IB Mathematics: Sampling with and Without Replacement | IB数学：有放回与不放回抽样

📚 IB Mathematics: Sampling with and Without Replacement | IB数学：有放回与不放回抽样

In the IB Mathematics curriculum, sampling methods form a core bridge between probability, combinatorics, and statistical inference. The distinction between sampling with replacement and sampling without replacement determines whether events are independent or dependent, and it dictates which probability distribution you must apply. This article provides a comprehensive, exam-focused guide covering definitions, tree diagrams, counting rules, binomial and hypergeometric distributions, expected values, and common pitfalls — all tailored to the IB Analysis & Approaches and Applications & Interpretation syllabi.

在IB数学课程中，抽样方法构成了连接概率、组合数学与统计推断的核心桥梁。有放回抽样与不放回抽样之间的区别，决定了事件究竟是独立的还是依赖的，也决定了你必须运用哪一种概率分布。本文将提供一份全面且紧扣考点（针对IB分析与方法及应用与解释两大方向）的指南，涵盖定义、树形图、计数法则、二项分布与超几何分布、期望值以及常见陷阱。

1. Definitions and Basic Framework | 定义与基本框架

Sampling with replacement means that after each draw, the selected item is returned to the population before the next draw. As a result, the population size remains constant, and every draw is an independent trial with the same probability of success.

有放回抽样是指每次抽取后，将选中的物品放回总体，再进行下一次抽取。因此，总体大小始终保持不变，每一次抽取都是具有相同成功概率的独立试验。

Sampling without replacement means that each drawn item is permanently removed from the population. The population shrinks by one after every draw, so the probability of a particular outcome changes from trial to trial. These trials are conditionally dependent.

不放回抽样意味着每次抽取后，被抽出的物品将永久地从总体中移除。每抽一次，总体就减少一个，因此每次试验中出现某一结果的概率都会发生变化。这些试验之间存在条件依赖性。

In IB examinations, you should begin every sampling problem by asking one key question: “Does the population stay the same or change between draws?” This single check determines whether you should use independent probability rules or conditional probability rules.

在IB考试中，解决任何抽样问题的第一步都应问一个关键问题：“总体在每次抽取之间是保持不变，还是发生了变化？” 这一个检查决定了你应该使用独立概率法则还是条件概率法则。

2. Probability Calculations: Independence vs Dependence | 概率计算：独立与依赖

When sampling with replacement, the multiplication rule for independent events applies directly. For example, if a bag contains 5 red and 3 blue balls, and we draw two balls with replacement, the probability of drawing red twice is simply the product of individual probabilities:

当采用有放回抽样时，可直接套用独立事件的乘法法则。例如，一个袋子中有5个红球和3个蓝球，有放回地抽取两个球，两次都抽到红球的概率就是各次概率的乘积：

P(R and R) = P(R) × P(R) = (5/8) × (5/8) = 25/64 ≈ 0.391

When sampling without replacement, we must use the conditional multiplication rule. After the first red ball is removed, the bag now contains 4 red and 3 blue balls (7 total), so the second probability changes:

当采用不放回抽样时，必须使用条件乘法法则。第一次抽出红球后，袋中剩下4个红球和3个蓝球（共7个），因此第二次的概率发生了变化：

P(R and R) = P(R) × P(R|R) = (5/8) × (4/7) = 20/56 = 5/14 ≈ 0.357

Notice that the two answers differ. The without-replacement probability is slightly lower because removing a red ball reduces the proportion of red balls remaining. This dependence is the fundamental reason why separate probability tools exist for each sampling method.

请注意，两个答案并不相同。不放回抽样的概率略低，因为抽出一个红球后，剩余总体中红球的比例下降了。这种依赖性正是两种抽样方法各自需要不同概率工具的根本原因。

3. Tree Diagrams for Multi-Stage Sampling | 多阶段抽样的树形图

Tree diagrams are the most intuitive tool for handling two or three draws, especially when sampling without replacement. Each branch represents a possible outcome, and the probabilities written on each branch must be updated after every stage to reflect the reduced population.

树形图是处理两到三次抽取最直观的工具，尤其适用于不放回抽样。每条分支代表一种可能的结果，每一层级分支上的概率都必须在抽取后更新，以反映总体的变化。

Consider a bag with 6 red and 4 white balls (N = 10). Suppose we draw twice without replacement. The first draw gives P(R) = 6/10 = 3/5 and P(W) = 4/10 = 2/5. The second-draw probabilities depend on the first outcome:

考虑一个装有6个红球和4个白球的袋子（N = 10）。假设我们不放回地抽取两次。第一次抽取的概率为P(红) = 6/10 = 3/5，P(白) = 4/10 = 2/5。第二次抽取的概率取决于第一次的结果：

If the first ball is red: P(R₂|R₁) = 5/9, P(W₂|R₁) = 4/9

如果第一次抽到红球：P(红₂|红₁) = 5/9，P(白₂|红₁) = 4/9
If the first ball is white: P(R₂|W₁) = 6/9 = 2/3, P(W₂|W₁) = 3/9 = 1/3

如果第一次抽到白球：P(红₂|白₁) = 6/9 = 2/3，P(白₂|白₁) = 3/9 = 1/3

The probability of any compound event, such as obtaining one red and one white in any order, is found by summing the probabilities of all paths that lead to that event:

任何一个复合事件（例如无论顺序如何，抽到一个红球和一个白球）的概率，等于所有通向该事件的分支路径概率之和：

P(exactly one red) = (6/10)(4/9) + (4/10)(6/9) = 24/90 + 24/90 = 48/90 = 8/15 ≈ 0.533

For with-replacement sampling, the tree diagram is simpler: every second-stage branch carries the exact same probabilities as the first-stage branches, since the population is restored before each draw.

对于有放回抽样，树形图更简单：第二阶段的每条分支都拥有与第一阶段完全相同的概率，因为每次抽取前总体都已被还原。

4. Counting Methods | 计数方法

Counting techniques provide an alternative route to probability, particularly for problems involving combinations without order. For with-replacement sampling of r items from a population of n items, the total number of ordered outcomes is nʳ, because every draw offers n independent choices.

计数技巧为概率计算提供了另一条途径，尤其适用于不涉及顺序的组合问题。对于从n个物品的总体中有放回地抽取r个物品，有序结果的总数为nʳ，因为每次抽取都有n个独立选择。

For without-replacement sampling, the number of ordered outcomes is the permutation ⁿPᵣ = n × (n−1) × (n−2) × … × (n−r+1). If order does not matter, we use the combination ⁿCᵣ = n! / [r!(n−r)!].

对于不放回抽样，有序结果的数量为排列数 ⁿPᵣ = n × (n−1) × (n−2) × … × (n−r−1+1)。如果顺序无关紧要，我们则使用组合数 ⁿCᵣ = n! / [r!(n−r)!]。

With replacement: nʳ outcomes | Without replacement: ⁿPᵣ ordered, ⁿCᵣ unordered

For example, drawing 3 cards from a standard deck of 52:
With replacement: 52³ = 140,608 possible ordered triples.
Without replacement: ⁵²P₃ = 52 × 51 × 50 = 132,600 possible ordered triples.

例如，从一副52张扑克牌中抽取3张：
有放回：52³ = 140,608种可能的有序三元组。
不放回：⁵²P₃ = 52 × 51 × 50 = 132,600种可能的有序三元组。

This difference in counting directly leads to different probabilities for identical-looking events. Many IB questions deliberately hide the sampling type inside phrases like “put the card back” or “without looking, the marble is taken and not returned.” Always scan the wording carefully.

这种计数差异直接导致看似相同的事件具有不同的概率。许多IB题目会故意将抽样方式隐藏在”将牌放回”或”不看地取出弹珠且不放回”等措辞中。务必仔细审题。

5. Binomial Distribution and With-Replacement Sampling | 二项分布与有放回抽样

When sampling with replacement, the four conditions of a binomial experiment are automatically satisfied: (1) a fixed number n of trials; (2) each trial has exactly two outcomes (success or failure); (3) the success probability p is identical for every trial; and (4) trials are independent.

当采用有放回抽样时，二项试验的四个条件自动满足：(1) 固定试验次数n；(2) 每次试验恰有两种结果（成功或失败）；(3) 每次试验的成功概率p完全相同；(4) 各次试验相互独立。

If X represents the number of successes in n draws with replacement, then X follows a binomial distribution: X ~ B(n, p). The probability mass function is:

若X表示有放回地抽取n次中成功的次数，则X服从二项分布：X ~ B(n, p)。其概率质量函数为：

P(X = x) = ⁿCₓ × pˣ × (1−p)ⁿ⁻ˣ, where x = 0, 1, 2, …, n

The expected value and variance are simple:

其期望与方差也非常简洁：

E(X) = np, Var(X) = np(1−p)

In IB exam problems, this distribution is appropriate whenever a process is repeated under unchanged conditions — e.g., rolling a die multiple times, spinning a fair spinner, or drawing a card and returning it each time.

在IB考试中，只要一个过程在不变条件下重复进行——例如多次掷骰子、旋转公平的转盘、或每次抽牌后放回——就该使用二项分布。

6. Hypergeometric Distribution and Without-Replacement Sampling | 超几何分布与不放回抽样

When sampling without replacement from a finite population, the binomial model collapses because the success probability p changes after every draw. Instead, the number of successes X follows a hypergeometric distribution.

当从有限总体中采用不放回抽样时，二项模型不再适用，因为每次抽取后成功概率p都会改变。此时，成功次数X服从超几何分布。

Suppose the population contains N items, of which K are classified as “successes.” We draw n items without replacement, and let X = the number of successes in the sample. The probability of exactly x successes is:

假设总体包含N个物品，其中K个被归类为”成功”。我们不放回地抽取n个物品，令X表示样本中成功的个数。恰好有x次成功的概率为：

P(X = x) = [C(K, x) × C(N−K, n−x)] / C(N, n)

Here C(K, x) counts the ways to choose x successes from the K available, C(N−K, n−x) counts the ways to choose the remaining failures from the N−K failures, and C(N, n) counts all possible samples.

其中C(K, x)表示从K个成功中选出x个的方式数，C(N−K, n−x)表示从N−K个失败中选出剩余n−x个的方式数，C(N, n)表示所有可能样本的总数。

The mean and variance of the hypergeometric distribution are:

超几何分布的期望与方差为：

E(X) = nK/N, Var(X) = n(K/N)(1 − K/N) × [(N−n)/(N−1)]

The extra factor (N−n)/(N−1) is called the finite population correction. It shrinks the variance when sampling a significant fraction of the population without replacement, reflecting the reduced variability caused by the shrinking population.

额外的因子(N−n)/(N−1)称为有限总体校正因子。当不放回地抽取总体的相当一部分时，它会压缩方差，反映出总体缩小时变异性降低的现象。

7. Expectation and Variance: A Side-by-Side Comparison | 期望与方差：并排比较

One of the most elegant results in sampling theory is that both distributions produce the same mean for the number of successes: E(X) = np = n(K/N). However, the variances differ, which is a favourite target for IB examiners.

抽样理论中最精妙的结果之一是：在成功次数方面，两种分布产生相同的期望值：E(X) = np = n(K/N)。然而，方差却不同，这正是IB考官最爱出题的方向。

Characteristic	With Replacement (Binomial)	Without Replacement (Hypergeometric)
Population size	Constant, N	Decreases each draw
Independence	Independent trials	Dependent trials
Success probability p	Constant	Changes after each draw
Number of ordered outcomes	Nⁿ	ᴺPₙ
Distribution of X (successes)	B(n, p) where p = K/N	Hypergeometric(N, K, n)
Mean E(X)	n × K/N	n × K/N
Variance Var(X)	n(K/N)(1 − K/N)	n(K/N)(1 − K/N) × (N−n)/(N−1)

Because the finite population correction factor (N−n)/(N−1) is always less than 1 (for n > 1), the variance without replacement is always smaller than the variance with replacement. This makes intuitive sense: without replacement, the sample cannot contain duplicate items, so extreme outcomes are less likely.

由于有限总体校正因子(N−n)/(N−1)总是小于1（当n > 1时），不放回抽样的方差总是小于有放回抽样的方差。这很符合直觉：不放回抽样中样本不可能出现重复物品，因此极端结果出现的概率更低。

Importantly, if the sample size n is tiny compared with the population N, then (N−n)/(N−1) ≈ 1, and the hypergeometric distribution is well approximated by the binomial distribution. Many IB questions explicitly ask you to state this approximation.

重要的是，当样本容量n相对于总体N来说非常小时，(N−n)/(N−1) ≈ 1，此时超几何分布可以很好地用二项分布来近似。许多IB题目会明确要求你说明这一近似关系。

8. Worked Example: Red and Blue Balls | 实例：红球与蓝球

Let us apply everything to a classic IB-style problem. A bag contains 5 red and 3 blue balls. Three balls are drawn. Find the probability of drawing exactly 2 red balls (a) with replacement and (b) without replacement.

让我们把以上知识应用到一个经典的IB题型中。一个袋子中装有5个红球和3个蓝球。现抽取3个球。分别求在(a)有放回和(b)不放回两种方式下，恰好抽到2个红球的概率。

Part (a): With replacement. Here X ~ B(3, 5/8). We need P(X = 2):

（a）有放回。此时X ~ B(3, 5/8)。我们需要计算P(X = 2)：

P(X = 2) = ³C₂ × (5/8)² × (3/8)¹ = 3 × (25/64) × (3/8) = 225/512 ≈ 0.439

Part (b): Without replacement. The easiest approach uses combinations. The total number of ways to choose 3 balls from 8 is ⁸C₃ = 56. The number of ways to choose 2 red balls from 5 red balls and 1 blue ball from 3 blue balls is ⁵C₂ × ³C₁ = 10 × 3 = 30. Therefore:

（b）不放回。最简单的方法是利用组合。从8个球中选3个的总方式数为⁸C₃ = 56。从5个红球中选2个、从3个蓝球中选1个的方式数为⁵C₂ × ³C₁ = 10 × 3 = 30。因此：

P(exactly 2 red) = 30/56 = 15/28 ≈ 0.536

Alternatively, using

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IB Mathematics: Sampling with and Without Replacement | IB数学：有放回与不放回抽样