📚 The negative binomial distribution | 负二项分布
In probability theory and statistics, the negative binomial distribution models the number of trials needed to achieve a fixed number of successes in a sequence of independent Bernoulli trials. Unlike the binomial distribution, which fixes the number of trials and counts the successes, the negative binomial distribution fixes the number of successes and counts the trials required. This distribution is particularly useful when we are interested in the waiting time until a certain event occurs a specified number of times, such as the number of attempts before securing a set number of sales, goals, or defective items in a production line. In Edexcel A-Level Further Mathematics, the negative binomial distribution is covered in the Further Statistics 1 (FS1) module and extends the ideas introduced by the geometric distribution.
在概率论与统计学中，负二项分布用于描述在一系列独立的伯努利试验中获得固定成功次数所需的总试验次数的概率分布。与固定试验次数、统计成功次数的二项分布不同，负二项分布固定成功次数，而对达到这些成功所需要的试验次数进行建模。当我们关心某个事件发生指定次数之前所需的等待时间时——例如在达成一定数量的销售、进球或检出次品之前所需尝试的次数——该分布有着广泛的应用。在爱德思 A-Level 进阶数学中，负二项分布属于进阶统计1（FS1）模块的内容，并且是对几何分布概念的延伸。
1. Conditions for a Negative Binomial Model | 负二项模型的条件
A negative binomial distribution arises when the following conditions are satisfied: (1) We perform a sequence of independent trials. (2) Each trial has only two possible outcomes, conventionally labelled 'success' and 'failure'. (3) The probability of success, denoted by p, remains constant from trial to trial. (4) The trials continue until a predetermined number of successes, r, has been observed. The random variable X represents the total number of trials conducted, so X can take values r, r+1, r+2, ... It is essential to recognise that the number of successes r is fixed, while the total number of trials is the random variable. If these conditions are met, we write X ~ NB(r, p) or sometimes X ~ Negative Binomial(r, p).
当满足以下条件时，就产生负二项分布：(1) 我们进行一系列独立的试验。(2) 每次试验只有两种可能的结果，通常被称为“成功”和“失败”。(3) 成功的概率 p 在每次试验中保持恒定。(4) 试验一直进行，直到观察到事先指定的 r 次成功为止。随机变量 X 表示所进行的总试验次数，因此 X 可以取 r, r+1, r+2, …… 必须认清的是，成功的次数 r 是固定的，而总试验次数是随机变量。如果这些条件满足，我们就记作 X ~ NB(r, p) 或 X ~ 负二项(r, p)。
2. Probability Mass Function (PMF) | 概率质量函数
The probability that exactly x trials are needed to obtain r successes is given by the formula:
P(X = x) = C(x − 1, r − 1) × pr × (1 − p)x − r, x = r, r+1, r+2, ...
Here C(x − 1, r − 1) is the binomial coefficient, also written as (x − 1 choose r − 1), which counts the number of ways to arrange the r − 1 successes in the first x − 1 trials. The final trial must always be a success, so the probability of the r-th success occurring on the x-th trial is pr(1 − p)x − r or any integer r ≥ 1 and 0 < p < 1. When r = 1, the formula reduces to P(X = x) = p(1 − p)x − 1, which is the geometric distribution.
恰好需要 x 次试验才能获得 r 次成功的概率由下式给出：P(X = x) = C(x − 1, r − 1) × pr × (1 − p)x − r, x = r, r+1, r+2, … 其中 C(x − 1, r − 1) 是二项式系数，也写作 (x − 1 choose r − 1)，它计算了在前 x − 1 次试验中排列 r − 1 次成功的方式数目。最后一次试验必须是成功，因此第 r 次成功发生在第 x 次试验的概率为 pr(1 − p)x − r 乘以有利序列的数量。该概率质量函数对于任意整数 r ≥ 1 且 0 < p < 1 均成立。当 r = 1 时，公式简化为 P(X = x) = p(1 − p)x − 1，即几何分布。
3. Mean and Variance | 期望与方差
The expected value (mean) and variance of a negative binomial random variable X ~ NB(r, p) are:
E(X) = r / p
Var(X) = r (1 − p) / p2
These results can be derived using the moment generating function or by treating X as a sum of r independent geometric random variables, each with parameter p. The mean r/p tells us the average number of trials required to achieve r successes. Notice that the variance is larger than the mean whenever p < 0.5, indicating that the distribution can be over-dispersed. In exam settings, you are expected to use these formulas directly without deriving them unless instructed otherwise.
负二项随机变量 X ~ NB(r, p) 的期望值（均值）与方差为：E(X) = r / p，Var(X) = r (1 − p) / p²。这些结果可以利用矩母函数推导得出，或者把 X 看作 r 个独立的几何随机变量（每个参数为 p）之和。均值 r/p 表明了获得 r 次成功所需试验次数的平均值。注意当 p < 0.5 时，方差大于均值，意味着该分布可能是过度离散的。在考试中，除非另行要求，你可以直接使用这些公式而无需推导。
4. Relationship with Other Distributions | 与其他分布的关系
distribution becomes the geometric distribution Geo(p), which models the number of trials until the first success. Conversely, if we fix the number of trials n, the number of successes follows a binomial distribution Bin(n, p). The negative binomial distribution swaps the roles: the number of successes r is fixed and the number of trials X is random. This duality is powerful in solving probability problems. Additionally, the negative binomial can be expressed as a sum of r independent geometric variables, which is often used to prove its mean and variance formulas.
负二项分布与几何分布和二项分布密切相关。如前所述，当 r = 1 时，负二项分布就变成了几何分布 Geo(p)，它描述的是直到第一次成功所需试验次数的分布。反之，如果我们固定试验次数 n，成功次数服从二项分布 Bin(n, p)。负二项分布则对调了角色：成功次数 r 固定，试验次数 X 随机。这种对偶性在求解概率问题时非常有力。此外，负二项分布可以表示为 r 个独立几何随机变量之和，常用于证明其均值和方差公式。
5. Using the PMF to Calculate Probabilities | 使用 PMF 计算概率
When calculating probabilities, always ensure you set up the correct ranges. For X ~ hat the r-th success occurs on the 10th trial, we compute P(X = 10) using the PMF. If you need the probability that it takes at most n trials to get r successes, you would sum P(X = k) for k = r to n. Because manual summation can be tedious, Edexcel exams often allow the use of a calculator or provide cumulative probabilities. However, you should still be comfortable using the PMF formula.
在计算概率时，务必要设置正确的取值范围。对于 X ~ NB(r, p)，最小可能的取值为 r。许多学生会错误地使用二项分布。例如，要计算第 r 次成功发生在第 10 次试验的概率，我们使用 PMF 求出 P(X = 10)。如果需要至多 n 次试验就能获得 r 次成功的概率，应对 k = r 到 n 累加 P(X = k)。由于手工求和往往繁琐，爱德思考试通常允许使用计算器或会提供累积概率。尽管如此，你仍然应当能够熟练运用 PMF 公式。
6. Worked Example 1: Free Throws | 例题1：罚球
A basketball player has a free throw success rate of 0.7. She keeps shooting until she has made 3 successful free throws. Find the probability that she needs exactly 5 attempts.
Here we have r = 3, p = 0.7, and x = 5. Using the PMF: P(X = 5) = C(5 − 1, 3 − 1) × (0.7)3 × (0.3)5 − 3 bability is approximately 0.185. It is always wise to check that the answer lies between 0 and 1 and is reasonable given the parameters.
某篮球运动员罚球命中率为 0.7。她持续投篮，直到投中 3 次为止。求她恰好需要 5 次尝试的概率。
此处 r = 3, p = 0.7, x = 5。用 PMF 计算：P(X = 5) = C(5 − 1, 3 − 1) × (0.7)³ × (0.3)⁵⁻³ = C(4, 2) × 0.7³ × 0.3²。C(4, 2) = 6, 0.7³ = 0.343, 0.3² = 0.09。相乘得 6 × 0.343 × 0.09 = 0.18522。因此概率约为 0.185。合理的检验是答案介于 0 到 1 之间，且与给定参数相符。
7. Worked Example 2: Defective Items Inspection | 例题2：次品检查
In a quality control process, the probability that an item is defective is 0.1. An inspector checks items one by one and stops when the second defective item is found. Calculate the expected number of items inspected and the standard deviation.
/ (0.1)² = 1.8 / 0.01 = 180. Hence the standard deviation is √180 ≈ 13.42. Thus, on average, the inspector examines 20 items, with a standard deviation of about 13.4 items. Such a large spread is typical of negative binomial distributions when p is small.
在质量控制流程中，某产品为次品的概率是 0.1。检查员逐件检查，直到发现第二个次品时停止。计算应检查的物件数量的期望值和标准差。
设 X 为发现第二个次品前检查的物件数量，因此 r = 2, p = 0.1。则 E(X) = r/p = 2/0.1 = 20。方差 Var(X) = r(1 − p)/p² = 2 × 0.9 / (0.1)² = 1.8 / 0.01 = 180。因此标准差为 √180 ≈ 13.42。所以，平均来说，检查员会检查 20 件物品，标准差约 13.4 件。当 p 较小时，这种较大的离散程度是负二项分布的典型特征。
8. Applications in Real-World Modelling | 现实建模应用
The negative binomial distribution is used extensively in ecology, insurance, and queueing theory. In ecology, it models the number of quadrats sampled before finding a certain number of rare species, accounting for aggregation. In actuarial science, it models the number of claims before a policyholder accumulates a given number of claims, allowing for over-dispersion in claim frequency. In sports analytics, it can model the number of attempts until a player scores a certain number of goals. Because the variance can exceed the mean, the negative binomial often fits real-world count data better than the Poisson distribution, especially when events are clustered.
负二项分布在生态学、保险学和排队论中有广泛应用。在生态学中，它用于模拟在找到一定数目的稀有物种前所取样的样方数目，能够体现物种聚集效应。在精算学中，它用于模拟保单持有人累计到给定次数索赔之前的总索赔次数，允许索赔频率的过度离散。在体育分析中，它可以模拟球员打进一定数量的进球所需的射门次数。由于方差可以大于均值，负二项分布往往比泊松分布更能贴合现实世界的计数数据，尤其是当事 件存在聚集现象时。
9. The Negative Binomial as a Pascal Distribution | 作为帕斯卡分布的负二项分布
mes made that the negative binomial counts the number of failures before the r-th success, while the Pascal distribution counts the total number of trials. In Edexcel FS1, we use the version that counts the total trials X, so X takes values r, r+1, ... It is important to read exam questions carefully to identify which random variable is being measured. If the variable Y represents the number of failures, then Y = X − r, and E(Y) = r(1 − p)/p, with Var(Y) = r(1 − p)/p².
有些教材将负二项分布称为帕斯卡分布，尤其是当 r 为整数时。有时会区分二者：负二项分布计算第 r 次成功之前的失败次数，而帕斯卡分布计算总试验次数。在爱德思 FS1 课程中，我们使用的是计算总试验次数 X 的版本，因此 X 的取值是 r, r+1, … 仔细阅读考题以明确所测的随机变量至关重要。如果变量 Y 表示失败的次数，则 Y = X − r，E(Y) = r(1 − p)/p，Var(Y) = r(1 − p)/p²。
10. Parameter Estimation Using Method of Moments | 用矩法估计参数
Given a sample mean x̄ distribution, we can estimate r and p by equating sample moments to theoretical moments: x̄ = r/p and s² = r(1 − p)/p². Solving these equations yields p̂ = 1 − (x̄ / s²) if x̄ < s², and r̂ = x̄p̂ / (1 − p̂) or directly r̂ = x̄² / (s² − x̄ e the estimates may be invalid. In exam contexts, you might be asked to find the parameters given summary statistics, assuming a negative binomial model is appropriate.
给定一组被认为服从负二项分布的数据，其样本均值为 x̄、方差为 s²，我们可以通过令样本矩等于理论矩来估计 r 和 p：x̄ = r/p 且 s² = r(1 − p)/p²。解这些方程可得，若 x̄ < s²，则 p̂ = 1 − (x̄ / s²)，进而 r̂ = x̄p̂ / (1 − p̂) 或直接用 r̂ = x̄² / (s² − x̄)。这种矩法估计简单直接，但要求样本方差大于均值（即过度离散），否则估计值可能无效。在考试情境中，你可能会被要求根据汇总统计量求出参数，前提是负二项模型适用。
11. Common Mistakes and Exam Tips | 常见错误与应试技巧
ixed or the number of successes is fixed. Another mistake is forgetting that the minimum value of X is r, not 1 or 0. When using the PMF, be careful with the binomial coefficient: it is C(x − 1, r − 1), not C(x, r). In longer questions, remember that the sum of probabilities over all possible x equals 1; this can be used to check your answers. Finally, if a question asks for the probability of 'more than n trials' X > n) = 1 − P(X ≤ n). Practising past paper questions will help reinforce these concepts.
一个常见的错误是将负二项分布与二项分布混淆。务必检查是试验次数固定还是成功次数固定。另一个错误是忘记 X 的最小值是 r，而不是 1 或 0。使用 PMF 时，要小心二项式系数：它是 C(x − 1, r − 1)，而非 C(x, r)。在长题目中，要记住所有可能 x 的概率之和为 1，这可以用来检查你的答案。最后，如果题目问“多于 n 次试验”的概率，请使用补集：P(X > n) = 1 − P(X ≤ n)。练习历年真题有助于巩固这些概念。
12. Summary of Key Results | 关键结果总结
The table below summarises the essential formulae for X ~ NB(r, p):
| Probability mass function | P(X = x) = C(x − 1, r − 1) pr (1 − p)x − r, x = r, r+1, ... |
| Mean | E(X) = r / p |
| Variance | Var(X) = r (1 − p) / p2 |
| Special case (r = 1) | Geometric distribution Geo(p) |
| Sum property | Sum of r independent Geo(p) variables |
Mastering the negative binomial distribution involves not only memorising these formulas but also understanding the context in which they apply. Recognising the fixed-success condition and the difference from the binomial distribution will give you confidence in tackling any A-Level question on this topic.
下表总结了 X ~ NB(r, p) 的核心公式：掌握负二项分布不仅需要熟记这些公式，还须理解其适用情境。认清成功次数固定的条件以及与二项分布的区别，将使你能够信心十足地应对任何与此主题相关的 A-Level 考题。
| 概率质量函数 | P(X = x) = C(x − 1, r − 1) pr (1 − p)x − r, x = r, r+1, ... |
| 期望 | E(X) = r / p |
| 方差 | Var(X) = r (1 − p) / p² |
| 特例（r = 1） | 几何分布 Geo(p) |
| 求和性质 | r 个独立 Geo(p) 变量之和 |
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