Probability – Tree Diagrams & Conditional Probability | 概率 – 树状图与条件概率

📚 Probability – Tree Diagrams & Conditional Probability | 概率 – 树状图与条件概率

Probability is one of the most practical topics in IGCSE Mathematics, appearing in nearly every Higher Tier paper. This article covers the key concepts of probability, focusing on tree diagrams, conditional probability, and the common pitfalls students face in the Edexcel IGCSE exams.

概率是 IGCSE 数学中最实用的板块之一，几乎每次 Higher Tier 试卷都会出现。本文将系统讲解概率的核心概念，重点梳理树状图、条件概率，以及 Edexcel IGCSE 考试中同学们最常犯的错误。

1. Basic Probability and the Probability Scale | 基础概率与概率标度

Probability measures how likely an event is to happen. It is always a number between 0 and 1, where 0 means impossible and 1 means certain.

概率用来衡量一件事发生的可能性大小。它的取值始终在 0 到 1 之间：0 表示不可能发生，1 表示必然发生。

P(A) = number of favourable outcomes ÷ total number of outcomes

P(A) = 有利结果数 ÷ 所有等可能结果总数

For example, when rolling a fair six-sided die, the probability of getting a 4 is 1 ÷ 6 = 1⁄6. This formula only works when all outcomes are equally likely.

例如，掷一枚均匀的六面骰子，得到点数 4 的概率为 1 ÷ 6 = 1⁄6。这个公式只有在所有结果等可能时才成立。

The probability scale is a line from 0 to 1. Common values include: 0 (impossible), ½ (even chance), and 1 (certain). On the IGCSE exam, you should be able to place probabilities on this scale and convert between fractions, decimals, and percentages.

概率标度是一条从 0 到 1 的线段。常见取值包括：0（不可能）、½（对半机会）和 1（必然发生）。在 IGCSE 考试中，你需要能把概率标到这条线上，并能在分数、小数和百分数之间互相转换。

2. Key Rules of Probability | 概率的基本法则

There are three fundamental rules you must know before drawing any tree diagram.

在画树状图之前，你必须掌握三条最基本法则。

The probability of an event NOT happening is 1 minus the probability of it happening: P(not A) = 1 − P(A).

事件不发生的概率等于 1 减去它发生的概率：P(非 A) = 1 − P(A)。
For mutually exclusive events (events that cannot happen at the same time), the probability of A or B occurring is the sum: P(A or B) = P(A) + P(B).

对互斥事件（不能同时发生的事件），A 或 B 发生的概率为二者之和：P(A 或 B) = P(A) + P(B)。
The probabilities of ALL possible outcomes of an experiment always add up to 1. This is the key idea behind “subtract from 1” in tree diagrams.

一次试验所有可能结果的概率之和总等于 1。这正是树状图中”用 1 去减”的核心依据。

In Edexcel IGCSE, the addition rule often appears in questions involving “at least one” or “neither”. Recognising mutually exclusive situations is essential for correct calculations.

在 Edexcel IGCSE 中，加法法则常出现在涉及”至少一次”或”两者都不”的题目里。正确识别互斥情形是计算不出错的前提。

3. Independent and Dependent Events | 独立事件与依赖事件

Two events are independent if the outcome of one does not affect the probability of the other. Tossing two coins is a classic example: the result of one toss has no effect on the other.

如果一个事件的结果不影响另一个事件发生的概率，则称这两个事件相互独立。抛两枚硬币就是典型例子：第一次抛掷的结果对第二次毫无影响。

In contrast, dependent events are those where the first outcome changes the probability of the second. Drawing marbles from a bag without replacement is the most common example in IGCSE.

相反，如果第一个结果改变了第二个事件的概率，那么这些事件就是依赖事件。IGCSE 中最常见的例子就是：从袋中取球且不放回。

For independent events: P(A and B) = P(A) × P(B)

独立事件：P(A 且 B) = P(A) × P(B)

This multiplication rule is the foundation of every tree diagram. If events are dependent, you must adjust the second probability to reflect the new total — for instance, when one marble has already been removed.

这条乘法法则是所有树状图的基础。若事件之间是依赖的，第二层分支的概率必须根据新的总数作出调整——例如已经取走一颗球之后。

4. Tree Diagrams for Independent Events | 独立事件的树状图

A tree diagram is a visual way to list all possible outcomes of two or more events. Each branch represents an outcome, and each branch is labelled with its probability.

树状图是一种直观列出多个事件所有可能结果的方法。每一条分支代表一种结果，每条分支上都标注其概率。

Example: A spinner has 3 red sectors and 2 blue sectors. It is spun twice. Draw a tree diagram and find P(two reds).

例：一个转盘有 3 个红色扇形和 2 个蓝色扇形，转动两次。画出树状图，求 P(两次都是红色)。

First spin: P(R) = 3⁄5, P(B) = 2⁄5. Since outcomes are independent (the spinner is reset), both branches for the second spin use the same probabilities.

第一次转动：P(红) = 3⁄5，P(蓝) = 2⁄5。由于两次转动相互独立（转盘每次复位），第二次分支使用相同概率即可。

P(RR) = (3⁄5) × (3⁄5) = 9⁄25

P(两次红) = (3⁄5) × (3⁄5) = 9⁄25

To find P(exactly one red), you add the two paths that contain one red: P(RB) + P(BR) = (3⁄5 × 2⁄5) + (2⁄5 × 3⁄5) = 6⁄25 + 6⁄25 = 12⁄25.

求 P(恰好一次红色)，需要把两条含一红的路径相加：P(红蓝) + P(蓝红) = (3⁄5 × 2⁄5) + (2⁄5 × 3⁄5) = 6⁄25 + 6⁄25 = 12⁄25。

Always multiply along the branches for “and”, and add across the paths for “or”. This single sentence summarises all tree diagram questions.

记住：”且”沿分支相乘，”或”跨路径相加。这一句话足以概括所有树状图题目。

5. Tree Diagrams for Dependent Events | 依赖事件的树状图

When objects are drawn without replacement, the total number of objects decreases, so the second-stage probabilities change.

当物体被抽出且不放回时，物体总数减少，因此第二层的概率会发生改变。

Example: A bag contains 4 red and 3 blue marbles. Two marbles are drawn without replacement. Find P(both red).

例：袋中有 4 颗红球、3 颗蓝球。不放回地抽取两颗，求 P(两颗都是红球)。

First draw: P(R) = 4⁄7. If the first marble is red, only 6 marbles remain and 3 are red, so the second draw has P(R) = 3⁄6 = ½.

第一次抽取：P(红) = 4⁄7。若第一颗是红球，袋中剩余 6 颗，其中 3 颗红球，因此第二次 P(红) = 3⁄6 = ½。

P(both red) = 4⁄7 × 3⁄6 = 12⁄42 = 2⁄7

P(两颗全红) = 4⁄7 × 3⁄6 = 12⁄42 = 2⁄7

The critical skill is updating the numerator AND denominator correctly. Misreading the remaining count is the number one reason marks are lost on this type of question.

关键技巧在于正确更新分子和分母。看错剩余数量是这类题丢分的头号原因。

6. Conditional Probability Formula | 条件概率公式

Conditional probability asks: given that one event has happened, what is the probability of another event? It is written as P(B | A), read as “the probability of B given A”.

条件概率研究的是：已知某事件已发生，另一事件发生的概率是多少？记作 P(B | A)，读作”在 A 发生的条件下 B 的概率”。

P(B | A) = P(A and B) ÷ P(A)

P(B | A) = P(A 且 B) ÷ P(A)

In Edexcel IGCSE, this formula is usually not needed in symbolic form; instead, you read the required probability directly from the branches of a tree diagram.

在 Edexcel IGCSE 中，通常不需要写出这个符号化公式，而是直接从树状图的分支上读出所求概率。

For example, using the marble bag above, P(second red | first red) = 3⁄6. This value is simply the label on the second branch after the first red branch. The formula becomes a useful check when questions link probability with data tables.

例如，沿用上面的取球问题，P(第二颗红 | 第一颗红) = 3⁄6。这个值就是第一层红球分支之后第二层分支上的标签。当题目将概率与数据表结合时，公式就是很好的检验工具。

7. Using Venn Diagrams for Probability | 用文氏图计算概率

Venn diagrams provide a clear way to visualise probabilities, especially for two overlapping events. Each region corresponds to a set operation: A only, B only, both, and neither.

文氏图能清晰地表现概率关系，尤其适合两个有重叠的事件。每个区域对应一种集合关系：仅 A、仅 B、二者皆有、二者皆无。

From a Venn diagram you can directly read:

从文氏图中可以直接读出：

P(A and B) is the probability in the intersection region.

P(A 且 B) 就是交集区域对应的概率。
P(A or B) is the total of all regions inside A or B, counted once.

P(A 或 B) 是 A 与 B 覆盖的所有区域概率之和（重叠部分只算一次）。
P(neither) comes directly from the outside region.

P(两者都不) 直接对应外部区域的概率。

Conditional probability from a Venn diagram is visual: given that the outcome is in A, the sample space shrinks to A, so P(B | A) = area of intersection ÷ area of A.

文氏图中的条件概率非常直观：已知结果落在了 A 中，样本空间就缩小为 A，因此 P(B | A) = 交集区域 ÷ A 区域。

8. With Replacement vs Without Replacement | 有放回与无放回的比较

Whether or not an item is replaced completely changes the tree diagram. The table below summarises the difference:

是否放回会完全改变树状图的构造。下表总结了二者的区别：

Aspect \| 比较项	With Replacement 有放回	Without Replacement 无放回
Events 事件关系	Independent 独立	Dependent 依赖
Second-stage probabilities 第二层概率	Same as first-stage 与第一层相同	Numerator and denominator both change 分子分母都变
Typical exam phrase 典型题干	“and replace it” “并放回”	“without replacement” “不放回”

Be alert: Edexcel sometimes phrases replacement indirectly, such as “the card is returned to the bag before the second draw”. Read the question carefully before writing any probability.

要警惕：Edexcel 有时会用间接方式表述放回，例如”第二次抽取前将卡片放回袋中”。在动笔写任何概率之前，务必仔细读题。

9. Common Exam Mistakes | 常见考试错误

Across past IGCSE papers, examiners repeatedly identify the same errors. Avoid these to secure full marks.

在历年 IGCSE 考卷中，考官反复指出同几类错误。避免它们就能稳拿满分。

Forgetting to add probabilities of different paths when the question asks for “at least one” — you must sum all qualifying paths.

题目问”至少一次”时忘记把多条不同路径相加——必须把所有符合条件的路径概率累加。
Using the same probability for the second draw in a no-replacement question — always reduce the denominator.

在无放回题中第二抽仍沿用原概率——切记分母要减少。
Leaving fractions unsimplified when the mark scheme expects simplest form — for example, 12⁄42 must be written as 2⁄7.

评分标准要求最简分数时却没有约分——例如 12⁄42 必须写成 2⁄7。
Not checking that all branch probabilities from one node sum to 1. If they do not, something is wrong.

没有检查同一节点所有分支概率之和是否为 1。若不为 1，则必有问题。

A quick self-check: after finishing a tree diagram calculation, ask yourself whether the final answer is between 0 and 1 and whether it matches your intuition.

一个快速自检：完成树状图计算后，问问自己最终答案是否在 0 到 1 之间、是否和直觉一致。

10. Worked Exam-Style Questions | 真题风格练习

Q1 (Independent events): A coin is tossed three times. Find the probability of getting exactly two heads.

问题 1（独立事件）：一枚硬币抛掷三次，求恰好出现两次正面的概率。

There are three paths: HHT, HTH, THH. Each has probability (½)³ = 1⁄8. Therefore P(exactly two heads) = 3 × 1⁄8 = 3⁄8.

共有三条路径：正正反、正反正、反正正。每条路径概率为 (½)³ = 1⁄8。因此 P(恰好两次正面) = 3 × 1⁄8 = 3⁄8。

Q2 (Dependent events): A box contains 5 green and 4 yellow counters. Two counters are drawn without replacement. Find the probability that the counters are of different colours.

问题 2（依赖事件）：盒中有 5 个绿色和 4 个黄色计数器。不放回地抽取两个，求两个计数器颜色不同的概率。

Different colours means one green and one yellow: P(GY) + P(YG) = (5⁄9 × 4⁄8) + (4⁄9 × 5⁄8) = 20⁄72 + 20⁄72 = 40⁄72 = 5⁄9.

颜色不同即一绿一黄：P(绿黄) + P(黄绿) = (5⁄9 × 4⁄8) + (4⁄9 × 5⁄8) = 20⁄72 + 20⁄72 = 40⁄72 = 5⁄9。

Q3 (Conditional probability): In a class, 60% of students play football and 35% play both football and tennis. Given that a student plays football, find the probability that they also play tennis.

问题 3（条件概率）：某班级中，60% 的学生踢足球，35% 的学生既踢足球又打网球。已知某学生踢足球，求他同时也打网球的概率。

P(T | F) = P(F and T) ÷ P(F) = 0.35 ÷ 0.60 = 7⁄12

P(网球 | 足球) = P(足球且网球) ÷ P(足球) = 0.35 ÷ 0.60 = 7⁄12

Practising these three styles covers most of what Edexcel can ask on probability in a calculator or non-calculator paper.

练好这三种题型，基本可以覆盖 Edexcel 在计算器或非计算器试卷中可能考查的大部分概率内容。

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Probability – Tree Diagrams & Conditional Probability | 概率 – 树状图与条件概率