Independent Events and Tree Diagrams | 独立事件与树状图应用

📚 Independent Events and Tree Diagrams | 独立事件与树状图应用

Probability is one of the most practical and frequently tested topics in the Edexcel IGCSE Mathematics syllabus. Among its core concepts, independent events and tree diagrams stand out as essential tools for solving multi-step probability problems. This article will guide you through the definitions, the multiplication rule, and the systematic use of tree diagrams, complete with worked examples aligned to the Edexcel examination style.

概率是 Edexcel IGCSE 数学考纲中最实用、最常考的主题之一。其中，独立事件与树状图是解决多步骤概率问题的核心工具。本文将带你系统掌握独立事件的定义、乘法法则以及树状图的规范用法，并配以符合 Edexcel 考试风格的精讲例题。

1. What Are Independent Events? | 什么是独立事件？

Two events A and B are said to be independent if the occurrence of one event does not affect the probability of the other event occurring. In other words, the probability of B happening remains the same whether or not A has already happened. For example, tossing a fair coin twice: the result of the first toss has no influence on the result of the second toss.

两个事件 A 和 B 被称为独立事件，是指其中一个事件的发生不会影响另一个事件发生的概率。换句话说，无论 A 是否已经发生，B 发生的概率都保持不变。例如，连续抛两次均匀硬币：第一次抛掷的结果对第二次抛掷的结果没有任何影响。

Mathematically, for independent events A and B:

P(A and B) = P(A) × P(B)

This is known as the multiplication rule for independent events. It is the single most important formula in this chapter, and it forms the basis for every tree diagram calculation you will perform.

数学上，对于独立事件 A 和 B：

P(A 且 B) = P(A) × P(B)

这就是独立事件的乘法法则。它是本章最重要的公式，也是所有树状图计算的基石。

2. The Multiplication Rule in Depth | 乘法法则深入理解

The multiplication rule can be extended to more than two independent events. For three independent events A, B, and C:

P(A and B and C) = P(A) × P(B) × P(C)

This extension is particularly useful when dealing with questions that involve three stages, such as selecting three balls from a bag with replacement, or throwing a die three times.

乘法法则可以推广到两个以上的独立事件。对于三个独立事件 A、B 和 C：

P(A 且 B 且 C) = P(A) × P(B) × P(C)

这一推广在处理三阶段问题时非常有用，例如从袋中放回地抽取三个球，或连续掷三次骰子。

It is crucial to remember that the multiplication rule for independent events requires the word ‘and’. When we calculate the probability of both events happening, we multiply. When the word ‘or’ is involved, we typically add, but we must first check whether the events are mutually exclusive.

务必记住：独立事件的乘法法则对应的是“且”的关系。当我们计算两个事件同时发生的概率时，用乘法。当涉及“或”时，通常用加法，但首先要判断事件是否互斥。

3. Tree Diagrams: The Basics | 树状图基础

A tree diagram is a visual representation of all possible outcomes of a sequence of events. Each branch represents a possible outcome, and the probability of that outcome is written on the branch. To find the probability of a particular path (a combination of outcomes), you multiply the probabilities along that path.

树状图是表示一系列事件所有可能结果的直观图形。每条分支代表一种可能的结果，分支上标注该结果的概率。要计算某条路径（结果的组合）的概率，将路径上所有分支的概率相乘即可。

Key conventions for drawing tree diagrams:

Each set of branches must sum to 1.
Multiply along branches for ‘and’ scenarios.
Add the results of different paths for ‘or’ scenarios.

绘制树状图的规范要点：

每组分支的概率之和必须等于 1。
“且”的情景沿分支相乘。
“或”的情景将不同路径的结果相加。

Let us begin with a classic example. A bag contains 5 red balls and 3 blue balls. A ball is drawn, its colour is recorded, and then it is replaced. This process is repeated twice. We can use a tree diagram to represent all possible outcomes.

我们先看一个经典例子。一个袋中有 5 个红球和 3 个蓝球。抽取一个球并记录颜色后放回，重复两次。我们可以用树状图表示所有可能的结果。

4. Example: Drawing with Replacement | 例题：有放回抽取

Consider the bag with 5 red (R) and 3 blue (B) balls. Since the ball is replaced after the first draw, the probabilities for the second draw are identical to the first: P(R) = 5/8 and P(B) = 3/8. The two draws are independent events.

考虑袋中有 5 个红球（R）和 3 个蓝球（B）。由于第一次抽取后放回，第二次抽取的概率与第一次完全相同：P(R) = 5/8，P(B) = 3/8。两次抽取是独立事件。

The tree diagram has four paths:

RR: P = 5/8 × 5/8 = 25/64
RB: P = 5/8 × 3/8 = 15/64
BR: P = 3/8 × 5/8 = 15/64
BB: P = 3/8 × 3/8 = 9/64

树状图共有四条路径：

RR：P = 5/8 × 5/8 = 25/64
RB：P = 5/8 × 3/8 = 15/64
BR：P = 3/8 × 5/8 = 15/64
BB：P = 3/8 × 3/8 = 9/64

Notice that the sum of all four probabilities is 25/64 + 15/64 + 15/64 + 9/64 = 64/64 = 1, confirming that we have accounted for all possible outcomes.

注意四条路径的概率之和为 25/64 + 15/64 + 15/64 + 9/64 = 64/64 = 1，说明我们已经涵盖了所有可能的结果。

5. Finding Probabilities from Tree Diagrams | 从树状图中求概率

Now let us use the tree diagram to answer typical examination questions. Suppose we want the probability of getting exactly one red ball. This corresponds to the paths RB and BR. Since these are mutually exclusive outcomes (they cannot both happen at the same time), we add their probabilities:

现在用树状图来解答典型考试问题。假设我们要求恰好抽到一个红球的概率。这对应路径 RB 和 BR。由于这两个结果互斥（不可能同时发生），我们将它们的概率相加：

P(exactly one red) = 15/64 + 15/64 = 30/64 = 15/32

If we want the probability of at least one red ball, we can either add the three paths containing R (RR, RB, BR) or use the complement: 1 − P(BB). The complement method is often quicker:

如果我们要求至少抽到一个红球的概率，可以相加三条包含 R 的路径（RR、RB、BR），也可以使用补集：1 − P(BB)。补集法通常更快：

P(at least one red) = 1 − 9/64 = 55/64

Both approaches are valid in the Edexcel examination, but the complement method reduces the number of calculations and minimises the risk of arithmetic errors.

这两种方法在 Edexcel 考试中均可行，但补集法减少了计算量，并最大限度降低了算术失误的风险。

6. Without Replacement: A Critical Distinction | 无放回抽取：关键区别

Many examination questions involve drawing without replacement. In this case, the events are not independent because the probability of the second draw depends on the outcome of the first draw. For instance, if a red ball is drawn first and not replaced, there are fewer red balls left, so the probability of drawing a red ball second changes.

许多考试题目涉及无放回抽取。此时事件不是独立的，因为第二次抽取的概率取决于第一次抽取的结果。例如，如果先抽到红球且不放回，袋中红球减少，第二次抽到红球的概率就会改变。

Consider the same bag of 5 red and 3 blue balls, but this time the first ball is NOT replaced before the second draw. The probabilities for the second draw now depend on the first outcome:

考虑同样的袋中有 5 个红球和 3 个蓝球，但这次第一次抽取后不放回。第二次抽取的概率现在取决于第一次的结果：

If first draw is R (probability 5/8), then for the second draw: P(R) = 4/7 and P(B) = 3/7.
If first draw is B (probability 3/8), then for the second draw: P(R) = 5/7 and P(B) = 2/7.

如果第一次抽到 R（概率 5/8），则第二次：P(R) = 4/7，P(B) = 3/7。
如果第一次抽到 B（概率 3/8），则第二次：P(R) = 5/7，P(B) = 2/7。

The tree diagram still works perfectly, but the probabilities on the second set of branches are conditional on the first outcome. The multiplication rule still applies: multiply along the branches. However, you must be extremely careful to read the question and determine whether replacement occurs.

树状图依然完全适用，但第二层分支上的概率是以第一次结果为条件的条件概率。乘法法则仍然成立：沿分支相乘。但是，你必须极其仔细地读题，判断是否有放回。

7. Worked Example: Without Replacement | 例题精讲：无放回抽取

A bag contains 4 green and 6 yellow sweets. Two sweets are taken at random without replacement. Find the probability that both sweets are green.

一个袋中有 4 颗绿色糖果和 6 颗黄色糖果。随机不放回地取两颗。求两颗都是绿色的概率。

On the first draw, P(G) = 4/10. If the first sweet is green, there are now 3 green and 6 yellow sweets left, a total of 9. Therefore, P(G on second | G on first) = 3/9 = 1/3.

第一次抽取时，P(G) = 4/10。如果第一颗是绿色，则剩下 3 颗绿色和 6 颗黄色，共 9 颗。因此，第二次抽到绿色的条件概率为 3/9 = 1/3。

P(both green) = 4/10 × 3/9 = 12/90 = 2/15

Now find the probability that the two sweets are of different colours. This can happen in two ways: green first then yellow, or yellow first then green.

再求两颗糖果颜色不同的概率。这有两种情况：先绿后黄，或先黄后绿。

P(G then Y) = 4/10 × 6/9 = 24/90

P(Y then G) = 6/10 × 4/9 = 24/90

P(different colours) = 24/90 + 24/90 = 48/90 = 8/15

This example shows that the tree diagram structure is identical whether or not replacement occurs; the difference lies entirely in the numerical values on the second set of branches.

这个例子说明，无论是否放回，树状图的结构完全相同；区别完全在于第二层分支上的数值。

8. Three-Stage Tree Diagrams | 三阶段树状图

For three-stage problems, the tree diagram has three sets of branches and eight paths in total. The same principles apply: multiply along each path, and add paths for mutually exclusive outcomes.

对于三阶段问题，树状图有三组分支，共八条路径。同样的原则适用：沿路径相乘，互斥结果相加。

Example: A fair coin is tossed three times. Find the probability of getting exactly two heads.

例题：一枚均匀硬币抛掷三次。求恰好出现两次正面的概率。

Each toss has P(H) = 1/2 and P(T) = 1/2. The paths that give exactly two heads are HHT, HTH, and THH. Each has probability:

每次抛掷有 P(H) = 1/2 和 P(T) = 1/2。恰好出现两次正面的路径为 HHT、HTH 和 THH。每条路径的概率为：

P(HHT) = P(HTH) = P(THH) = 1/2 × 1/2 × 1/2 = 1/8

P(exactly two heads) = 1/8 + 1/8 + 1/8 = 3/8

When drawing a three-stage tree diagram, be sure to label every branch clearly and keep the probabilities aligned so that the examiner can follow your logic without difficulty.

绘制三阶段树状图时，务必清晰标注每条分支并使概率对齐，让阅卷者能够轻松理解你的逻辑。

9. Conditional Probability Notation | 条件概率符号

In tree diagrams, the probabilities on the second and later sets of branches are often conditional probabilities. The notation P(B|A) means ‘the probability of B given that A has occurred’. When multiplying along branches, you calculate P(A) × P(B|A), which gives P(A and B).

在树状图中，第二层及之后分支上的概率通常是条件概率。符号 P(B|A) 表示“在 A 已发生的条件下 B 发生的概率”。沿分支相乘时，你计算的是 P(A) × P(B|A)，即 P(A 且 B)。

For independent events, P(B|A) = P(B), which is why the multiplication rule simplifies to P(A) × P(B). Understanding this notation helps you interpret tree diagrams correctly and avoid confusing independent and dependent events.

对于独立事件，P(B|A) = P(B)，这就是乘法法则简化为 P(A) × P(B) 的原因。理解这一符号有助于你正确解读树状图，避免混淆独立事件与相关事件。

In the Edexcel IGCSE examination, conditional probabilities are explicitly tested. You may be asked to calculate P(B|A) from a tree diagram or from a two-way table.

在 Edexcel IGCSE 考试中，条件概率会被明确考查。题目可能要求你从树状图或双向表中计算 P(B|A)。

10. Real-World Applications | 实际应用

Tree diagrams are not merely an examination technique; they model real-world situations across many fields. In medicine, doctors use decision trees to estimate the probability of a patient having a disease given a positive test result. In manufacturing, quality control engineers use tree diagrams to calculate the probability of defective products in a batch. In finance, analysts model the probability of market outcomes using sequential decision trees.

树状图不仅仅是应试技巧；它在众多领域建模真实情境。医学中，医生使用决策树来估计患者检测结果为阳性时患病的概率。制造业中，质量控制工程师用树状图计算整批产品中出现次品的概率。金融领域中，分析师用序贯决策树对市场结果的概率进行建模。

For IGCSE students, the ability to construct and interpret tree diagrams demonstrates a mature understanding of uncertainty and risk, which is a key mathematical competency assessed in the Edexcel syllabus under the strand of probability.

对于 IGCSE 学生而言，能够构建并解读树状图，体现了对不确定性和风险的成熟理解，这正是 Edexcel 考纲中概率部分所评估的核心数学能力。

Whenever you encounter a problem involving a sequence of trials with ‘success’ and ‘failure’ outcomes, a tree diagram should be your first instinct.

每当你遇到涉及一系列具有“成功”和“失败”结果的试验问题时，树状图应该成为你的第一反应。

11. Common Mistakes and How to Avoid Them | 常见错误与规避方法

Students often lose marks in probability questions due to a small number of recurring errors. Below are the most common pitfalls and how to avoid them:

学生在概率题中丢分往往源于几个反复出现的错误。以下是最常见的陷阱及规避方法：

Forgetting to check for replacement: Always read the question carefully. The words ‘with replacement’ and ‘without replacement’ change the probabilities on the second set of branches completely.
Adding instead of multiplying: Remember: ‘and’ means multiply, ‘or’ means add (provided the events are mutually exclusive).
Forgetting that branch probabilities sum to 1: Before starting any calculation, check that each pair of branches sums to 1.
Not simplifying fractions: Edexcel mark schemes often require answers in their simplest form.

忘记判断是否有放回：务必仔细读题。“有放回”和“无放回”会完全改变第二层分支上的概率。
把加法当乘法：记住：“且”用乘法，“或”用加法（前提是事件互斥）。
忘记分支概率之和为 1：开始任何计算前，检查每组分支概率之和是否为 1。
未化简分数：Edexcel 评分标准通常要求答案化为最简分数。

You can avoid these errors by taking 30 seconds to sketch a labelled tree diagram before performing any calculations. This small investment of time yields substantial dividends in accuracy.

你可以通过先花 30 秒绘制一个带标签的树状图再进行计算来避免这些错误。这小小的时投入会在准确性上带来丰厚的回报。

12. Exam-Style Practice Questions | 真题风格练习

Let us end with a set of practice questions in the Edexcel style. Attempt each question before reading the solution.

最后我们做一组 Edexcel 风格的练习题。请先尝试作答，再对照解答。

Question 1: A spinner has 3 equal sections labelled A, B, and C. It is spun twice. Find the probability that both spins land on the same letter.

题目 1：一个转盘有 3 个等分区域，分别标为 A、B、C。转动两次。求两次落在同一字母上的概率。

Solution: P(same) = P(AA) + P(BB) + P(CC) = 1/9 + 1/9 + 1/9 = 1/3.

解答：P(相同) = P(AA) + P(BB) + P(CC) = 1/9 + 1/9 + 1/9 = 1/3。

Question 2: A box contains 2 black pens and 5 blue pens. Two pens are drawn without replacement. Find the probability that both pens are blue.

题目 2：一个盒中有 2 支黑色笔和 5 支蓝色笔。不放回地抽取两支。求两支都是蓝色的概率。

Solution: P(BB) = 5/7 × 4/6 = 20/42 = 10/21.

解答：P(蓝蓝) = 5/7 × 4/6 = 20/42 = 10/21。

Question 3: The probability that a train is late on any given day is 0.2. Assuming that lateness on different days is independent, find the probability that the train is late on exactly one of the next two days.

题目 3：某列火车任意一天晚点的概率是 0.2。假设不同日期的晚点相互独立，求接下来两天中恰好一天晚点的概率。

Solution: P(late then on time) = 0.2 × 0.8 = 0.16. P(on time then late) = 0.8 × 0.2 = 0.16. Total = 0.32.

解答：P(晚点后准时) = 0.2 × 0.8 = 0.16。P(准时后晚点) = 0.8 × 0.2 = 0.16。合计 = 0.32。

In summary, independent events and tree diagrams are inseparable partners in the IGCSE probability curriculum. Master the multiplication rule, practice drawing tree diagrams methodically, and pay careful attention to whether replacement occurs. With consistent practice, you will find these questions some of the most rewarding in the examination, as they reward clarity of thought and systematic method. Good luck with your revision.

总而言之，独立事件与树状图是 IGCSE 概率章节中密不可分的伙伴。掌握乘法法则，练习规范地绘制树状图，并密切注意是否有放回。通过持续练习，你会发现这类题目是考试中最能回报你清晰思维与系统性方法的题型。祝你复习顺利。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Independent Events and Tree Diagrams | 独立事件与树状图应用