval of time or space, when events happen independently and at a constant average rate \(\lambda\). For example, the number of calls received by a call centre in one hour, the number of bacteria in a sample of water, or the number of accidents at a junction in one week.

泊松分布用于描述在固定时间或空间区间内事件发生的次数，前提是事件相互独立且以恒定的平均速率 \(\lambda\) 发生。例如，呼叫中心一小时内接到的电话数、水样中的细菌数，或一个路口一周内的事故数。

P(X = x) = e^-λ λ^x / x!, x = 0, 1, 2, …

Here \(e \approx 2.71828\), and \(x!\) denotes the factorial of \(x\).

其中 \(e \approx 2.71828\)，\(x!\) 表示 \(x\) 的阶乘。

One remarkable property of the Poisson distribution is that its mean and variance are equal:

泊松分布的一个显著特征是均值与方差相等：

E(X) = λ, Var(X) = λ

This equality is a quick check for whether a Poisson model might be suitable for a given data set.

这一等式可以用来快速判断泊松模型是否适合某一组数据。

7. When Can We Use the Poisson Approximation? | 何时可用泊松近似？

The Poisson distribution can approximate the binomial distribution when the number of trials \(n\) is large and the success probability \(p\) is small. In this case, the average number of successes is \(\lambda = np\). A commonly used rule of thumb is that the approximation is good when \(n \geq 50\) and \(p \leq 0.1\), or when \(n \geq 20\) and \(p \leq 0.05\).

当二项分布的试验次数 \(n\) 很大而成功概率 \(p\) 很小时，泊松分布可用来近似二项分布。此时平均成功次数为 \(\lambda = np\)。常用的经验法则是：当 \(n \geq 50\) 且 \(p \leq 0.1\)，或 \(n \geq 20\) 且 \(p \leq 0.05\) 时，近似效果较好。

Why does this work? For small \(p\), the binomial probability of any particular number of successes can be shown to approach the Poisson formula as \(n \to \infty\). Intuitively, rare events occurring over many independent trials behave like random events in time.

为什么这样近似可行？当 \(p\) 很小时，可以证明在 \(n \to \infty\) 时，二项分布中某个成功次数的概率趋近于泊松公式。直观地说，大量独立试验中的稀有事件，其行为很像时间中的随机事件。

However, the Poisson approximation is not appropriate for sampling without replacement. In that case, the hypergeometric distribution should be used, unless the population is very large.

然而，泊松近似不适用于无放回抽样。在这种情况下应使用超几何分布，除非总体非常大。

8. Choosing the Right Discrete Model | 选择正确的离散模型

When faced with an exam problem, ask yourself three questions:

面对一道考题时，请问自己三个问题：

Are we counting trials or counting events? Trials suggest geometric/negative binomial; events over time/space suggest Poisson.
中文：我们在统计试验次数还是事件次数？ 试验次数暗示几何/负二项分布；时间/空间中的事件次数暗示泊松分布。
Is sampling with or without replacement? With replacement or independent trials gives binomial/geometric; without replacement gives hypergeometric.
中文：抽样是有放回还是无放回？ 有放回或独立试验用二项/几何分布；无放回用超几何分布。
Are we looking for the first success, the r-th success, or a fixed number of successes? First success means geometric; r-th success means negative binomial; fixed number of successes in a fixed number of trials means binomial.
中文：我们在寻找第一次成功、第 r 次成功，还是固定次数中的成功数？ 第一次成功用几何分布；第 r 次成功用负二项分布；固定试验次数中的成功数用二项分布。

Always read the wording carefully. Phrases like “until”, “on the 5th attempt” or “before the first success” signal a waiting-time distribution.

务必仔细阅读题干措辞。“直到”“在第5次尝试时”或“在第一次成功之前”等表述，提示使用等待时间类分布。

9. Worked Example 1: Geometric or Hypergeometric? | 例题1：几何还是超几何？

Problem. A box contains 10 red and 90 blue balls. Balls are drawn one at a time with replacement. Find the probability that the first red ball appears on the 5th draw.

题目。一个盒子中有 10 个红球和 90 个蓝球。每次抽取一个球，有放回。求第一次出现红球发生在第 5 次抽取的概率。

Solution. With replacement, each draw has \(p = 10/100 = 0.1\) for red. The waiting time until the first red follows a geometric distribution. Thus:

解答。因为有放回，每次抽到红球的概率为 \(p = 10/100 = 0.1\)。直到第一次出现红球的等待时间服从几何分布。因此：

P(First red on 5th draw) = (0.9)⁴ × (0.1) ≈ 0.0656

If instead the draws were without replacement, the distribution would be hypergeometric, and the probability would be \(\frac{C(10,1) \times C(90,4)}{C(100,5)}\).

如果改为无放回抽取，则该分布为超几何分布，概率为 \(\frac{C(10,1) \times C(90,4)}{C(100,5)}\)。

10. Worked Example 2: Poisson in Context | 例题2：泊松分布的应用

Problem. Calls at a service centre arrive at an average rate of 3 per 10 minutes, following a Poisson distribution. Find the probability that exactly 5 calls arrive in a 10-minute period.

题目。某客服中心每 10 分钟平均接到 3 个电话，电话数服从泊松分布。求在 10 分钟内恰好接到 5 个电话的概率。

Solution. Here \(\lambda = 3\). Using the Poisson PMF:

解答。此处 \(\lambda = 3\)。利用泊松概率质量函数：

P(X = 5) = e^-3 × 3⁵ / 5! ≈ 0.1008

To find the probability that at most 2 calls arrive in 5 minutes, first adjust the rate: in 5 minutes, \(\lambda = 1.5\). Then \(P(X \leq 2) = e^{-1.5}(1 + 1.5 + 1.5²/2) ≈ 0.8088\). Remember to scale \(\lambda\) to the interval being considered.

若要计算 5 分钟内最多接到 2 个电话的概率，需要先将速率调整为每 5 分钟 \(\lambda = 1.5\)。于是 \(P(X \leq 2) = e^{-1.5}(1 + 1.5 + 1.5²/2) ≈ 0.8088\)。记住要根据所考虑的区间按比例调整 \(\lambda\)。

11. Common Exam Pitfalls and Tips | 常见考点与陷阱

Students often make the following mistakes:

同学们经常犯以下错误：

Using the binomial distribution when trials are not independent, such as sampling without replacement.
中文：在试验并不独立时使用二项分布，例如无放回抽样。
Confusing the two versions of the geometric distribution: trials until first success versus failures before first success. Check the support of the variable.
中文：混淆几何分布的两种定义：直到首次成功的试验次数，与首次成功前的失败次数。注意变量的取值集合。
Forgetting to adjust \(\lambda\) when the time interval changes in Poisson problems.
中文：在泊松问题中改变时间区间时忘记调整 \(\lambda\)。
Using \(C(x, r)\) instead of \(C(x-1, r-1)\) in the negative binomial formula.
中文：在负二项分布公式中误用 \(C(x, r)\) 而不是 \(C(x-1, r-1)\)。

To avoid these pitfalls, always write down the distribution family, its parameters, and the required probability expression before plugging in numbers.

为避免这些陷阱，在代入数值之前，务必先写出分布类型、参数以及所需概率的表达式。

12. Summary | 总结

In this article we have reviewed the geometric, negative binomial, hypergeometric and Poisson distributions. The geometric distribution counts trials to the first success, and the negative binomial counts trials to the \(r\)-th success. The hypergeometric distribution describes sampling without replacement from a finite population, while the Poisson distribution models rare events occurring at a constant average rate.

本文我们复习了几何分布、负二项分布、超几何分布和泊松分布。几何分布统计直到首次成功的试验次数，负二项分布统计直到第 \(r\) 次成功的试验次数。超几何分布描述有限总体中的无放回抽样，而泊松分布则用于以恒定平均速率发生的稀有事件。

Remember that each distribution answers a different question. Recognising the underlying process is more important than memorising formulas. With practice, you will quickly identify the correct model and apply it confidently in your IB exams.

请记住，每种分布回答的是不同的问题。识别背后的过程比记忆公式更重要。通过练习，你将能快速识别正确的模型，并在 IB 考试中自信地应用。

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Other Common Discrete Distributions | 其他常见离散分布概览