📚 Geometric and Negative Binomial Distributions | 几何分布与负二项分布
In A-Level Further Statistics, the geometric and negative binomial distributions model the number of independent Bernoulli trials needed to achieve a specified number of successes. Both distributions are built upon the concept of repeated, identical trials with only two outcomes – success and failure – and they provide powerful tools for analysing waiting times, reliability, and sequential experiments. A deep understanding of their probability mass functions, expected values, variances and unique properties such as memorylessness is essential for Edexcel examinations and real-world statistical modelling.
在A-Level进阶统计中，几何分布和负二项分布用于描述在一系列独立伯努利试验中达到指定成功次数所需的试验次数。这两种分布都建立在重复、相同且只有成功与失败两种结果的试验之上，为分析等待时间、可靠性以及顺序实验提供了有力工具。深刻理解它们的概率质量函数、期望值、方差以及无记忆性等独特性质，对于应对Edexcel考试和实际统计建模至关重要。
1. Defining the Geometric Distribution | 几何分布的定义
A geometric distribution arises when we count the number of trials X up to and including the first success in a sequence of independent Bernoulli trials, each with constant probability of success p. Formally, we write X ~ Geo(p), where the support is x = 1, 2, 3, … . The trial at which the first success occurs is the random variable of interest.
当我们在每次试验成功概率恒为 p 的独立伯努利试验中，计数直到并包含首次成功在内的试验次数 X 时，就产生了几何分布。形式上记为 X ~ Geo(p)，其取值范围为 x = 1, 2, 3, … 。首次成功发生所在的那次试验就是我们关注的随机变量。
The probability that the first success happens on the x-th trial is obtained by requiring that the first x – 1 trials are failures (each with probability 1 – p) and the x-th trial is a success. Hence the probability mass function (PMF) is
P(X = x) = p(1 – p)x–1 for x = 1, 2, 3, … .
首次成功发生在第 x 次试验的概率由前 x – 1 次均为失败（每次概率 1 – p）且第 x 次为成功这一要求得出。因此概率质量函数为
P(X = x) = p(1 – p)x–1 其中 x = 1, 2, 3, … 。
In Edexcel, the geometric distribution is defined with support starting at 1. It is important not to confuse this with an alternative definition that counts failures before the first success; examination questions will specify which definition is being used, but the majority of Edexcel problems assume the 'trial count' version shown here.
在Edexcel考试中，几何分布的定义以1为起点。千万不要把这个定义与另一种计数首次成功前失败次数的定义相混淆；考题会明确说明采用哪种定义，不过绝大多数Edexcel题目都采用这里给出的“试验次数”版本。
2. Geometric PMF and Cumulative Probabilities | 几何分布的PMF与累积概率
The PMF P(X = x) = p(1 – p)x–1 can be used to compute individual probabilities. For cumulative probabilities, it is often easier to work with the complement. In particular, the probability that more than n trials are needed to obtain the first success is P(X > n) = (1 – p)n, because the first n trials must all be failures.
利用PMF P(X = x) = p(1 – p)x–1 可以计算单个概率。对于累积概率，通常利用补集会更加简便。特别是，需要多于 n 次实验才能获得首次成功的概率为 P(X > n) = (1 – p)n，因为前 n 次试验必然全部失败。
Similarly, the probability that the first success occurs by the n-th trial is P(X ≤ n) = 1 – (1 – p)n. These complementary forms are extremely useful for solving problems involving inequalities.
类似地，首次成功在第 n 次试验或之前发生的概率为 P(X ≤ n) = 1 – (1 – p)n。这些补集形式在求解涉及不等式的题目时极为有用。
>> Example: A die is rolled repeatedly until a six appears. Find the probability that the first six occurs on the 4th roll, and the probability that at least 5 rolls are needed. Here p = 1/6. P(X = 4) = (1/6)(5/6)³ ≈ 0.0965. P(X ≥ 5) = P(X > 4) = (5/6)⁴ ≈ 0.4823.
>> 例子：反复投掷一粒骰子直到出现六点。求首次六点恰好在第4次投掷出现的概率，以及至少需要投掷5次的概率。这里 p = 1/6。P(X = 4) = (1/6)(5/6)³ ≈ 0.0965。P(X ≥ 5) = P(X > 4) = (5/6)⁴ ≈ 0.4823。
3. Expectation and Variance of Geometric Distribution | 几何分布的期望与方差
For X ~ Geo(p), the expected number of trials to get the first success is E(X) = 1/p. This makes intuitive sense: if p = 0.2, on average you need 5 trials.
对于 X ~ Geo(p)，获得首次成功所需的试验次数期望值为 E(X) = 1/p。这从直观上很好理解：如果 p = 0.2，那么平均需要5次试验。
The variance is given by Var(X) = (1 – p)/p² . This quantifies the spread of the distribution and shows that as p decreases, the variance increases rapidly.
方差由 Var(X) = (1 – p)/p² 给出。它衡量了分布的离散程度，且表明随着 p 减小，方差会急剧增大。
Edexcel students are expected to be able to use these formulae directly, but also to understand them through derived results and, in some specifications, sum-to-infinity derivations involving arithmetico-geometric series.
Edexcel学生不仅要能直接使用这些公式，还需要透过推导结果来理解它们；在某些大纲中还会涉及利用等差-等比级数求和至无穷的推导。
4. The Memoryless Property | 无记忆性质
A unique feature of the geometric distribution is its memoryless property. For positive integers s and t, the conditional probability that more than s + t trials are needed, given that more than s trials have already been performed, depends only on the number of remaining trials t:
几何分布的一个独特性质就是无记忆性。对于正整数 s 和 t，已知已经进行了超过 s 次试验仍未成功的条件下，还需要超过 s + t 次试验的条件概率仅取决于剩余试验次数 t：
P(X > s + t | X > s) = P(X > t) = (1 – p)t
In words, the process does not 'remember' past failures; the future waiting time behaves exactly as if the experiment had just started.
也就是说，这一过程不会“记忆”过去的失败；未来的等待时间就如同实验刚刚开始时一样。
This property is often tested using conditional probability questions. For example, if a light bulb fails with probability 0.01 each hour, and it has survived 200 hours, the probability it survives a further 50 hours is the same as the unconditional probability that a new bulb survives 50 hours, namely (0.99)⁵⁰.
这个性质经常通过条件概率题来考查。例如，一个灯泡每小时损坏的概率为0.01，在已经正常工作了200小时后，它再继续工作50小时的概率与一个全新灯泡工作50小时的概率相同，即为 (0.99)⁵⁰。
5. Geometric Distribution in Context: Worked Examples | 几何分布应用实例
Consider a biased coin with p = 0.3 for heads. Let X be the number of tosses to get the first head. Then P(X 720. The expected number of tosses is 1/0.3 ≈ 3.33, and the variance is 0.7/(0.3²) ≈ 7.78.
考虑一枚正面概率为 p = 0.3 的偏倚硬币。记 X 为投掷出首次正面的次数。则 P(X = 5) = 0.3 × 0.7⁴ ≈ 0.0720。投掷次数的期望值为 1/0.3 ≈ 3.33，方差为 0.7/(0.3²) ≈ 7.78。
A typical exam question might ask: 'Find the probability that the first head occurs after the 3rd toss.' Using the complement, P(X > 3) = 0.7³ = 0.343. It is far quicker than summing individual probabilities P(4) + P(5) + … .
一个典型的考题可能会问：“求首次正面出现在第3次投掷之后的概率。”利用补集，P(X > 3) = 0.7³ = 0.343。这比单独求和 P(4) + P(5) + … 要迅速得多。
6. Introducing the Negative Binomial Distribution | 负二项分布简介
When we wish to model the number of trials needed to achieve not just the first success but the r-th success, we use the negative binomial distribution. The random variable X counts the total number of independent Bernoulli trials required to obtain r successes, where each trial has probability p of success. We write X ~ NB(r, p) or sometimes X ~ Negative B(r, p). The support is x = r, r + 1, r + 2, ….
当我们要建模的不是首次成功，而是第 r 次成功所需的试验次数时，就使用负二项分布。随机变量 X 计数了为获得 r 次成功所需进行的独立伯努利试验的总次数，其中每次试验的成功概率为 p。我们记为 X ~ NB(r, p) 或 X ~ Negative B(r, p)，其取值范围为 x = r, r + 1, r + 2, … 。
The probability mass function is
P(X = x) = x–1Cr–1 pr (1 – p)x–r , x = r, r+1, …
概率质量函数为
P(X = x) = x–1Cr–1 pr (1 – p)x–r , x = r, r+1, …
The combinatorial term x–1Cr–1 counts the number of ways to arrange r – 1 successes among the first x – 1 trials, since the x-th trial must be the r-th success.
组合项 x–1Cr–1 计算了在前 x – 1 次试验中安排 r – 1 次成功的方法数，因为第 x 次试验必须是第 r 次成功。
7. Relationship Between Geometric and Negative Binomial | 几何分布与负二项分布的关系
When r 1, the negative binomial distribution reduces exactly to the geometric distribution: NB(1, p) ≡ Geo(p). Indeed, substituting r = 1 into the negative binomial PMF gives P(X = x) = x–1C0 p¹ (1 – p)x–1 = p(1 – p)x–1, which is exactly the geometric PMF.
当 r = 1 时，负二项分布完全退化为几何分布：NB(1, p) ≡ Geo(p)。事实上，将 r = 1 代入负二项PMF便可得到 P(X = x) = x–1C0 p¹ (1 – p)x–1 = p(1 – p)x–1，这正是几何分布的PMF。
This nested relationship means that the geometric distribution is a special case of the negative binomial, and many properties of the
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