IGCSE Mathematics: Probability Tree Diagrams and Combined Events | IGCSE 数学：概率树图与组合事件

📚 IGCSE Mathematics: Probability Tree Diagrams and Combined Events | IGCSE 数学：概率树图与组合事件

Probability at IGCSE level often feels straightforward until questions combine several events, draw branches, or remove items from a bag. This article explains the essential ideas behind sample spaces, mutually exclusive events, independent events, tree diagrams, conditional probability, and the rules for combined events, so you can answer harder questions with confidence.

在 IGCSE 阶段，概率看起来并不难，但一旦题目涉及多个事件、画出树状图或从袋子中取出物品，很多同学就会开始丢分。本文讲解样本空间、互斥事件、独立事件、树状图、条件概率以及组合事件的运算规则，帮助你更有把握地解决难题。

1. Basic Probability Language | 基础概率术语

A probability is a number between 0 and 1 that measures how likely an event is to occur. A probability of 0 means the event is impossible, while a probability of 1 means the event is certain.

概率是介于 0 和 1 之间的一个数，用来衡量事件发生的可能性。概率为 0 表示事件不可能发生，概率为 1 表示事件必然发生。

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

P(事件) = 有利结果数 ÷ 总结果数

In IGCSE questions, the universal set, outcomes, events, and the idea of fair dice or equally likely outcomes are assumed unless the question states otherwise. Be careful to count outcomes only when every possible result has the same chance.

在 IGCSE 题目中，如果没有特别说明，通常默认骰子是公平的，每个基本结果出现的可能性相同。你在计数时必须确保所有结果的机会相等，否则不能直接使用这个公式。

2. Sample Space and Equally Likely Outcomes | 样本空间与等可能结果

The sample space is the set of all possible outcomes of an experiment. For a fair six-sided die, the sample space is {1, 2, 3, 4, 5, 6}, and each number has probability 1/6.

样本空间是一个试验所有可能结果的集合。对于一个公平的六面骰子，样本空间为 {1, 2, 3, 4, 5, 6}，每个数字出现的概率都是 1/6。

For two dice, there are 36 equally likely ordered pairs. A systematic table helps avoid missing outcomes. If two dice are rolled, the probability of a total of 5 is 4/36 because the favourable pairs are (1,4), (2,3), (3,2), and (4,1).

掷两枚骰子时共有 36 个等可能的有序结果。系统列表或表格可以避免遗漏。若两枚骰子的点数和为 5，则有 4 个有利结果，分别是 (1,4)、(2,3)、(3,2) 和 (4,1)，因此概率为 4/36。

List outcomes systematically in tables or lists.
用表格或列表系统地列出所有结果。
Check that the total number of outcomes matches the product rule of counting.
检查总结果数是否符合乘法计数原理。

3. Mutually Exclusive Events | 互斥事件

Two events are mutually exclusive if they cannot happen at the same time. For example, when rolling one die, the events ‘rolling a 2’ and ‘rolling an odd number’ are mutually exclusive, because 2 is not odd.

如果两个事件不能同时发生，它们就是互斥事件。例如，掷一颗骰子时，“掷出 2”和“掷出奇数”是互斥事件，因为 2 不是奇数。

If A and B are mutually exclusive, P(A ∪ B) = P(A) + P(B)

如果 A 与 B 互斥，则 P(A ∪ B) = P(A) + P(B)

This addition rule is often used for ‘or’ questions. If the events overlap, you must subtract the overlap once. The general addition rule is P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

这个加法法则常用于“或”的问题。如果事件有重叠部分，就必须减去一次重叠部分。一般加法法则为 P(A ∪ B) = P(A) + P(B) − P(A ∩ B)。

4. Independent Events | 独立事件

Two events are independent if the occurrence of one event does not affect the probability of the other. For example, rolling a die and tossing a coin are independent events.

如果两个事件中一个事件发生与否不影响另一个事件发生的概率，它们就是独立事件。例如，掷骰子和抛硬币是独立事件。

If A and B are independent, P(A ∩ B) = P(A) × P(B)

如果 A 与 B 独立，则 P(A ∩ B) = P(A) × P(B)

Students sometimes use this multiplication rule incorrectly when events are not independent. Later, tree diagrams show how to multiply probabilities along branches even when probabilities change, such as when items are taken from a bag without replacement.

有些同学会错误地在事件不独立时使用这个乘法规则。后文树状图会展示即使概率发生变化，例如从袋中不放回取球时，也可以沿分支相乘。

5. Tree Diagrams for Two-Stage Events | 两阶段事件树状图

A tree diagram is a visual way to display all outcomes of a multi-stage event. It is extremely useful when probabilities change after the first outcome, such as removing a counter from a bag and not replacing it.

树状图是一种直观展示多阶段试验所有结果的方法。当第一次结果之后概率发生变化时，例如从袋中取球且不放回，树状图非常有用。

For example, a bag contains 3 red balls and 2 blue balls. If one ball is drawn, the probability of red is 3/5 and the probability of blue is 2/5. If the ball is not replaced, then for the second draw the probabilities change depending on what was removed.

例如，一个袋子里有 3 个红球和 2 个蓝球。如果取出一球，红球的概率是 3/5，蓝球的概率是 2/5。如果不放回，第二次抽取时概率会根据第一次取出的球而变化。

First branch: P(R) = 3/5, P(B) = 2/5

第一层分支：P(R) = 3/5，P(B) = 2/5

If red is drawn first, the bag now has 2 red and 2 blue, so P(second red) = 2/4 and P(second blue) = 2/4. If blue is drawn first, the bag has 3 red and 1 blue, so P(second red) = 3/4 and P(second blue) = 1/4.

如果第一次取出红球，袋中剩下 2 红 2 蓝，因此第二次红球概率为 2/4，第二次蓝球概率为 2/4。如果第一次取出蓝球，袋中剩下 3 红 1 蓝，因此第二次红球概率为 3/4，第二次蓝球概率为 1/4。

6. Tree Diagrams: Multiplication Along Branches | 树状图：沿分支相乘

To find the probability of two specific outcomes happening in sequence, multiply the probabilities along the path of the tree. This works because the probability on the second branch is already conditioned on the first outcome.

要计算两个特定结果按顺序发生的概率，就沿着树状图的路径相乘。这是因为第二层分支上的概率已经以第一层结果作为条件。

P(A and B) = P(A) × P(B given A)

P(A 且 B) = P(A) × P(在 A 发生下 B 的概率)

Using the bag example with 3 red and 2 blue balls, the probability of drawing red then red without replacement is:

仍以 3 红 2 蓝的袋子为例，不放回抽取时，先红后红的概率为：

P(R and R) = 3/5 × 2/4 = 6/20 = 3/10

P(红且红) = 3/5 × 2/4 = 6/20 = 3/10

Similarly, the probability of red then blue is 3/5 × 2/4 = 6/20 = 3/10. The probability of blue then red is 2/5 × 3/4 = 6/20 = 3/10, and blue then blue is 2/5 × 1/4 = 2/20 = 1/10. The four final probabilities should add to 1.

同理，先红后蓝的概率为 3/5 × 2/4 = 6/20 = 3/10。先蓝后红的概率为 2/5 × 3/4 = 6/20 = 3/10，先蓝后蓝的概率为 2/5 × 1/4 = 2/20 = 1/10。这四个最终概率之和应为 1。

7. Addition Rule for Combined Events | 组合事件的加法法则

When a question asks for the probability of ‘event A or event B’, you add the probabilities of the relevant final paths. However, you must not double-count any overlap.

当题目询问“事件 A 或事件 B”的概率时，你需要把相关最终路径的概率相加。不过要注意不要重复计算重叠部分。

P(A or B) = P(A) + P(B) − P(A and B)

P(A 或 B) = P(A) + P(B) − P(A 且 B)

In tree diagram questions, ‘at least one red’ is often best calculated by finding all paths with at least one red and adding them, or by using the complement: 1 − P(no red). Using the complement is usually faster and reduces errors.

在树状图题目中，“至少一个红球”通常可以找出所有包含红球的路径并相加，也可以使用补集：1 − P(没有红球)。使用补集通常更快，也能减少计算错误。

From the previous no-replacement example, P(no red) = P(blue then blue) = 1/10, so P(at least one red) = 1 − 1/10 = 9/10.

根据前面不放回例子，P(没有红球) = P(先蓝后蓝) = 1/10，所以 P(至少一个红球) = 1 − 1/10 = 9/10。

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

One of the most common IGCSE mistakes is ignoring whether an item is replaced. With replacement, the probabilities on the second stage stay the same as the first stage. Without replacement, the probabilities change because the total number of items decreases.

IGCSE 中最常见的错误之一就是忽略是否放回。有放回时，第二阶段的概率与第一阶段相同。不放回时，由于总数减少，概率会发生变化。

Consider a bag containing 4 red and 2 blue balls. The table below compares the probability of drawing two red balls.

考虑一个装有 4 红 2 蓝的袋子。下表比较了取出两个红球的概率。

Situation	First draw	Second draw	P(red then red)
With replacement	4/6	4/6	4/6 × 4/6 = 16/36 = 4/9
Without replacement	4/6	3/5	4/6 × 3/5 = 12/30 = 2/5

Always check the wording: ‘replaced’, ‘returned’, ‘with replacement’ mean probabilities stay constant. ‘Not replaced’, ‘removed’, ‘kept out’, or ‘without replacement’ mean you must adjust the second branch.

一定要检查题干的措辞：“放回”“归还”“有放回”表示概率保持不变。“不放回”“取出后不再放回”“拿走”表示必须调整第二层分支的概率。

9. Conditional Probability | 条件概率

Conditional probability is the probability of an event occurring given that another event has already occurred. It is denoted by P(A | B), meaning ‘the probability of A given B’.

条件概率是指在已知另一个事件已经发生的条件下，某事件发生的概率。它记作 P(A | B)，表示“在 B 发生下 A 的概率”。

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

P(A | B) = P(A ∩ B) ÷ P(B)，其中 P(B) > 0

In tree diagrams, the second-stage probability is already a conditional probability. For example, in the bag without replacement, P(second red | first red) = 2/4. If the question asks for this type of probability, simply read it from the tree or use the formula.

在树状图中，第二阶段的概率本身就是一个条件概率。例如，在不放回袋子中，P(第二次红 | 第一次红) = 2/4。如果题目要求这类概率，可以直接从树状图读取，也可以使用公式。

10. Worked Exam-Style Example | 考试风格例题

Example: A box contains 5 pens, of which 2 are defective. Two pens are drawn at random without replacement. Find the probability that exactly one of the two pens is defective.

例题：一个盒子里有 5 支笔，其中 2 支是缺陷品。随机抽取 2 支且不放回。求恰好有 1 支是缺陷品的概率。

Step 1: Let D represent defective and G represent good. Initially, P(D) = 2/5 and P(G) = 3/5.

第一步：设 D 表示缺陷品，G 表示良品。最初 P(D) = 2/5，P(G) = 3/5。

Step 2: If the first pen is defective, then 1 defective and 3 good pens remain, so P(second D) = 1/4 and P(second G) = 3/4.

第二步：如果第一支是缺陷品，则剩下 1 支缺陷品和 3 支良品，因此 P(第二次 D) = 1/4，P(第二次 G) = 3/4。

Step 3: If the first pen is good, then 2 defective and 2 good pens remain, so P(second D) = 2/4 and P(second G) = 2/4.

第三步：如果第一支是良品，则剩下 2 支缺陷品和 2 支良品，因此 P(第二次 D) = 2/4，P(第二次 G) = 2/4。

Step 4: Exactly one defective can occur as D first then G, or G first then D.

第四步：恰好 1 支缺陷品可以表现为先 D 后 G，或先 G 后 D。

P(D and G) + P(G and D) = (2/5 × 3/4) + (3/5 × 2/4) = 6/20 + 6/20 = 12/20 = 3/5

P(先 D 后 G) + P(先 G 后 D) = (2/5 × 3/4) + (3/5 × 2/4) = 6/20 + 6/20 = 12/20 = 3/5

The probability that exactly one pen is defective is 3/5. Always add the probabilities of all possible paths that satisfy the condition.

恰好有 1 支缺陷品的概率为 3/5。一定要把所有满足条件的路径概率相加。

11. Common Misconceptions and Tips | 常见误区与备考建议

Many mistakes come from confusing ‘and’ with ‘or’, multiplying when you should add, or forgetting to update probabilities without replacement. Use these checks in every probability question.

很多错误来自混淆“且”和“或”、在该相加时相乘，或者忘记在不放回时更新概率。每条概率题都可以用以下要点检查。

‘And’ usually means multiply along the tree path; ‘or’ usually means add valid paths.
“且”通常沿树状图路径相乘；“或”通常将有效路径相加。
With replacement, probabilities stay the same; without replacement, they do not.
有放回时概率保持不变；不放回时概率会改变。
Check that all final probabilities add to 1 for a complete tree.
检查完整树状图中所有最终概率之和是否为 1。
Use 1 − P(opposite) for ‘at least one’ questions to save time.
对于“至少一个”的问题，用 1 − P(对立事件) 可以节省时间。
Always write conditional probabilities on second and later branches, not just the same probabilities as the first stage.
始终在第二层及之后的分支写条件概率，而不是照抄第一阶段的概率。

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IGCSE Mathematics: Probability Tree Diagrams and Combined Events | IGCSE 数学：概率树图与组合事件