G-4 Topic 10: Combined Events | G-4 主题10：合并事件

📚 G-4 Topic 10: Combined Events | G-4 主题10：合并事件

In IGCSE Mathematics, the topic of combined events sits at the heart of probability. When you toss two coins, roll a die and spin a spinner, or choose two cards from a deck, you are dealing with combined events. This article will guide you through the essential rules — the addition rule, the multiplication rule, tree diagrams, Venn diagrams, and conditional probability — with clear examples and exam-style advice.

在 IGCSE 数学中，合并事件是概率板块的核心内容。当你抛两枚硬币、掷一颗骰子并旋转转盘、或从一副牌中抽取两张牌时，你就在处理合并事件。本文将带你掌握关键规则——加法规则、乘法规则、树状图、维恩图以及条件概率——并配以清晰的例题和考试技巧。

1. What Are Combined Events? | 什么是合并事件

A combined event involves the outcome of two or more separate events happening together. For example, rolling a die and flipping a coin simultaneously produces combined outcomes such as (Head, 3) or (Tail, 5). Each pair is a single outcome in the combined sample space.

合并事件是指两个或多个独立事件同时发生所产生的结果。例如，同时掷一颗骰子和抛一枚硬币，会产生诸如（正面，3）或（反面，5）这样的合并结果。每一对结果都构成合并样本空间中的一个单一结果。

To list the sample space systematically, we often use:

为了系统地列出样本空间，我们常用以下方法：

Two-way tables (tables with rows and columns for each event);
Tree diagrams (each branch represents a possible outcome);
Venn diagrams (showing relationships between sets).

二维表格（行与列分别代表每个事件的可能结果）；
树状图（每条分支代表一种可能结果）；
维恩图（展示集合之间的关系）。

Total number of outcomes = n(A) × n(B)

where n(A) is the number of outcomes for event A and n(B) for event B. This multiplication principle is fundamental to counting combined outcomes.

其中 n(A) 表示事件 A 的可能结果数量，n(B) 表示事件 B 的可能结果数量。这个乘法原理是计算合并结果数量的基础。

2. The Addition Rule | 加法规则

The addition rule calculates the probability that at least one of two events occurs. For any two events A and B:

加法规则用于计算两个事件中至少有一个发生的概率。对于任意两个事件 A 和 B：

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

The term P(A ∩ B) counts the outcomes that belong to both events; subtracting it avoids double-counting.

其中 P(A ∩ B) 表示同时属于两个事件的结果数量，减去它是为了避免重复计算。

Example: A card is drawn from a standard 52-card deck. What is the probability of drawing a heart or a queen?

示例：从一副 52 张的标准扑克牌中抽取一张，求抽到红心或 Q 的概率。

Let A = “draw a heart”, P(A) = 13/52. Let B = “draw a queen”, P(B) = 4/52. The card that is both a heart and a queen is the Queen of Hearts, so P(A ∩ B) = 1/52.

设 A = “抽到红心”，P(A) = 13/52。设 B = “抽到 Q”，P(B) = 4/52。既是红心又是 Q 的牌是红心 Q，因此 P(A ∩ B) = 1/52。

P(A ∪ B) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13

3. Mutually Exclusive Events | 互斥事件

When two events cannot happen at the same time, they are mutually exclusive. For example, “rolling a 3” and “rolling an even number” on a single die cannot both occur. In this case, P(A ∩ B) = 0, so the addition rule simplifies to:

当两个事件不可能同时发生时，它们就是互斥事件。例如，掷一颗骰子时，”掷出 3″ 与 “掷出偶数” 不可能同时发生。在这种情况下，P(A ∩ B) = 0，因此加法规则简化为：

P(A ∪ B) = P(A) + P(B)

This simplified form is often called the “OR rule” for mutually exclusive events. In IGCSE questions, you will often be asked to identify whether events are mutually exclusive before applying the formula.

这个简化形式常被称为互斥事件的”或规则”。在 IGCSE 题目中，你通常需要先判断事件是否互斥，再选择恰当的公式。

Example: In a bag there are 6 red, 4 blue, and 2 green marbles. One marble is picked at random. Find P(red or blue).

示例：袋中有 6 颗红珠、4 颗蓝珠和 2 颗绿珠。随机取出一颗，求取到红色或蓝色的概率。

Since a marble cannot be both red and blue, the events are mutually exclusive:

由于一颗珠子不可能同时是红色和蓝色，这两个事件是互斥的：

P(red or blue) = 6/12 + 4/12 = 10/12 = 5/6

4. The Multiplication Rule | 乘法规则

The multiplication rule calculates the probability that two events both occur. For any two events A and B:

乘法规则用于计算两个事件同时发生的概率。对于任意两个事件 A 和 B：

P(A ∩ B) = P(A) × P(B | A)

where P(B | A) is the probability of B given that A has already occurred. If A and B are independent, then P(B | A) = P(B), and the rule simplifies to:

其中 P(B | A) 表示在 A 已发生的条件下 B 发生的概率。如果 A 与 B 相互独立，则 P(B | A) = P(B)，规则简化为：

P(A ∩ B) = P(A) × P(B)

This is known as the “AND rule”.

这就是著名的”与规则”。

5. Independent Events | 独立事件

Two events are independent if the occurrence of one does not affect the probability of the other. Common examples include:

如果两个事件中一个的发生不影响另一个发生的概率，则称这两个事件相互独立。常见例子包括：

Tossing a coin twice — the second toss is independent of the first;
Rolling a die and spinning a spinner — results do not affect each other;
Choosing a marble from a bag with replacement.

连续抛两次硬币——第二次抛掷与第一次无关；
掷骰子并旋转转盘——两者互不影响；
从袋中取球并放回。

Example: A coin is tossed and a fair die is rolled. Find the probability of getting a head and an odd number.

示例：抛一枚硬币并掷一颗公平的骰子，求出现正面且掷出奇数的概率。

The events are independent:

这两个事件相互独立：

P(head and odd) = 1/2 × 3/6 = 1/2 × 1/2 = 1/4

Note that ‘without replacement’ scenarios are NOT independent, because the probability changes after each draw.

注意：”不放回”场景下事件不是独立的，因为每次抽取后概率会发生变化。

6. Tree Diagrams | 树状图

Tree diagrams are the most powerful visual tool for combined events, especially when events occur in sequence. Each branch is labelled with its probability; multiply along branches for AND, and add between branches for OR.

树状图是处理合并事件最有力的可视化工具，尤其在事件按顺序发生时。每条分支都标注其概率；沿分支相乘用于”且”，分支之间相加用于”或”。

Example: A bag contains 3 red and 5 blue balls. Two balls are drawn without replacement. Draw a tree diagram and find P(both red).

示例：袋中有 3 个红球和 5 个蓝球。不放回地连取两个球。画出树状图并求两球都是红色的概率。

First draw: P(R) = 3/8, P(B) = 5/8. Second draw (given first outcome): if first was red, P(R) = 2/7, P(B) = 5/7; if first was blue, P(R) = 3/7, P(B) = 4/7.

第一次抽取：P(红) = 3/8，P(蓝) = 5/8。第二次抽取（在第一次结果已知的条件下）：若第一次为红，则 P(红) = 2/7，P(蓝) = 5/7；若第一次为蓝，则 P(红) = 3/7，P(蓝) = 4/7。

P(both red) = 3/8 × 2/7 = 6/56 = 3/28

To find P(one red and one blue), add the probabilities of the two relevant paths: R then B, or B then R.

要求 P(一红一蓝)，则需将两条相关路径的概率相加：先红后蓝，或先蓝后红。

P(RB) = 3/8 × 5/7 = 15/56, P(BR) = 5/8 × 3/7 = 15/56

P(one red, one blue) = 15/56 + 15/56 = 30/56 = 15/28

7. Venn Diagrams | 维恩图

Venn diagrams help visualise the relationships between combined events. The rectangle represents the sample space, circles represent events, and overlapping regions represent intersections.

维恩图能直观展示合并事件之间的关系。矩形代表样本空间，圆代表事件，重叠区域代表交集。

In IGCSE problems, you may be given a diagram with numbers and asked to calculate:

在 IGCSE 题目中，给定带数字的维恩图后，常要求计算：

P(A) only — numbers inside circle A only;
P(A ∩ B) — numbers in the overlapping region;
P(A ∪ B) — all numbers inside either circle;
P(A’ ∩ B) — numbers inside B but outside A;
P(A’ ∩ B’) — numbers outside both circles.

P(A) —— 仅圆 A 内的数字；
P(A ∩ B) —— 重叠区域的数字；
P(A ∪ B) —— 任一圆内的所有数字；
P(A’ ∩ B) —— 在圆 B 内但在圆 A 外的数字；
P(A’ ∩ B’) —— 两个圆外部的数字。

Example: In a class of 30 students, 18 play football (F), 14 play tennis (T), and 8 play both. How many play neither?

示例：某班 30 名学生中，18 人踢足球（F），14 人打网球（T），8 人两项都参加。求两项都不参加的人数。

Only football: 18 − 8 = 10. Only tennis: 14 − 8 = 6. Neither:

只踢足球：18 − 8 = 10。只打网球：14 − 8 = 6。两项都不参与的人数：

30 − (10 + 8 + 6) = 30 − 24 = 6

8. Conditional Probability | 条件概率

Conditional probability is the probability of an event B occurring given that event A has already occurred. It is written P(B | A) and calculated using the formula:

条件概率是指在事件 A 已发生的前提下，事件 B 发生的概率。它记作 P(B | A)，计算公式为：

P(B | A) = P(A ∩ B) / P(A)

This formula is essential when dealing with sequential events without replacement, or when information about one event gives us new information about another.

这个公式在处理不放回的序列事件、或当一个事件的信息能提供关于另一个事件的新信息时至关重要。

Example: A bag contains 4 red and 6 blue balls. A ball is drawn at random and not replaced. Then a second ball is drawn. Given that the first ball was red, find the probability that the second ball is blue.

示例：袋中有 4 个红球和 6 个蓝球。随机取出一球不放回，然后再取一球。已知第一次取出的是红球，求第二次取到蓝球的概率。

After removing one red ball, there are 3 red and 6 blue balls left, 9 in total:

除去一个红球后，剩下 3 个红球和 6 个蓝球，共 9 个：

P(blue | red) = 6/9 = 2/3

Notice how conditional probability is simply read directly from the tree diagram branch in many cases — this saves time in the exam.

请注意，在许多情况下，条件概率可直接从树状图的分支上读取——这在考试中非常省时。

9. Common Exam Mistakes | 常见考试错误

Below are the most frequent errors students make when tackling combined events questions in IGCSE:

以下是学生在完成 IGCSE 合并事件题目时最常犯的错误：

Mistake \| 错误	Correction \| 改正
Using P(A)×P(B) when draws are without replacement.	Adjust the second probability based on the first outcome.
Forgetting to subtract P(A∩B) in the addition rule.	Always check whether the events overlap.
Adding branch probabilities instead of multiplying along a path.	Multiply along branches (AND); add between branches (OR).
Confusing P(A∪B) with P(A∩B).	∪ means “at least one”; ∩ means “both”.
Giving final answers as unsimplified fractions.	Always simplify fractions to their lowest terms.

Mistake | 错误: using P(A)×P(B) when draws are without replacement. Correction | 改正: adjust the second probability based on the first outcome.

10. Worked Exam Question | 真题演练

Let us now solve a typical IGCSE-style question step by step.

下面我们逐步解一道典型的 IGCSE 风格题目。

Question: A bag contains 5 red and 3 green counters. Two counters are drawn at random without replacement. Find the probability that the two counters are the same colour.

题目：袋中有 5 个红色和 3 个绿色计数器。不放回地随机取出两个，求两次取出的计数器颜色相同的概率。

Solution:

解答：

Same colour means either both red OR both green. These are mutually exclusive final outcomes, so we add their probabilities.

颜色相同意味着两个都是红色或两个都是绿色。这两种最终结果互斥，因此我们将它们的概率相加。

P(RR) = 5/8 × 4/7 = 20/56

P(GG) = 3/8 × 2/7 = 6/56

Adding these:

将二者相加：

P(same colour) = 20/56 + 6/56 = 26/56 = 13/28

Always check: the probabilities of all possible final outcomes must sum to 1. Here, P(RG) + P(GR) = 15/56 + 15/56 = 30/56, and 26/56 + 30/56 = 56/56 = 1. ✓

务必检查：所有可能最终结果的概率之和必须等于 1。此处 P(红绿) + P(绿红) = 15/56 + 15/56 = 30/56，且 26/56 + 30/56 = 56/56 = 1。✓

11. Practice Questions | 练习题

Test your understanding with these quick questions. Try to solve them before checking the answers below.

用以下快速练习题检验你的理解。请先尝试解答，再对照下方的答案。

A fair die is rolled twice. Find P(both results are greater than 4).
Two cards are drawn from a deck with replacement. Find P(two aces).
In a group of 20 students, 12 study Biology, 9 study Chemistry, and 4 study both. Find P(student studies neither subject).
Two coins are tossed. Find P(at least one head).

掷一颗公平骰子两次，求两次结果都大于 4 的概率。
从一副牌中有放回地抽取两张，求两张都是 A 的概率。
一组 20 名学生中，12 人学生物，9 人学化学，4 人两科都学。求随机选一名学生两科都不学的概率。
抛两枚硬币，求至少出现一个正面的概率。

Answers:

参考答案：

1. P(>4) = 2/6 = 1/3 on each roll. Both → (1/3)² = 1/9.

1. 每次掷出大于 4 的概率为 2/6 = 1/3。两次都满足 → (1/3)² = 1/9。

2. P(Ace) = 4/52 = 1/13. With replacement → (1/13)² = 1/169.

2. 抽到 A 的概率为 4/52 = 1/13。有放回 → (1/13)² = 1/169。

3. Biology only: 12 − 4 = 8. Chemistry only: 9 − 4 = 5. Both: 4. Neither: 20 − (8+5+4) = 3. P(neither) = 3/20.

3. 只选生物：12 − 4 = 8。只选化学：9 − 4 = 5。两科都选：4。都不选：20 − (8+5+4) = 3。概率为 3/20。

4. P(at least one head) = 1 − P(no heads) = 1 − (1/2 × 1/2) = 1 − 1/4 = 3/4.

4. P(至少一个正面) = 1 − P(没有正面) = 1 − (1/2 × 1/2) = 1 − 1/4 = 3/4。

The last question nicely introduces a KEY technique: the complement rule. Sometimes it is easier to find the probability of the complementary event and subtract from 1. This is especially useful for “at least one” problems.

最后一道题很好地引出了一个关键技巧：补事件法。有时求补事件的概率再用 1 减会更简单。这在处理”至少一个”类问题时尤其有效。

12. Summary and Final Tips | 总结与最终建议

The key takeaway from this topic is knowing which rule to apply. Ask yourself two questions: “Are the events sequential or simultaneous?” and “Does the probability of one event depend on the other?”

本主题的核心在于学会判断该用哪条规则。时刻问自己两个问题：”事件是按顺序发生还是同时发生？”以及”一个事件发生的概率是否依赖于另一个事件？”

Use the addition rule for “at least one” (OR) situations.
Use the multiplication rule for “both” (AND) situations.
Adjust probabilities for “without replacement” scenarios.
Draw a tree diagram for any sequential event question.
Draw a Venn diagram for overlapping set questions.
Use the complement rule for “at least one” problems.

对”至少一个”（或）的情况使用加法规则。
对”两者都”（且）的情况使用乘法规则。
对”不放回”的情况调整相应概率。
遇到按顺序发生的事件，画树状图。
遇到集合重叠的问题，画维恩图。
遇到”至少一个”问题，使用补事件法。

Practising past paper questions is the most effective way to master combined events. Recognise the patterns, and the exam becomes far less daunting.

练习历届真题是掌握合并事件的最有效方法。识别题目中的模式，考试就会变得轻松许多。

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G-4 Topic 10: Combined Events | G-4 主题10：合并事件