Combined Events and Probability Rules | 组合事件与概率规则

📚 Combined Events and Probability Rules | 组合事件与概率规则

In A-Level Mathematics, understanding how to combine events is essential for solving complex probability problems. Whether you are dealing with mutually exclusive events, independent trials, or conditional probabilities, a solid grasp of the rules that govern combined events allows you to model real-world situations and tackle exam questions with confidence. This article breaks down the key concepts, formulas, and techniques needed for Edexcel A-Level Probability.

在A-Level数学中，理解如何组合事件是解决复杂概率问题的关键。无论你处理的是互斥事件、独立试验还是条件概率，扎实掌握组合事件的规则都能让你建立真实世界模型，并自信地应对考试题目。本文详细解析Edexcel A-Level概率部分的核心概念、公式和解题方法。

1. Introduction to Combined Events | 组合事件简介

A combined event occurs when we consider the outcome of two or more events happening together. For example, tossing a coin and rolling a die is a combined event because we look at the result of both actions simultaneously. In probability, we often need to find the chance of event A or event B occurring, or event A and event B occurring. These are known as ‘union’ and ‘intersection’ respectively.

当我们将两个或多个事件的结果结合起来考虑时，就产生了组合事件。例如，同时抛一枚硬币和掷一个骰子就是一个组合事件，因为我们要同时考察两个动作的结果。在概率中，我们经常需要计算事件A 或事件B发生的概率，以及事件A 且事件B发生的概率，这分别称为“并集”和“交集”。

The sample space for combined events can be represented using a grid, a tree diagram, or a Venn diagram. The total number of outcomes is the product of the individual numbers of outcomes if the stages are independent. Understanding how to set up the sample space correctly is the first step in solving any combined event problem.

组合事件的样本空间可以用表格、树形图或维恩图来表示。如果各阶段相互独立，总结果数就是各阶段结果数的乘积。正确建立样本空间是解决所有组合事件问题的第一步。

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2. Mutually Exclusive Events | 互斥事件

Two events are mutually exclusive if they cannot occur at the same time. For example, when rolling a die, the events ‘rolling a 3’ and ‘rolling an even number’ are not mutually exclusive because a 3 is not even, but ‘rolling a 3’ and ‘rolling a 5’ are mutually exclusive, as a single roll cannot be both a 3 and a 5.

如果两个事件不可能同时发生，则它们是互斥事件。例如，掷骰子时，“掷出3点”和“掷出偶数”并不互斥，因为3不是偶数；但“掷出3点”和“掷出5点”就是互斥的，因为一次掷骰不可能同时是3和5。

For mutually exclusive events, the probability of either event A or event B occurring is simply the sum of their individual probabilities: P(A or B) = P(A) + P(B). There is no overlap to subtract. This is the simplest form of the addition rule.

对于互斥事件，事件A或事件B发生的概率直接就是它们各自概率的和：P(A 或 B) = P(A) + P(B)。这里没有重叠部分需要减去。这是加法法则的最简单形式。

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3. The Addition Rule for Any Two Events | 任意两事件的加法法则

When events are not mutually exclusive, they have outcomes in common. If we simply add P(A) and P(B), the intersection P(A ∩ B) is counted twice. Therefore, the general addition rule is:

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

当事件不互斥时，它们存在共同的结果。如果直接把P(A)和P(B)相加，交集P(A ∩ B)就会被重复计算。因此，通用的加法法则为：

P(A ∪ B) = P(A) + P(B) – P(A ∩ B)

This formula is fundamental and works for all events, whether mutually exclusive or not. When A and B are mutually exclusive, P(A ∩ B) = 0, and the formula reduces to the simple addition rule.

这个公式是根本性的，适用于所有事件，无论是否互斥。当A和B互斥时，P(A ∩ B) = 0，公式就退化为简单的加法法则。

Example: In a class, the probability that a student studies Mathematics is 0.7, the probability they study Physics is 0.5, and the probability they study both is 0.3. The probability that a student studies Mathematics or Physics is 0.7 + 0.5 – 0.3 = 0.9.

例子：一个班级中，学生学习数学的概率是0.7，学习物理的概率是0.5，两门都学的概率是0.3。那么学生学习数学或物理的概率是0.7 + 0.5 – 0.3 = 0.9。

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4. Independent Events | 独立事件

Two events are independent if the occurrence of one does not affect the probability of the other occurring. For instance, flipping a fair coin twice: the outcome of the first flip does not change the probability of getting heads on the second flip. Independence is a crucial concept when combining events using multiplication.

如果一件事的发生不影响另一件事发生的概率，则二者为独立事件。例如两次抛掷公平硬币：第一次的结果不会改变第二次出现正面的概率。在使用乘法组合事件时，独立性是一个核心概念。

To test for independence, check whether P(A ∩ B) = P(A) × P(B). If this equality holds, the events are independent. Be careful not to assume independence unless the problem states it or the experiment makes it clear (e.g. repeated trials with replacement).

检验独立性的方法是检查P(A ∩ B) = P(A) × P(B)是否成立。如果这个等式成立，则事件是独立的。注意不要轻易假设独立性，除非题目明确指出或实验条件保证（如有放回重复试验）。

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5. The Multiplication Rule for Independent Events | 独立事件的乘法法则

If A and B are independent, the probability that both events occur is the product of their individual probabilities:

P(A and B) = P(A) × P(B)

如果A和B独立，那么两个事件同时发生的概率是它们各自概率的乘积：

P(A 且 B) = P(A) × P(B)

This rule extends to any number of independent events. For example, the probability of tossing three heads in a row with a fair coin is (1/2) × (1/2) × (1/2) = 1/8.

该规则可以推广到任意多个独立事件。例如，用公平硬币连续掷出三个正面的概率是(1/2) × (1/2) × (1/2) = 1/8。

Always check that events are independent before applying this rule. If events are not independent, a different multiplication rule involving conditional probability must be used.

在应用此规则前务必要确认事件是独立的。如果事件不是独立的，就必须使用涉及条件概率的另一种乘法规则。

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6. Tree Diagrams for Combined Events | 组合事件的树形图

Tree diagrams are a powerful tool for visualising the outcomes of combined events, especially when an experiment involves several stages. Each branch represents a possible outcome at that stage, and the probability is written along the branch. The probabilities on the branches from a single point must sum to 1.

树形图是可视化组合事件结果的有力工具，当试验包含多个阶段时尤为有用。每条分支代表该阶段的一种可能结果，概率写在分支线上。从同一点发出的各分支概率之和必须等于1。

To find the probability of a particular sequence of events, multiply the probabilities along the corresponding path. To find the total probability of an event that occurs via multiple paths, add the probabilities of those paths. Tree diagrams are particularly useful for handling conditional probabilities and dependent events in multi-stage experiments.

要计算某一特定事件序列的概率，只需将对应路径上的概率相乘。要计算可通过多条路径发生的事件总概率，则将各路径概率相加。在处理多阶段试验中的条件概率和相依事件时，树形图尤为实用。

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7. Conditional Probability and Combined Events | 条件概率与组合事件

Conditional probability is the probability of event A occurring given that event B has already occurred, written as P(A|B). It is defined as:

P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0

条件概率是指在事件B已经发生的条件下事件A发生的概率，记作P(A|B)。其定义为：

P(A|B) = P(A ∩ B) / P(B)，其中 P(B) > 0

This formula rearranges to give a general multiplication rule valid for any events, independent or not:

P(A ∩ B) = P(B) × P(A|B) or P(A) × P(B|A)

通过移项，这个公式可转化为对所有事件都通用的乘法规则（无论独立与否）：

P(A ∩ B) = P(B) × P(A|B) 或 P(A) × P(B|A)

Understanding conditional probability is essential for analysing situations where the sample space is reduced by the given condition. It underpins Bayes’ theorem and many exam problems involving two-way tables and tree diagrams.

理解条件概率对于分析样本空间因给定条件而缩减的情况至关重要。它是贝叶斯定理以及许多涉及双向表格和树形图的考题的基础。

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8. Dependent Events and the General Multiplication Rule | 相依事件与通用乘法法则

Events are dependent if the probability of one occurring depends on whether the other has occurred. In such cases, we cannot simply multiply unconditional probabilities. Instead, we use:

P(A and B) = P(A) × P(B|A)

如果一个事件发生的概率依赖于另一事件是否发生，则它们是相依事件。这时，我们不能简单地乘以无条件概率，而应使用：

P(A 且 B) = P(A) × P(B|A)

This formula says the probability of both A and B occurring is the probability of A times the probability of B given that A has happened. It is critical to identify whether events are dependent, often signalled by ‘without replacement’ in selection problems.

该公式表示A和B同时发生的概率等于A的概率乘以在A已发生条件下B的概率。准确识别事件是否相依非常关键，通常选题中的“不放回”抽取就是典型标志。

Example: A bag contains 3 red and 2 blue balls. Two balls are drawn without replacement. The probability both are red is P(1st red) × P(2nd red | 1st red) = (3/5) × (2/4) = 6/20 = 0.3.

例子：一个袋子里有3个红球和2个蓝球。不放回地抽取两个球。两个都是红球的概率为 P(第一个红) × P(第二个红|第一个红) = (3/5) × (2/4) = 6/20 = 0.3。

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9. Using Venn Diagrams for Combined Events | 利用维恩图分析组合事件

Venn diagrams show the relationships between different events within a sample space. They are particularly useful for visualising unions, intersections, and complements. When given percentages or probabilities, you can place numbers in the appropriate regions to solve for unknowns.

维恩图能展示同一样本空间中不同事件间的关系。它们对于可视化并集、交集和补集特别有用。当题目给出百分比或概率时，可以将数字填入相应区域以求解未知量。

A typical exam question might provide P(A), P(B), and P(A ∪ B), then ask you to find P(A ∩ B) or P(neither A nor B). By labelling the overlapping region and applying the addition rule, Venn diagrams help organise your working clearly.

典型的考题可能会给出P(A)、P(B)和P(A ∪ B)，然后要求计算P(A ∩ B)或P(既非A也非B)。通过标注重叠区域并应用加法法则，维恩图能帮助你清晰地组织解题过程。

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10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students often confuse mutually exclusive with independent. Remember: mutually exclusive events have no overlap (P(A ∩ B)=0), while independent events have P(A ∩ B)=P(A)×P(B). Two events can be independent and not mutually exclusive, unless one of them has probability zero.

学生经常将互斥与独立混淆。请牢记：互斥事件没有重叠（P(A ∩ B)=0），而独立事件满足P(A ∩ B)=P(A)×P(B)。两个事件可以既独立又不互斥，除非其中一个事件概率为零。

Another common error is forgetting to adjust probabilities for conditional branches in tree diagrams, especially in ‘without replacement’ scenarios. Always check whether the sample space changes after each stage. Also, never add probabilities for events connected by ‘and’ unless they are mutually exclusive outcomes of a single experiment (which would be incorrect anyway; ‘and’ requires multiplication).

另一个常见错误是在树形图中忘记调整条件分支的概率，特别是在“不放回”场景中。务必检查每阶段后样本空间是否发生变化。同时，决不要将用“且”连接的事件的概率相加，除非它们是一次试验中互斥的结果（这本身也是错误的；“且”应该用乘法）。

In exam questions, write down the formula you are using, substitute values carefully, and keep probabilities as fractions where possible to maintain accuracy. A clear working is rewarded even if the final answer is slightly off.

在考试题目中，写下所用的公式，仔细代入数值，并尽可能保留分数形式以保持准确。即使最终答案略有偏差，清晰的解题过程也能获得认可。

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11. Worked Example: Combined Events in Context | 实例讲解：情境中的组合事件

Question: A factory produces light bulbs on two machines, A and B. Machine A produces 60% of the bulbs, and 2% of its bulbs are defective. Machine B produces the remaining 40%, with a 5% defect rate. A bulb is selected at random. What is the probability it is defective? Given that the bulb is defective, what is the probability it was produced by Machine B?

题目：某工厂有两台机器A和B生产灯泡。机器A生产60%的灯泡，其产品中有2%为次品。机器B生产剩余的40%，次品率为5%。随机抽取一个灯泡。它恰为次品的概率是多少？在已知灯泡为次品的条件下，它由机器B生产的概率是多少？

We first construct a tree diagram or use law of total probability. P(defective) = P(A and defective) + P(B and defective) = (0.60 × 0.02) + (0.40 × 0.05) = 0.012 + 0.020 = 0.032.

我们先构建树形图或应用全概率公式。P(次品) = P(A且次品) + P(B且次品) = (0.60 × 0.02) + (0.40 × 0.05) = 0.012 + 0.020 = 0.032。

For the second part, we need P(B|defective) = P(B and defective) / P(defective) = 0.020 / 0.032 = 0.625. So, given that a bulb is defective, there is a 62.5% chance it came from Machine B.

对于第二问，需要求 P(B|次品) = P(B且次品) / P(次品) = 0.020 / 0.032 = 0.625。因此，已知灯泡是次品时，它有62.5%的概率来自机器B。

This example illustrates how to combine conditional probabilities with the addition and multiplication rules, a very common exam task.

这个例子展示了如何将条件概率与加法、乘法法则结合使用，这是考试中非常常见的题型。

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12. Summary of Key Formulas | 核心公式总结

• Addition Rule (general): P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
• Mutually Exclusive Events: P(A ∪ B) = P(A) + P(B) [since P(A ∩ B)=0]
• Multiplication Rule for Independent Events: P(A ∩ B) = P(A) × P(B)
• General Multiplication Rule: P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
• Conditional Probability: P(A|B) = P(A ∩ B) / P(B)
• Test for Independence: P(A ∩ B) = P(A) × P(B)
• Complement Rule: P(not A) = 1 – P(A) (useful for combined events via Venn diagrams)

• 通用加法法则： P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
• 互斥事件： P(A ∪ B) = P(A) + P(B) [因为 P(A ∩ B)=0]
• 独立事件乘法法则： P(A ∩ B) = P(A) × P(B)
• 通用乘法法则： P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)
• 条件概率： P(A|B) = P(A ∩ B) / P(B)
• 独立性检验： P(A ∩ B) = P(A) × P(B)
• 补集法则： P(非A) = 1 – P(A) （通过维恩图在组合事件中很有用）

Memorise these rules and practise applying them to different contexts, such as tree diagrams, two-way tables, and word problems. Mastery of combined events will greatly strengthen your overall probability skills for the Edexcel A-Level exam.

熟记这些法则，并练习将它们应用到不同情境中，如树形图、双向表格和文字题。掌握组合事件将极大提升你在Edexcel A-Level考试中的整体概率解题能力。

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