Edexcel A-Level Maths: Combined Events and Conditional Probability | Edexcel A-Level 数学：组合事件与条件概率

📚 Edexcel A-Level Maths: Combined Events and Conditional Probability | Edexcel A-Level 数学：组合事件与条件概率

In Edexcel A-Level Mathematics, combined events lie at the heart of the Statistics component. They appear in AS and A2 papers whenever a question asks about two or more outcomes happening together, either with replacement, without replacement, or under a given condition. Mastering this topic is not just about memorising formulas; it is about learning to translate a real-world scenario into a probability model, then choosing the correct rule for the job.

在 Edexcel A-Level 数学中，组合事件是统计学部分的核心内容。无论是在 AS 还是 A2 试卷中，只要题目涉及两个或多个结果同时发生、有放回、无放回或在给定条件下发生，都会考查这一知识点。掌握该专题不仅仅是背公式，而是要学会将现实情境转化为概率模型，然后选择正确的计算规则。

1. Understanding Combined Events | 理解组合事件

A combined event is any event formed by joining two or more simple events using the words “and”, “or” or “given that”. For example, rolling a die and flipping a coin produces a combined event: “rolling a 6 and getting heads”. In Edexcel questions, combined events are often hidden inside everyday language, so the first skill is to identify whether the question is asking for an intersection or a union.

组合事件是指通过“和”“或”“已知”等词将两个或多个简单事件连接起来的事件。例如掷骰子和抛硬币会产生组合事件：“掷出 6 并且正面朝上”。在 Edexcel 的考题中，组合事件常常藏在日常语言里，因此第一项技能就是判断题目问的是交集还是并集。

Combined events can be represented using set notation. The intersection A ∩ B means both A and B happen, while the union A ∪ B means A happens, or B happens, or both happen. The complement A′ means A does not happen. These symbols are not just shorthand; they help you organise conditional probability and tree diagram problems correctly.

组合事件可以用集合符号表示。交集 A ∩ B 表示 A 和 B 同时发生，而并集 A ∪ B 表示 A 发生、B 发生或两者都发生。补集 A′ 表示 A 不发生。这些符号不只是简写，它们能帮助你正确地组织条件概率和树图问题。

2. Mutually Exclusive vs Independent Events | 互斥事件与独立事件

Two events are mutually exclusive if they cannot happen at the same time. For example, when rolling a fair six-sided die, the events “getting an odd number” and “getting a 6” are mutually exclusive because 6 is not odd. Mathematically, mutually exclusive events satisfy P(A ∩ B) = 0, so the addition rule simplifies to P(A ∪ B) = P(A) + P(B).

如果两个事件不能同时发生，则它们互斥。例如掷一个均匀的六面骰子时，“得到奇数”和“得到 6”是互斥事件，因为 6 不是奇数。数学上，互斥事件满足 P(A ∩ B) = 0，因此加法法则简化为 P(A ∪ B) = P(A) + P(B)。

Independent events are different. Independence means the outcome of one event does not affect the probability of the other. For example, rolling a die and flipping a coin are independent because the die result cannot change the chance of heads. For independent events, P(A ∩ B) = P(A) × P(B). A common exam mistake is to treat mutually exclusive and independent as the same idea; they are not.

独立事件则不同。独立意味着一个事件的结果不影响另一个事件的概率。例如掷骰子和抛硬币是独立的，因为骰子的结果不会改变正面朝上的概率。对于独立事件，P(A ∩ B) = P(A) × P(B)。考试中常见的错误是把互斥和独立混为一谈；它们并不相同。

Feature	Mutually Exclusive	Independent
Meaning	Cannot occur together	Do not affect each other
Intersection	P(A ∩ B) = 0	P(A ∩ B) = P(A) × P(B)
Key test	Can both happen at once?	Does one probability change given the other?

Use this table to decide which formula to apply. In Edexcel past papers, many candidates lose marks by using the multiplication rule for non-independent events or by adding probabilities that overlap.

使用该表可判断该用哪个公式。在 Edexcel 历年真题中，许多考生因对非独立事件误用乘法法则，或因重复叠加概率而失分。

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

The general addition rule for any two events A and B is written as

任意两个事件 A 和 B 的一般加法法则写作

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

This formula subtracts the intersection once so that the overlap is not double-counted. If A and B are mutually exclusive, the intersection is zero, and the formula reduces to the simple addition rule.

该公式减去一次交集，避免重叠部分被重复计算。如果 A 和 B 互斥，交集为零，公式就简化为简单加法法则。

For example, a bag contains 5 red, 3 blue and 2 green balls. If one ball is drawn at random, the probability that it is red or blue is P(red) + P(blue) = 5/10 + 3/10 = 8/10 = 4/5. Since a ball cannot be both red and blue, these events are mutually exclusive, so no subtraction is needed.

例如一个袋子中有 5 个红球、3 个蓝球和 2 个绿球。随机抽出一个球，它是红色或蓝色的概率为 P(红) + P(蓝) = 5/10 + 3/10 = 8/10 = 4/5。由于一个球不可能既是红色又是蓝色，这些事件互斥，因此无需减去交集。

When events overlap, you must subtract the intersection. For example, if P(A) = 0.6, P(B) = 0.5 and P(A ∩ B) = 0.3, then P(A ∪ B) = 0.6 + 0.5 − 0.3 = 0.8. Forgetting this subtraction is one of the most common errors in probability questions.

当事件有重叠时，必须减去交集。例如若 P(A) = 0.6，P(B) = 0.5 且 P(A ∩ B) = 0.3，则 P(A ∪ B) = 0.6 + 0.5 − 0.3 = 0.8。忘记这次减法运算是概率题中最常见的错误之一。

4. Multiplication Rule and Conditional Probability | 乘法法则与条件概率

Conditional probability is the probability of one event given that another event has already occurred. It is written as P(A|B), read as “the probability of A given B”. The core formula is

条件概率是指在已知另一个事件已经发生的情况下，某一事件发生的概率。它写作 P(A|B)，读作“在 B 发生的条件下 A 的概率”。核心公式为

P(A|B) = P(A ∩ B) / P(B)

provided that P(B) > 0. Rearranging this gives the general multiplication rule:

其中要求 P(B) > 0。将公式变形即可得到一般乘法法则：

P(A ∩ B) = P(A|B) × P(B)

If A and B are independent, then P(A|B) = P(A), so the multiplication rule becomes P(A ∩ B) = P(A) × P(B). In many Edexcel questions, you are asked to test independence by checking whether P(A|B) equals P(A), or whether P(A ∩ B) equals P(A) × P(B).

如果 A 和 B 独立，则 P(A|B) = P(A)，因此乘法法则变为 P(A ∩ B) = P(A) × P(B)。在许多 Edexcel 考题中，你需要通过检验 P(A|B) 是否等于 P(A)，或 P(A ∩ B) 是否等于 P(A) × P(B) 来判断独立。

Conditional probability questions often involve “given that” phrasing. For example, a class has 12 boys and 18 girls. Of the boys, 4 wear glasses; of the girls, 6 wear glasses. The probability of wearing glasses given that the student is a boy is P(glasses|boy) = 4/12 = 1/3. Here the sample space has been reduced to boys only, which is the key idea of conditional probability.

条件概率题目常常使用“已知……”的表述。例如一个班级有 12 名男生和 18 名女生。男生中有 4 人戴眼镜，女生中有 6 人戴眼镜。已知学生是男生时戴眼镜的概率为 P(戴眼镜|男生) = 4/12 = 1/3。此时样本空间已缩小到仅限男生，这正是条件概率的核心思想。

5. Probability Tree Diagrams | 概率树图

Tree diagrams are an essential tool for combined events that happen in sequence. They are especially useful for conditional probability and for sampling with or without replacement. Each branch represents a possible outcome, and the probabilities on the branches must sum to 1 at every node.

树图是处理按顺序发生的组合事件的重要工具。它们特别适用于条件概率以及有放回或无放回抽样。每条分支代表一种可能结果，且每个节点上的分支概率之和必须为 1。

For two events A and B, the probability of following one path is found by multiplying the probabilities along that path. For example, if the probability of rain tomorrow is 0.3 and the probability of rain the day after, given rain tomorrow, is 0.6, then the probability that it rains on both days is 0.3 × 0.6 = 0.18. This is a direct application of the general multiplication rule.

对于两个事件 A 和 B，沿某一路径发生的概率等于该路径上各分支概率相乘。例如明天下雨的概率为 0.3，在已知明天下雨的条件下后天下雨的概率为 0.6，则两天都下雨的概率为 0.3 × 0.6 = 0.18。这是对一般乘法法则的直接应用。

When a question involves “without replacement”, the probabilities on the second set of branches change. Suppose a bag contains 4 red and 6 blue counters. If one counter is removed and not replaced, the probability of drawing red then red is (4/10) × (3/9) = 12/90 = 2/15. The denominator changes from 10 to 9 because one counter has been removed.

当题目涉及“无放回”时，第二组分支上的概率会发生变化。假设一个袋子中有 4 个红色和 6 个蓝色计数片。若取出一个后不放回，则先红后红的概率为 (4/10) × (3/9) = 12/90 = 2/15。分母从 10 变为 9，因为已经取走了一个计数片。

Always label the end of each path with the combined event, such as RR, RB, BR or BB. Then add the probabilities of all paths that satisfy the question. If asked for at least one red, you would add P(RR) + P(RB) + P(BR), or use the complement 1 − P(BB), which is often faster.

始终在每条路径的末端标注组合事件，如 RR、RB、BR 或 BB。然后加总所有满足题目条件的路径概率。如果求至少一个红球，可加总 P(RR) + P(RB) + P(BR)，或使用补集 1 − P(BB)，后者通常更快。

6. Venn Diagrams and Set Notation | 维恩图与集合符号

Venn diagrams are used to show overlaps between events. A rectangle represents the sample space S, while circles represent events A and B. The overlapping region is A ∩ B, the region in A but not B is A ∩ B′, and the region outside both circles is A′ ∩ B′.

维恩图用于展示事件之间的重叠关系。矩形代表样本空间 S，圆圈代表事件 A 和 B。重叠区域为 A ∩ B，A 中不在 B 中的区域为 A ∩ B′，两个圆圈之外的区域为 A′ ∩ B′。

From a Venn diagram you can read off probabilities directly and answer questions such as “Find P(A ∪ B)” or “Find P(A′ ∩ B)”. The addition rule also follows visually: adding P(A) and P(B) counts the overlap twice, so you subtract P(A ∩ B) once.

从维恩图中你可以直接读出概率，并回答诸如“求 P(A ∪ B)”或“求 P(A′ ∩ B)”的问题。加法法则也可通过图形直观理解：将 P(A) 和 P(B) 相加会重复计算重叠部分，因此需要减去一次 P(A ∩ B)。

Venn diagram questions often require algebraic reasoning. For example, if P(A) = 0.5, P(B) = 0.4 and P(A ∪ B) = 0.7, then P(A ∩ B) = 0.5 + 0.4 − 0.7 = 0.2. Fill this into the overlap first, then subtract to find the remaining regions. This systematic approach prevents errors in algebra-based Edexcel problems.

维恩图题目常常需要代数推理。例如若 P(A) = 0.5，P(B) = 0.4 且 P(A ∪ B) = 0.7，则 P(A ∩ B) = 0.5 + 0.4 − 0.7 = 0.2。先将此值填入重叠区域，再逐一相减求出其余区域。这种系统方法可避免基于代数的 Edexcel 问题中出现错误。

7. Two-Way Tables and Combined Outcomes | 双向表与组合结果

Two-way tables organise data for two categorical variables, such as gender and subject choice, or colour and size. They are particularly useful for conditional probability because each row or column total can act as a reduced sample space.

双向表用于整理两个分类变量的数据，例如性别与选课、颜色与尺寸。它们对条件概率特别有用，因为每一行或每一列的总计都可以作为缩小后的样本空间。

For example, a survey of 100 people records eye colour and handedness. The table shows 60 are right-handed, 40 are left-handed, 70 have brown eyes, 30 have blue eyes, and the cells give the joint frequencies. To find P(brown eyes | right-handed), divide the number who are both brown-eyed and right-handed by the total number right-handed, not by 100.

例如一项 100 人的调查记录眼睛颜色和惯用手。表格显示 60 人惯用右手，40 人惯用左手，70 人是棕色眼睛，30 人是蓝色眼睛，各单元格给出联合频数。要求 P(棕色眼睛 | 惯用右手)，应用同时具有棕色眼睛且惯用右手的人数除以惯用右手总人数，而不是除以 100。

Two-way tables are also useful for listing combined outcomes in probability experiments, such as rolling two dice. The sample space contains 36 equally likely ordered pairs. To find the probability of a sum of 7, count the pairs (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), giving 6/36 = 1/6.

双向表也可用于列出概率实验中的组合结果，例如掷两个骰子。样本空间包含 36 个等可能的有序数对。要求点数和为 7 的概率，数出数对 (1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1)，得到 6/36 = 1/6。

8. Worked Exam-Style Example | 考试风格例题解析

A bag contains 3 red counters, 4 blue counters and 5 green counters. Two counters are drawn at random without replacement. Find the probability that:

一个袋子中有 3 个红色计数片、4 个蓝色计数片和 5 个绿色计数片。随机无放回地抽取两个计数片。求以下概率：

both counters are red
at least one counter is green
the second counter is red given that the first was blue

两个都是红色
至少一个是绿色
已知第一个是蓝色时第二个是红色

First, note the total number of counters is 3 + 4 + 5 = 12. For both red, the probability is (3/12) × (2/11) = 6/132 = 1/22. The first fraction is 3 red out of 12, and after removing one red there are 2 red out of 11.

首先，注意计数片总数为 3 + 4 + 5 = 12。对于两个都是红色，概率为 (3/12) × (2/11) = 6/132 = 1/22。第一个分数是 12 个中有 3 红，取走一个红后剩下 11 个中有 2 红。

For at least one green, it is easier to use the complement: P(at least one green) = 1 − P(no green). No green means both counters come from the 7 non-green counters. So P(no green) = (7/12) × (6/11) = 42/132 = 7/22. Therefore P(at least one green) = 1 − 7/22 = 15/22.

对于至少一个绿色，更简便的方法是使用补集：P(至少一个绿) = 1 − P(没有绿)。没有绿意味着两个计数片都来自 7 个非绿色计数片。所以 P(没有绿) = (7/12) × (6/11) = 42/132 = 7/22。因此 P(至少一个绿) = 1 − 7/22 = 15/22。

For the conditional probability, we want P(second red | first blue). If the first counter is blue, there are 11 counters left, of which 3 are red. Therefore P(second red | first blue) = 3/11. The key is to update the composition of the bag after the first draw.

对于条件概率，我们要求 P(第二个红 | 第一个蓝)。如果第一个计数片是蓝色，剩下 11 个计数片，其中 3 个是红色。因此 P(第二个红 | 第一个蓝) = 3/11。关键是在第一次抽取后更新袋子中的计数片组成。

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

One common mistake is multiplying probabilities for events that are not independent. For example, if two cards are drawn without replacement from a standard deck, the probability of drawing two aces is (4/52) × (3/51), not (4/52) × (4/52).

一个常见错误是对非独立事件使用概率相乘。例如从一副标准扑克牌中无放回地抽两张牌，抽到两张 A 的概率是 (4/52) × (3/51)，而不是 (4/52) × (4/52)。

Another mistake is adding probabilities without checking for overlap. When events are not mutually exclusive, such as “drawing a club” and “drawing a queen”, the simple addition P(club) + P(queen) would count the queen of clubs twice. Always subtract P(club ∩ queen) = 1/52.

另一个错误是不检查重叠就相加概率。当事件不互斥时，例如“抽到梅花”和“抽到 Q”，简单相加 P(梅花) + P(Q) 会把梅花 Q 计算两次。必须减去 P(梅花 ∩ Q) = 1/52。

Students also misuse conditional probability notation. Remember that P(A|B) is not the same as P(B|A). For example, P(defective | from machine A) is different from P(from machine A | defective). Use the formula rather than guessing, and keep the condition in the denominator.

学生还容易误用条件概率符号。记住 P(A|B) 与 P(B|A) 不相同。例如 P(次品 | 来自机器 A) 与 P(来自机器 A | 次品) 不同。应使用公式而不是猜测，并将条件放在分母中。

10. Exam Tips for Edexcel Statistics | Edexcel 统计学备考技巧

In the Edexcel A-Level Statistics exam, you should show your working clearly, especially when using tree diagrams or two-way tables. Even if you make an arithmetic slip, the method marks can be awarded if your diagram or table is correctly labelled.

在 Edexcel A-Level 统计学考试中，应清晰地展示解题过程，尤其是在使用树图或双向表时。即使算术出现小错，只要图表标注正确，方法分仍然可以获得。

When answering a combined events question, first write down the given probabilities using proper notation. For conditional probability, identify the reduced sample space or use the formula P(A|B) = P(A ∩ B) / P(B). If the question asks for independence, show a clear test such as P(A|B) = P(A) or P(A ∩ B) = P(A) × P(B).

回答组合事件问题时，应先用正确的符号写出已知概率。对于条件概率，找出缩小的样本空间或使用公式 P(A|B) = P(A ∩ B) / P(B)。如果题目要求判断独立，应展示明确的检验过程，例如 P(A|B) = P(A) 或 P(A ∩ B) = P(A) × P(B)。

Finally, check that your final probability lies between 0 and 1, and that the probabilities of all mutually exclusive outcomes in a tree diagram or table sum to 1. These quick checks can catch errors before you move on.

最后，检查最终概率是否在 0 到 1 之间，以及树图或表格中所有互斥结果的概率之和是否为 1。这些快速检查可以在继续作答前发现错误。

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Edexcel A-Level Maths: Combined Events and Conditional Probability | Edexcel A-Level 数学：组合事件与条件概率