📚 6 Other Discrete Distributions | 6 其他离散分布
While the binomial and Poisson distributions often dominate introductory probability, IB Mathematics also requires fluency with a broader family of discrete models. These include the geometric, negative binomial, hypergeometric, and discrete uniform distributions, each designed for specific counting scenarios. Understanding their probability mass functions, conditions, expected values, and variances is essential for tackling unseen problems and modelling real-world situations where trials are performed until an event occurs or sampling takes place without replacement.
虽然二项分布和泊松分布在概率入门中占据主导，但 IB 数学还要求熟练掌握更广泛的离散分布家族。这包括几何分布、负二项分布、超几何分布和离散均匀分布，每一种都针对特定的计数场景而设计。理解它们的概率质量函数、适用条件、期望值和方差，对于应对陌生问题和模拟诸如试验进行到事件发生或无放回抽样等现实情境至关重要。
1. Bernoulli Distribution Revisited | 伯努利分布回顾
and failure (0) with probability 1-p. Its probability mass function is simply P(X = x) = pˣ (1-p)¹⁻ˣ for x ∈ {0,1}. The mean is E(X) = p and the variance is Var(X) = p(1-p).
伯努利分布是许多其他离散模型的基石。一次伯努利试验只有两种结果：成功（通常编码为 1），概率为 p；失败（编码为 0），概率为 1-p。其概率质量函数简单表示为 P(X = x) = pˣ (1-p)¹⁻ˣ，其中 x ∈ {0,1}。期望 E(X) = p，方差 Var(X) = p(1-p)。
When independent Bernoulli trials are repeated, we obtain binomial, geometric, and negative binomial distributions depending on what we choose to count. Mastering the Bernoulli distribution therefore helps you quickly derive the properties of its extensions.
当独立的伯努利试验被重复进行时，根据我们所选择计数的对象，我们会得到二项分布、几何分布和负二项分布。因此，掌握伯努利分布有助于快速推导其扩展分布的性质。
2. Geometric Distribution – Counting Trials to First Success | 几何分布——计数到首次成功的试验次数
A geometric distribution models the number of independent Bernoulli trials needed to obtain the first ss. success. Two common variants exist: one that counts the total number of trials (support {1,2,3,…) and another that counts the number of failures before the first success (support {0,1,2,… ily uses the version where X represents the trial on which the first success occurs. For X ~ Geo(p), the probability that the first success happens on the kᵗʰ trial is P(X = k) = (1-p)ᵏ⁻¹ p, for k = 1,2,3,…
几何分布对获得第一次成功所需的独立伯努利试验次数进行建模。常见有两种变体：一种计算总试验次数（支撑集为 {1,2,3,…}），另一种计算首次成功前的失败次数（支撑集为 {0,1,2,…}）。IB 教学大纲主要使用 X 表示首次成功发生时的试验次数的版本。对于 X ~ Geo(p)，首次成功发生在第 k 次试验的概率为 P(X = k) = (1-p)ᵏ⁻¹ p，其中 k = 1,2,3,…
The conditions for a geometric distribution are: each trial is independent, each trial produces a success with constant probability p, and the experiment continues until the first success is observed. A typical example is repeatedly rolling a fair die until a six is rolled: then X ~ Geo(1/6).
几何分布的适用条件是：每次试验相互独立，每次试验以恒定概率 p 产生成功，且实验一直进行到观察到第一次成功为止。一个典型示例是反复掷一枚均匀骰子直到出现六点：此时 X ~ Geo(1/6)。
3. Mean and Variance of a Geometric Distribution | 几何分布的均值与方差
For X ~ Geo(p) counting the trial of the first success, the expected value and variance are standard results that should be memorised.
E(X) = 1/p
Var(X) = (1-p)/p²
20. The standard deviation √(Var(X)) can be comparably large, reflecting high uncertainty.
这些公式表明，随着成功概率降低，所需的平均试验次数会迅速增加。例如，若 p = 0.2，平均需要 5 次试验，方差为 0.8/0.04 = 20。标准差 √(Var(X)) 可能相对较大，反映了较大的不确定性。
The derivation of these formulas uses infinite series, but for the exam you are mainly required to apply them and understand their meaning. You should also be able to calculate expectations of functions of X, such as E(2X+3), using the standard linearity rule.
这些公式的推导使用无穷级数，但在考试中你主要需要会应用并理解其含义。你也应能使用标准的线性性质计算 X 的函数的期望，例如 E(2X+3)。
4. Memoryless Property of the Geometric Distribution | 几何分布的无记忆性
y failures have already occurred. Formally, for a geometric random variable X counting total trials, P(X > m+n | X > m) = P(X > n). Equivalently, P(X > m+n | X > m) = (1-p)ⁿ.
几何分布的一个独特性质是其无记忆性：需要额外试验的概率与已经发生了多少次失败无关。严格地说，对于计数总试验次数的几何随机变量 X，有 P(X > m+n | X > m) = P(X > n)。等价地，P(X > m+n | X > m) = (1-p)ⁿ。
This property means that a geometric distribution is the only discrete distribution with this constant conditional probability structure. It explains why, if you have already rolled a die ten times without a six, the probability that you will need at least five more rolls is exactly the same as when you started.
这一性质意味着几何分布是唯一具有这种恒定条件概率结构的离散分布。它解释了为什么如果你已经掷了十次骰子都没有出现六点，你还需要至少再掷五次的概率与你刚开始时完全一样。
5. Negative Binomial Distribution – Counting Trials to rᵗʰ Success | 负二项分布——计数到第 r 次成功
The negative binomial distribution generalises the geometric distribution. Instead of stopping at the first success, we continue until the rᵗʰ success is obtained (r is a positive integer). Let Y be the number of independent Bernoulli trials required to achieve r successes. For Y ~ NB(r,p), the probability mass function is
P(Y = k) = C(k-1, r-1) pʳ (1-p)ᵏ⁻ʳ, k = r, r+1, r+2, …
where C(k-1, r-1) denotes the binomial coefficient "k-1 choose r-1". Note that some textbooks define the negative binomial differently (counting failures only), so always read the problem statement carefully. The IB typically uses the “total trials” version described here.
负二项分布是对几何分布的推广。我们不等到第一次成功就停止，而是继续试验直到获得第 r 次成功（r 为正整数）。设 Y 为获得 r 次成功所需的独立伯努利试验次数。对于 Y ~ NB(r,p)，其概率质量函数为
P(Y = k) = C(k-1, r-1) pʳ (1-p)ᵏ⁻ʳ, k = r, r+1, r+2, …
其中 C(k-1, r-1) 表示二项式系数 “k-1 选 r-1”。注意有些教科书对负二项分布的定义不同（只计数失败次数），因此务必仔细阅读题目陈述。IB 通常使用这里描述的总试验次数版本。
6. Mean and Variance of a Negative Binomial Distribution | 负二项分布的均值与方差
For Y ~ NB(r,p) counting the total trials to achieve r successes, the expected value and variance are natural extensions of the geometric ones:
E(Y) = r/p
Var(Y) = r(1-p)/p²
p)/p². The negative binomial model is useful when you have a sequence of independent Bernoulli trials and want to predict how long it will take to collect a certain number of successes, for example in quality control where you continue sampling until finding r defective items.
对于计数获得 r 次成功所需总试验次数的 Y ~ NB(r,p)，其期望和方差是几何分布的自然推广：
E(Y) = r/p
Var(Y) = r(1-p)/p²
当 r = 1 时，这两个公式就退化为几何分布的 E(X) = 1/p 和 Var(X) = (1-p)/p²。负二项模型在你有一系列独立伯努利试验并希望预测收集到一定数量成功所需的时间时非常有用，例如在质量控制中，你持续抽样直至发现 r 件缺陷品。
7. Hypergeometric Distribution – Sampling Without Replacement | 超几何分布——无放回抽样
The hypergeometric distribution arises when we sample without replacement from a finite population. Consider a population of N items, of which M are classified as “successes” and N-M as “failures”. We draw a sample of size n without replacement and let X denote the number of successes in the sample. Then X follows a hypergeometric distribution, with probability mass function
P(X = k) = C(M, k) C(N-M, n-k) / C(N, n), max(0, n-N+M) ≤ k ≤ min(n, M)
l relative to N, the binomial distribution with p = M/N can serve as a good approximation.
超几何分布出现在我们从一个有限总体中无放回抽样时。考虑一个包含 N 个元素的总体，其中有 M 个被归类为“成功”，N-M 个为“失败”。我们无放回地抽取一个容量为 n 的样本，令 X 表示样本中成功的个数。那么 X 服从超几何分布，其概率质量函数为
P(X = k) = C(M, k) C(N-M, n-k) / C(N, n), max(0, n-N+M) ≤ k ≤ min(n, M)
与二项分布的关键区别在于，试验之间不是独立的，因为每次抽取后总体的构成都会改变。然而，如果 n 相对于 N 非常小，那么可以用 p = M/N 的二项分布作为良好近似。
8. Hypergeometric Expected Value and Variance | 超几何分布的期望与方差
For X ~ Hypergeometric(N, M, n), the mean and variance formulas reflect the finite population correction factor.
E(X) = n × (M/N)
Var(X) = n × (M/N) × (1 - M/N) × (N - n)/(N - 1)
but the variance is smaller by the factor (N - n)/(N - 1). This factor approaches 1 when the population size is huge relative to the sample size, explaining why the binomial approximation works well in large populations. A classic example is drawing cards from a deck without replacement and counting the number of hearts.
对于 X ~ Hypergeometric(N, M, n)，均值和方差公式反映了有限总体校正因子。
E(X) = n × (M/N)
Var(X) = n × (M/N) × (1 - M/N) × (N - n)/(N - 1)
均值正如同你从 p = M/N 的二项分布中所期望的那样，但方差缩小了因子 (N - n)/(N - 1)。当总体容量相对于样本容量极大时，该因子趋近于 1，这就解释了为什么二项分布在总体很大时能作为很好近似。一个经典的例子是从一副扑克牌中无放回地抽牌并计算红心牌的数量。
9. Discrete Uniform Distribution | 离散均匀分布
The discrete uniform distribution is the simplest of all: every outcome in a finite set is equally likely. If a random variable X takes values x₁, x₂, …, xₙ with equal probability 1/n, it is said to follow a discrete uniform distribution. A common special case is when X takes consecutive integer values from 1 to n, denoted X ~ Uniform{1,…,n}. For this case:
E(X) = (n + 1)/2
Var(X) = (n² - 1)/12
离散均匀分布是所有分布中最简单的：有限集合中的每个结果等可能。如果一个随机变量 X 等可能地取值 x₁, x₂, …, xₙ，概率均为 1/n，则称其服从离散均匀分布。一个常见的特例是当 X 取从 1 到 n 的连续整数值，记作 X ~ Uniform{1,…,n}。对于这种情况：
E(X) = (n + 1)/2
Var(X) = (n² - 1)/12
) or selecting a random day of the month. Recognising it allows you to immediately write down probabilities and moments without lengthy calculations.
尽管该分布本身非常简单，但它出现在无数情境中，如掷一枚均匀骰子（n=6）或者随机选取一个月中的某一天。识别出均匀分布能使你立即写出概率和矩而无需冗长计算。
10. Relationships Among the Distributions | 分布之间的关系
These discrete distributions form an interconnected family. Understanding their links helps you choose the correct model in an exam problem.
- Bernoulli is the basis; binomial sums independent Bernoulli trials.
- Geometric is a special case of negative binomial with r=1.
- Hypergeometric converges to binomial as N → ∞ with fixed proportion M/N.
- Poisson can be derived as a limit of binomial distributions, though not covered here.
A useful exercise is to identify which distribution corresponds to a given scenario: Is there replacement? Is there a fixed number of trials? Are we waiting for the first or the rᵗʰ success? These questions quickly narrow down the choice.
这些离散分布构成了一个相互关联的家族。理解它们之间的联系有助于你在考试题目中选择正确的模型。
- 伯努利是基础；二项分布是独立伯努利试验的加总。
- 几何分布是负二项分布在 r=1 时的特例。
- 超几何分布在 N → ∞ 且比例 M/N 固定时收敛于二项分布。
- 泊松分布可作为二项分布的极限导出，但此处不展开讨论。
一个有用的练习是识别给定情境对应哪种分布：是否存在放回？是否有固定的试验次数？我们是在等待第一次成功还是第 r 次成功？这些问题能迅速缩小选择范围。
11. Common Pitfalls and Exam Tips | 常见陷阱与考试技巧
In geometric problems, students often confuse the parameterisation (counting failures vs total trials). Always define your random variable explicitly and check whether the smallest possible value is 0 or 1.
在几何分布问题中，学生常常混淆参数化方式（计数失败次数还是总试验次数）。务必明确地定义你的随机变量，并检查最小可能取值是 0 还是 1。
1 we should recover the geometric probability (1-p)ᵏ⁻¹p, which works only with C(k-1,0)=1.
对于负二项分布，组合系数是 C(k-1, r-1)，而不是 C(k, r)。快速检验：当 r=1 时，我们应能恢复几何概率 (1-p)ᵏ⁻¹p，而这仅在 C(k-1,0)=1 时才成立。
With the hypergeometric distribution, do not forget the finite population correction when computing variance. Also, ensure that your combinatorial expressions are well-defined: k must lie between max(0, n-N+M) and min(n, M).
使用超几何分布时，计算方差不要忘记有限总体校正因子。此外，要确保组合表达式的定义域合理：k 必须介于 max(0, n-N+M) 和 min(n, M) 之间。
Finally, for discrete uniform distributions, do not overcomplicate. The mean of the first n integers is simply the midpoint (n+1)/2, and the variance squared is a useful constant to know.
最后，对于离散均匀分布，不要过度复杂化。前 n 个整数的均值就是中点 (n+1)/2，而方差也是一个值得记住的有用常数。
12. Summary Table of Key Distributions | 关键分布总结表
The following table provides a quick reference for the discrete distributions discussed, including their notation, support, mean, and variance. Use it as a revision tool before the exam.
下表为所讨论的离散分布提供了快速参考，包括它们的符号、支撑集、均值和方差。可用作考前复习工具。
| Distribution | Notation | Mean E(X) | Variance Var(X) |
|---|---|---|---|
| Bernoulli | X ~ Bern(p) | p | p(1-p) |
| Geometric (trials to 1ˢᵗ success) | X ~ Geo(p) | 1/p | (1-p)/p² |
| Negative Binomial (trials to rᵗʰ success) | Y ~ NB(r, p) | r/p | r(1-p)/p² |
| Hypergeometric | X ~ Hyp(N, M, n) | n(M/N) | n(M/N)(1-M/N)×(N-n)/(N-1) |
| Discrete Uniform (1 to n) | X ~ U{1,…,n} | (n+1)/2 | (n²-1)/12 |
Remember that the support and notation may vary slightly across textbooks. In an IB examination, always clarify the version being used by defining the random variable clearly. Practising mixed exercises involving at least two of these distributions is an excellent way to solidify your understanding and speed up problem solving.
请记住，支撑集和符号可能因教科书而略有不同。在 IB 考试中，始终通过明确定义随机变量来澄清所使用的版本。练习至少涉及其中两种分布的混合题是巩固理解和提升解题速度的绝佳方式。
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