📚 G - Sampling with and without replacement | 放回与不放回抽样
Sampling is a fundamental process in statistics and probability where we select a subset of individuals from a population to observe or measure. The choice between sampling with replacement and sampling without replacement has profound consequences on the independence of trials and the probability distributions that describe the outcomes. In IB Mathematics, understanding this distinction is essential for tackling problems in combinatorics, discrete probability distributions, and inferential statistics. This article explores both methods in depth, providing clear definitions, worked examples, and connections to key distributions like the binomial and hypergeometric distributions.
抽样是统计学与概率论中的一个基本过程，即从总体中选取一部分个体进行观测或测量。选择放回抽样还是不放回抽样，会深刻影响试验的独立性以及描述结果的概率分布。在 IB 数学中，理解这一区别对于解决组合计数、离散概率分布和推断统计问题至关重要。本文将深入探讨这两种方法，提供清晰的定义、详细示例，并阐明它们与二项分布、超几何分布等重要分布的联系。
1. Basic Concepts of Sampling | 抽样的基本概念
Sampling involves selecting elements from a defined population. The population could be finite (e.g., the 30 students in a classroom) or conceptually infinite (e.g., all possible outcomes of rolling a die). When we draw a sample, two fundamental protocols exist: with replacement, where a selected element is placed back into the population before the next draw, and without replacement, where it is set aside. The sampling method determines whether successive draws are independent or dependent events.
抽样涉及从明确的总体中选取元素。总体可以是有限的（例如教室里的 30 名学生），也可以是概念上无限的（例如掷骰子的所有可能结果）。当我们抽取样本时，存在两种基本操作方式：放回抽样，即在下一次抽取前将选中的元素放回总体；以及不放回抽样，即将其放在一边。抽样方法决定了连续抽取是独立事件还是相关事件。
2. Sampling with Replacement | 放回抽样
In sampling with replacement, every time an item is selected, it is returned to the population, so the population remains unchanged for each draw. This means that each selection is independent of the previous ones, and the probability of obtaining any particular outcome remains constant across trials. For example, if you draw a card from a deck, record it, and put it back before shuffling, the chance of drawing an ace is always 4/52 on each draw.
在放回抽样中，每次选中一个项目后，都会将其放回总体，因此总体在每次抽取时保持不变。这意味着每次选取相互独立，且在多次试验中获得某一特定结果的概率恒定不变。例如，从一副牌中抽一张牌，记录后放回并洗牌，每次抽到 A 的概率始终为 4/52。
3. Sampling without Replacement | 不放回抽样
When sampling without replacement, selected items are not returned to the population. Consequently, the composition of the population changes after each draw, and the probabilities for subsequent draws are altered. The events are no longer independent; they become conditionally dependent. Consider picking two marbles from a bag containing 5 red and 3 blue marbles without putting any back. The probability of picking a red marble on the second draw depends on what was picked first.
在不放回抽样中，被选中的项目不放回总体。因此，每次抽取后总体的组成发生变化，后续抽取的概率也随之改变。这些事件不再独立，而变成了条件依赖。考虑从一个装有 5 个红球和 3 个蓝球的袋子里不放回地取两个球：第二次取到红球的概率取决于第一次取到了什么。
4. Probability Trees and Systematic Counting | 概率树与系统计数
Visualising the sampling process with tree diagrams helps to correctly assign probabilities at each stage. For with-replacement sampling, the probabilities on the branches remain the same across all levels. For without-replacement sampling, the branch probabilities must be updated to reflect the reduced population size and changed composition. Systematic counting using combinations (nCr) is a powerful tool: with replacement, we often use ordered selections with repetition allowed; without replacement, we use combinations where order does or does not matter depending on the problem context.
使用树状图可视化抽样过程有助于在每个阶段正确分配概率。对于放回抽样，各层分支上的概率始终保持不变。对于不放回抽样，必须更新分支概率以反映总体数量减少和构成变化。利用组合数 (nCr) 进行系统计数是一种强大的工具：放回抽样时常使用允许重复的有序选取；不放回抽样时则使用组合，其中顺序重要与否取决于问题的具体情境。
5. Probability Calculations: With Replacement | 放回抽样的概率计算
r) is (6/10) × (4/10) × (6/10) = 144/1000 = 0.144. The number of green balls in a fixed number of draws follows a binomial distribution with parameters n (number of trials) and p (success probability).
由于试验是独立的，一串事件序列的概率就是它们各自恒定概率的乘积。例如，如果从一个装有 6 个绿球和 4 个黄球的袋子里放回地抽取三个球，那么按顺序得到绿、黄、绿的概率就是 (6/10) × (4/10) × (6/10) = 144/1000 = 0.144。在固定次数的抽取中，绿球出现的次数服从二项分布，参数为 n（试验次数）和 p（成功概率）。
6. Probability Calculations: Without Replacement | 不放回抽样的概率计算
.1667. Notice that the probability is different from the with-replacement case. More generally, the probability of obtaining exactly k successes in n draws without replacement is given by the hypergeometric distribution.
不放回抽样时，计算必须考虑变化的分母。同样用装有 6 个绿球和 4 个黄球的袋子，如果不放回地抽取三个球，得到绿、黄、绿的概率为 (6/10) × (4/9) × (5/8) = 120/720 = 1/6 ≈ 0.1667。请注意这个概率与放回情况不同。更一般地，不放回地抽取 n 次恰好获得 k 次成功的概率由超几何分布给出。
7. Conditional Probability and Independence | 条件概率与独立性
tured by the definition of independence: events A and B are independent if and only if P(A ∩ es not. The probability of the second draw is conditional on the outcome of the first: P(B|A) = P(A ∩ B) / P(A), and since P(B| ditional reasoning is essential when solving problems involving draws without replacement.
两种抽样的关键区别可通过独立性的定义体现：当且仅当 P(A ∩ B) = P(A) × P(B) 时，事件 A 和 B 独立。在放回抽样下，从有限总体中抽取时该等式成立。不放回抽样则不成立。第二次抽取的概率以第一次结果为条件：P(B|A) = P(A ∩ B) / P(A)，而由于 P(B|A) 通常不等于 P(B)，事件是相关的。解决涉及不放回抽取的问题时，这种条件推理至关重要。
8. Expectation and Variance | 期望与方差
E(X) = n(K/N), which is identical to the with-replacement case, but the variance is Var(X) = n(K/N)(1 - K/N) × (N - n)/(N - 1). The extra factor (N - n)/(N - 1), known as the finite population correction factor, reduces the variance because sampling without replacement provides more precise information about the population.
抽样方法的选择会影响样本计数的期望值和方差。对于放回抽取 n 次，成功次数 X 满足 E(X) = np，Var(X) = np(1-p)。对于从包含 K 个成功个体的总体大小为 N 中不放回地抽取 n 次，超几何分布的期望是 E(X) = n(K/N)，与放回情况相同；但方差是 Var(X) = n(K/N)(1 - K/N) × (N - n)/(N - 1)。多出的因子 (N - n)/(N - 1) 称为有限总体校正因子，它减小了方差，因为不放回抽样能提供更精确的总体信息。
9. Relationship with the Binomial Distribution | 与二项分布的关系
Sampling with replacement leads directly to the binomial setting: a fixed number of independent trials, each with two possible outcomes (success/failure) and a constant probability of success. Counts of successes follow a binomial distribution B(n, p). This assumption is valid even for sampling without replacement if the population size N is very large compared to the sample size n (commonly, if n/N < 0.1), because the change in probabilities becomes negligible.
放回抽样直接满足二项分布的条件：固定次数的独立试验，每次只有两种可能结果（成功/失败），且成功概率恒定。成功次数服从二项分布 B(n, p)。即使是不放回抽样，当总体大小 N 远大于样本量 n 时（通常 n/N < 0.1），这一近似也成立，因为概率变化可以忽略不计。
10. Relationship with the Hypergeometric Distribution | 与超几何分布的关系
K, n-k)] / C(N, n), where C(a, b) denotes the binomial coefficient. This is the go-to distribution for small populations, such as selecting a committee from a group or quality inspection from a small batch without replacement. The table below compares the two distributions.
超几何分布精确地描述了从包含恰好 K 个成功个体的有限总体中不放回地抽取 n 次时，成功次数的分布。其概率质量函数为 P(X = k) = [C(K, k) × C(N-K, n-k)] / C(N, n)，其中 C(a, b) 表示二项式系数。对于小总体（如从群体中选拔委员会或对小批量产品进行不放回质检），这是首选的分布模型。下表对两种分布进行了比较。
| Property | Binomial (with replacement) | Hypergeometric (without replacement) |
|---|---|---|
| Trials | Independent | Dependent |
| Probability of success | Constant, p | Changes after each draw |
| Population size | Infinite or very large relative to n | Finite, N |
| Variance | np(1-p) | n(K/N)(1 - K/N) × (N - n)/(N - 1) |
11. Real-World Applications | 实际应用
These concepts are widely used in quality control, ecological capture-recapture studies, opinion polls, and card games. For example, a factory inspector testing light bulbs without replacement from a small batch uses the hypergeometric distribution to find the probability of discovering a certain number of defective items. In contrast, simulating a simple random sample from a very large voter registry with replacement can be approximated by the binomial method. Understanding which sampling scheme applies prevents errors in inference and decision-making.
这些概念广泛应用于质量控制、生态标记重捕研究、民意调查和纸牌游戏。例如，工厂质检员对小批次灯泡进行不放回检测时，使用超几何分布来计算出发现一定数量次品的概率。相反，从超大型选民登记册中进行简单的随机抽样放回，可以用二项分布近似。理解适用哪种抽样方案可以避免推断和决策中的错误。
12. Common Errors and Exam Tips | 常见错误与考试技巧
Students often mix up the two models: they use binomial nCr pk (1-p)n-k for a without-replacement problem, or they incorrectly assume independence when sampling without replacement. Always check whether the population is finite and whether the item is replaced. When N is large and n is small, the binomial approximation is a time-saving shortcut, but be prepared to justify its use. Pay close attention to wording in IB questions such as 'without returning the ball' or 'two cards drawn successively with replacement' to decide the correct approach. Practice drawing tree diagrams and writing conditional probabilities step by step.
学生经常混淆两种模型：他们对不放回问题误用二项式 nCr pk (1-p)n-k，或者在不放回抽样时错误地假设独立性。务必检查总体是否有限以及物品是否被放回。当 N 很大而 n 很小时，二项近似是省时的捷径，但要准备好说明使用理由。密切注意 IB 考题中的表述，如“不将球放回”或“有放回地连续抽取两张牌”，以决定正确的方法。通过练习逐步绘制树状图和书写条件概率来提升能力。
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