Mastering Combined Events in Probability | 掌握概率中的组合事件

📚 Mastering Combined Events in Probability | 掌握概率中的组合事件

In Edexcel A-Level Mathematics, combined events are situations where two or more events are considered together, such as drawing cards, rolling dice, or selecting students from a group. Mastering the rules for combining probabilities is essential for success in both the AS and A2 statistics papers.

在 Edexcel A-Level 数学中，组合事件是指同时考虑两个或多个事件的情形，例如抽牌、掷骰子或从群体中选取学生。掌握组合概率的运算法则对 AS 和 A2 统计试卷的成功至关重要。

1. Sample Space and Basic Terminology | 样本空间与基本术语

A combined event is any event formed by combining two or more simple events using ‘and’ or ‘or’. The sample space is the set of all possible outcomes of an experiment.

组合事件是通过“和”或“或”将两个或多个简单事件组合而成的任何事件。样本空间是试验所有可能结果的集合。

For example, when rolling a fair six-sided die, the sample space is {1, 2, 3, 4, 5, 6}. If A is ‘rolling an even number’ and B is ‘rolling a number greater than 3’, then A and B can be combined in different ways to form new events.

例如，掷一枚公平的六面骰子时，样本空间为 {1, 2, 3, 4, 5, 6}。若 A 表示“掷出偶数”，B 表示“掷出大于 3 的数”，那么 A 与 B 可以以不同方式组合形成新事件。

A ∩ B = {4, 6} and A ∪ B = {2, 4, 5, 6}

A ∩ B 表示“偶数且大于 3”，A ∪ B 表示“偶数或大于 3”。符号 ∩ 读作交集，表示 and；符号 ∪ 读作并集，表示 or。

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

Venn diagrams help visualise combined events. Each circle represents an event, and the overlapping region represents the intersection where both events occur.

维恩图有助于直观展示组合事件。每个圆表示一个事件，重叠区域表示两个事件同时发生的交集。

The key set notation used in Edexcel exams is shown below.

Edexcel 考试中常用的集合符号如下表所示。

Notation	Meaning	中文含义
A ∩ B	Outcomes in both A and B	A 与 B 的交集
A ∪ B	Outcomes in A or B or both	A 与 B 的并集
A′	Not A, the complement of A	A 的补集
(A ∪ B)′	Neither A nor B	既非 A 也非 B

Drawing a Venn diagram from a word problem often makes the structure of a combined event much clearer before you apply probability rules.

在应用概率法则之前，根据文字题绘制维恩图通常能使组合事件的结构更加清晰。

3. Mutually Exclusive Events and the Addition Rule | 互斥事件与加法法则

Two events are mutually exclusive if they cannot happen at the same time. In this case P(A ∩ B) = 0, so the events have no overlapping outcomes.

如果两个事件不能同时发生，则它们互斥。此时 P(A ∩ B) = 0，因此这两个事件没有重叠的结果。

For mutually exclusive events, the addition rule simplifies to:

对于互斥事件，加法法则简化为：

P(A ∪ B) = P(A) + P(B)

For events that are not mutually exclusive, we must subtract the intersection to avoid double counting.

对于非互斥事件，必须减去交集以避免重复计算。

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

For example, if P(A) = 0.4, P(B) = 0.5 and P(A ∩ B) = 0.2, then P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7.

例如，若 P(A) = 0.4，P(B) = 0.5 且 P(A ∩ B) = 0.2，则 P(A ∪ B) = 0.4 + 0.5 − 0.2 = 0.7。

4. Independent Events and the Multiplication Rule | 独立事件与乘法法则

Two events A and B are independent if the occurrence of one does not affect the probability of the other. For independent events, the probability of both A and B occurring is given by the multiplication rule.

如果事件 A 和 B 的发生互不影响对方的概率，则它们相互独立。对于独立事件，A 和 B 同时发生的概率由乘法法则给出。

P(A ∩ B) = P(A) × P(B)

For example, if a coin is tossed and a die is rolled, the probability of getting a head and a 6 is (1/2) × (1/6) = 1/12. These events are independent because the coin toss does not affect the die roll.

例如，若同时抛一枚硬币并掷一个骰子，得到正面和 6 的概率为 (1/2) × (1/6) = 1/12。这些事件是独立的，因为抛硬币不会影响掷骰子。

Independence must be established from the context or by checking P(A ∩ B) = P(A) × P(B). Do not assume independence unless the question states it.

独立性必须根据情境确定，或通过检验 P(A ∩ B) = P(A) × P(B) 来判断。除非题目明确说明，否则不要假设事件相互独立。

5. Conditional Probability and Dependent Events | 条件概率与相关事件

When events are not independent, we use conditional probability. The probability of A given that B has already occurred is written P(A | B).

当事件不相互独立时，需要使用条件概率。在 B 已经发生的条件下 A 发生的概率写作 P(A | B)。

P(A | B) = P(A ∩ B) ÷ P(B)

This formula can be rearranged to give the general multiplication rule for any two events.

该公式可变形为任意两个事件的一般乘法法则。

P(A ∩ B) = P(B) × P(A | B)

For example, if a bag contains 4 red and 6 blue balls and two balls are drawn without replacement, the probability both are red is P(R₁ ∩ R₂) = (4/10) × (3/9) = 12/90 = 2/15. The second probability is conditional because the first ball is not replaced.

例如，若一个袋子中有 4 个红球和 6 个蓝球，无放回地先后取出两个球，则两个都是红球的概率为 P(R₁ ∩ R₂) = (4/10) × (3/9) = 12/90 = 2/15。第二个概率是条件概率，因为第一个球没有被放回。

6. Tree Diagrams for Multi-Stage Events | 多阶段事件的树形图

Tree diagrams are ideal for combined events that happen in stages, such as drawing two balls from a bag or taking two tests. Branch probabilities multiply along a path, and probabilities of different paths are added.

树形图非常适合分阶段发生的组合事件，例如从袋子中取两个球或参加两次测试。沿路径的分支概率相乘，不同路径的概率相加。

Each set of branches from a node must sum to 1. For example, if the probability of success on a test is 0.7 and failure is 0.3, then for two independent tests the probability of exactly one success is:

每个节点分出的一组分支概率之和必须为 1。例如，若一次测试成功的概率为 0.7，失败的概率为 0.3，则两次独立测试中恰好一次成功的概率为：

P(success then failure) + P(failure then success) = (0.7 × 0.3) + (0.3 × 0.7) = 0.42

Tree diagrams are especially useful when probabilities change after each stage, such as in sampling without replacement.

当每一阶段之后概率发生变化时，例如无放回抽样，树形图尤其有用。

7. Sampling With and Without Replacement | 有放回与无放回抽样

In sampling with replacement, the items are put back before the next draw, so the probabilities stay the same and the draws are independent.

在有放回抽样中，抽取的物品在下次抽取之前被放回，因此概率保持不变，各次抽取相互独立。

In sampling without replacement, the probabilities change after each draw because the number of items in the bag decreases. These events are dependent, so conditional probabilities must be used.

在无放回抽样中，由于袋中物品数量减少，每次抽取后概率都会发生变化。这些事件是相关的，因此必须使用条件概率。

For example, with 5 green and 3 yellow counters, the probability of drawing two green counters with replacement is (5/8) × (5/8) = 25/64. Without replacement, it is (5/8) × (4/7) = 20/56 = 5/14.

例如，有 5 个绿色计数器和 3 个黄色计数器，有放回地抽到两个绿色计数器的概率为 (5/8) × (5/8) = 25/64。无放回时则为 (5/8) × (4/7) = 20/56 = 5/14。

8. The ‘At Least One’ Shortcut | “至少一个”快捷方法

In many combined event problems, the easiest way to find the probability of at least one success is to use the complement rule. Instead of adding many mutually exclusive cases, subtract the probability of no success from 1.

在许多组合事件问题中，求至少一次成功概率的最简单方法是使用补集规则。与其添加许多互斥情形，不如用 1 减去没有成功的概率。

P(at least one) = 1 − P(none)

For example, if a biased die has a 0.2 chance of landing on a 5, the probability of getting at least one 5 in three rolls is:

例如，若一枚有偏骰子掷出 5 的概率为 0.2，则掷三次中至少出现一次 5 的概率为：

1 − (0.8)³ = 1 − 0.512 = 0.488

This shortcut saves time and reduces the chance of arithmetic errors in Edexcel exam questions.

这一快捷方法节省时间，并能减少 Edexcel 考试题中计算错误的机会。

9. Combining Rules in Exam Problems | 在考试题中综合运用法则

Edexcel exam questions often require you to combine addition, multiplication and conditional probability within one problem. It is important to identify the correct rule for each part of the question.

Edexcel 考试题经常要求在一个问题中综合运用加法、乘法和条件概率法则。必须为问题的每一部分确定正确的法则。

For example, a bag contains 4 red, 5 blue and 1 green ball. Two balls are drawn without replacement. The probability they are the same colour is found by adding three mutually exclusive cases: both red, both blue, or both green.

例如，一个袋中有 4 个红球、5 个蓝球和 1 个绿球。无放回地取出两个球。两个球颜色相同的概率通过将三个互斥情形相加得到：两个红球、两个蓝球或两个绿球。

P(same colour) = (4/10 × 3/9) + (5/10 × 4/9) + (1/10 × 0/9) = 12/90 + 20/90 + 0 = 32/90 = 16/45

Notice that the green case contributes zero because there is only one green ball. This is a common style of Edexcel problem.

注意绿色情形的贡献为零，因为只有一个绿球。这是 Edexcel 常见的一类问题。

10. Common Pitfalls and Exam Tips | 常见误区与应试技巧

Students often lose marks by confusing similar rules or misreading the events in a question. The list below highlights the most common pitfalls and how to avoid them.

学生常常因为混淆相似的法则或误读题目中的事件而失分。下表列出了最常见的误区及避免方法。

Confusing independence with mutual exclusivity. Independent events can happen together; mutually exclusive events cannot.

混淆独立性与互斥性。独立事件可以同时发生，而互斥事件不能同时发生。
Forgetting to subtract P(A ∩ B) when using the general addition rule.

在使用一般加法法则时忘记减去 P(A ∩ B)。
Using the multiplication rule P(A) × P(B) without checking independence.

未经检验独立性就使用乘法法则 P(A) × P(B)。
Misreading ‘at least one’ as ‘exactly one’. Always use the complement shortcut for at least one.

将“至少一个”误读为“恰好一个”。处理“至少一个”时务必使用补集快捷方法。
Not updating probabilities after draws when sampling without replacement.

在无放回抽样中未在每次抽取后更新概率。

Before applying any formula, write down the events and their probabilities clearly. A tree diagram or Venn diagram can prevent many careless mistakes.

在应用任何公式之前，清楚地写出事件及其概率。树形图或维恩图可以防止许多粗心错误。

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Mastering Combined Events in Probability | 掌握概率中的组合事件