The independence of the Court | 法院的独立性

📚 The independence of the Court | 法院的独立性

The title of this article uses the word ‘Court’ as supplied in the brief; however, the mathematical content below focuses on the concept of independence, which is a core topic in the Edexcel A-Level Mathematics specification. Independence of events lies at the heart of probability theory and underpins many statistical models, including the binomial distribution and hypothesis testing.

本文标题保留了题目中给出的 ‘Court’ 一词；但下面的数学内容重点讨论独立性这一概念，它是 Edexcel A-Level 数学大纲中的核心主题。事件的独立性是概率论的核心，并为许多统计模型（包括二项分布和假设检验）奠定了基础。

1. Definition of Independent Events | 独立事件的定义

Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, this is written as P(A | B) = P(A) and P(B | A) = P(B). Equivalently, the joint probability satisfies the multiplication rule:

两个事件 A 和 B 独立，是指一个事件的发生不影响另一个事件的概率。数学上写作 P(A | B) = P(A) 且 P(B | A) = P(B)。等价地，联合概率满足乘法法则：

P(A ∩ B) = P(A) × P(B)

For Edexcel, you must know both forms and be able to switch between them. Independence is a symmetric property: if A is independent of B, then B is independent of A. It only makes sense when both events have non-zero probability; otherwise conditional probabilities are undefined.

在 Edexcel 考试中，你必须掌握这两种形式并能够相互转换。独立性是对称的性质：如果 A 与 B 独立，那么 B 也与 A 独立。只有当两个事件的概率都非零时，独立性才有意义；否则条件概率无定义。

2. Multiplication Rule | 乘法法则

The multiplication rule for independent events states that the probability that both A and B occur is the product of their individual probabilities. For three independent events A, B and C, this extends to:

独立事件的乘法法则指出，A 和 B 同时发生的概率等于它们各自概率的乘积。对于三个独立事件 A、B、C，推广为：

P(A ∩ B ∩ C) = P(A) × P(B) × P(C)

This rule is frequently used in probability questions involving repeated independent trials. A common error is to apply the multiplication rule to events that are not independent. Always check the context: drawing cards without replacement or selecting students without replacement usually creates dependence.

这一法则常用于涉及重复独立试验的概率问题。一个常见错误是把乘法法则用于不独立的事件。务必检查情境：不放回抽牌或不放回抽取学生通常会产生相关性。

In contrast, when sampling with replacement, the selections are independent. For example, drawing a card, recording it, and returning it to the deck before the next draw keeps the probabilities unchanged. Questions often hide this cue in phrases such as ‘the card is replaced’ or ‘the token is put back’.

相反，有放回抽样时，各次选取是独立的。例如，抽一张牌、记录后放回牌堆再抽下一张，概率保持不变。题目常将这些线索隐藏在 ‘卡片被放回’ 或 ‘筹码被放回’ 等短语中。

3. Conditional Probability and Independence | 条件概率与独立性

The definition of conditional probability is P(A | B) = P(A ∩ B) / P(B), provided P(B) > 0. If A and B are independent, then P(A | B) = P(A), which means knowing B occurred gives no extra information about A. This link is often tested by asking you to show that two events are independent or to find a missing probability that makes them independent.

条件概率的定义为 P(A | B) = P(A ∩ B) / P(B)，其中 P(B) > 0。如果 A 与 B 独立，则 P(A | B) = P(A)，这意味着知道 B 发生对 A 没有提供额外信息。考试常要求你证明两个事件独立，或求使它们独立的缺失概率。

For instance, if P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, then P(A | B) = 0.2 / 0.5 = 0.4 = P(A), so A and B are independent. If the joint probability were 0.3 instead, then P(A | B) = 0.6, which differs from P(A), so the events would be dependent.

例如，若 P(A) = 0.4，P(B) = 0.5 且 P(A ∩ B) = 0.2，则 P(A | B) = 0.2 / 0.5 = 0.4 = P(A)，因此 A 与 B 独立。若联合概率改为 0.3，则 P(A | B) = 0.6，与 P(A) 不同，因此事件相关。

4. Mutually Exclusive vs Independent | 互斥事件与独立事件的区别

Mutually exclusive events cannot occur at the same time, so P(A ∩ B) = 0. Independent events can occur together, and their joint probability is the product of their individual probabilities. Two non-trivial events cannot be both mutually exclusive and independent unless one of them has probability zero. This distinction is tested in multiple-choice and structured questions.

互斥事件不可能同时发生，因此 P(A ∩ B) = 0。独立事件可以同时发生，其联合概率等于各自概率的乘积。两个非平凡事件不可能既互斥又独立，除非其中一个概率为零。这一区别常出现在选择题和结构题中。

Use a Venn diagram to visualise: mutually exclusive events have no overlap, while independent events may overlap but the overlap area is exactly P(A)P(B). The table below summarises the key differences.

用维恩图可视化：互斥事件没有重叠，而独立事件可以重叠，但重叠区域的面积恰好是 P(A)P(B)。下表总结了关键区别。

Property	Mutually Exclusive \| 互斥	Independent \| 独立
Joint probability	P(A ∩ B) = 0	P(A ∩ B) = P(A)P(B)
Conditional probability	P(A \| B) = 0 (if P(B)>0)	P(A \| B) = P(A)
Venn diagram	No overlap	Overlap of size P(A)P(B)

5. Testing for Independence | 独立性的检验

To test whether two events A and B are independent, compare P(A ∩ B) with P(A) × P(B). If these are equal, the events are independent; if not, they are dependent. Another method is to compare P(A | B) with P(A). Both approaches are accepted in Edexcel mark schemes as long as you show clear working.

要检验两个事件 A 和 B 是否独立，可比较 P(A ∩ B) 与 P(A) × P(B)。若相等，事件独立；若不相等，则相关。另一种方法是比较 P(A | B) 与 P(A)。只要解题步骤清晰，Edexcel 评分标准接受这两种方法。

In a two-way table question, first find the row and column totals, then calculate the expected joint probability under independence and compare it with the observed joint probability. For example, a table showing 30 boys and 20 girls, of whom 18 boys and 10 girls like football, gives P(boy ∩ likes football) = 18/50 = 0.36 while P(boy) × P(likes football) = 30/50 × 28/50 = 0.336, so the events are not independent.

在双向表题目中，先求行和列的总计，然后计算独立假设下的期望联合概率，并与观测到的联合概率进行比较。例如，一张表显示 30 名男生和 20 名女生，其中 18 名男生和 10 名女生喜欢足球，则 P(男生 ∩ 喜欢足球) = 18/50 = 0.36，而 P(男生) × P(喜欢足球) = 30/50 × 28/50 = 0.336，因此事件不独立。

6. Tree Diagrams and Independent Trials | 树状图与独立试验

Tree diagrams are especially useful for modelling independent trials, such as tossing a coin twice or rolling a die multiple times. Along each branch, probabilities are multiplied because the events are independent. At the end, add probabilities of branches that lead to the same outcome. Remember that the probabilities on the second set of branches are the same as the first only if the trials are independent.

树状图特别适合对独立试验建模，例如掷硬币两次或多次掷骰子。沿每条分支，概率相乘，因为事件独立。最后，将导致同一结果的分支概率相加。请记住，只有当试验独立时，第二组分支上的概率才与第一组相同。

For example, if a fair coin is tossed twice, the probability of getting two heads is (1/2) × (1/2) = 1/4. The tree diagram shows four equally likely outcomes, each with probability 1/4. If a biased coin with P(head) = 0.6 is tossed twice, the outcomes are still independent, so P(two heads) = 0.6 × 0.6 = 0.36.

例如，若公平硬币抛掷两次，得到两个正面的概率为 (1/2) × (1/2) = 1/4。树状图显示四种等可能结果，每种概率为 1/4。如果一枚有偏硬币 P(正面) = 0.6 抛掷两次，结果仍然独立，因此 P(两个正面) = 0.6 × 0.6 = 0.36。

7. Binomial Distribution and Independence | 二项分布与独立性

The binomial distribution B(n, p) requires a fixed number n of independent trials, each with the same probability p of success. If the trials are not independent, the binomial model is not valid. Edexcel questions often ask you to justify why a situation can be modelled by a binomial distribution; independence of trials is one of the key conditions.

二项分布 B(n, p) 要求固定次数 n 次独立试验，每次成功概率相同为 p。如果试验不独立，二项模型无效。Edexcel 考试常要求你说明为什么某个情境可以用二项分布建模；试验的独立性是关键条件之一。

In practice, when sampling without replacement from a large population, independence is approximately satisfied if the sample size is less than 5% of the population, but in A-Level questions this is usually stated explicitly. The binomial probability formula P(X = r) = ⁿCᵣ pʳ (1-p)ⁿ⁻ʳ relies on the multiplication rule for independent trials.

实际中，从大总体中不放回抽样时，如果样本量小于总体的 5%，独立性近似成立，但在 A-Level 题目中通常会明确说明。二项概率公式 P(X = r) = ⁿCᵣ pʳ (1-p)ⁿ⁻ʳ 依赖于独立试验的乘法法则。

8. Common Mistakes | 常见错误

Mistake 1: Assuming events are independent just because they are described in separate sentences. Always look for words like ‘replaced’, ‘returned’, ‘fair’, ‘random’ or ‘independent’ in the question. If the question says ‘without replacement’, events are not independent. Mistake 2: Confusing P(A ∩ B) with P(A | B). Mistake 3: Adding probabilities when you should multiply for independent events.

错误 1：仅因为事件在不同句子中描述就假设它们独立。务必寻找题目中的关键词，如 ‘放回’、’返还’、’公平’、’

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

The independence of the Court | 法院的独立性