More Compound Probability | 复合概率进阶

📚 More Compound Probability | 复合概率进阶

In IGCSE Edexcel Mathematics, compound probability explores the likelihood of two or more events occurring together. This topic builds on basic probability and introduces essential rules that allow you to calculate combined probabilities systematically.

在 IGCSE Edexcel 数学课程中，复合概率研究两个或多个事件同时发生的可能性。该主题建立在基础概率之上，引入了关键法则，使你能够系统地计算组合概率。

1. The Addition Rule (OR Rule) | 加法法则（OR 法则）

The addition rule is used when you want to find the probability that at least one of two events occurs. For two events A and B, the probability of A or B occurring is given by: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). This formula subtracts the overlap to avoid double-counting outcomes that belong to both events.

加法法则用于计算两个事件中至少一个发生的概率。对于事件 A 和 B，A 或 B 发生的概率为：P(A ∪ B) = P(A) + P(B) − P(A ∩ B)。该公式减去重叠部分，避免重复计算同时属于两个事件的结果。

Example: A card is drawn from a standard deck. What is the probability of drawing a heart or a king? There are 13 hearts and 4 kings, but the king of hearts is counted twice, so P = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.

示例：从一副标准扑克牌中抽一张牌，抽到红心或老 K 的概率是多少？共有 13 张红心和 4 张老 K，但红心老 K 被重复计算了一次，因此 P = 13/52 + 4/52 − 1/52 = 16/52 = 4/13。

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

2. The Multiplication Rule for Independent Events | 独立事件的乘法法则

The multiplication rule applies when events are independent — the occurrence of one event does not affect the probability of the other. For independent events A and B, the probability of both occurring is: P(A ∩ B) = P(A) × P(B).

乘法法则适用于独立事件——即一个事件的发生不影响另一个事件的概率。对于独立事件 A 和 B，两者同时发生的概率为：P(A ∩ B) = P(A) × P(B)。

Example: Rolling a fair die and flipping a fair coin. The probability of rolling a 6 and getting heads is 1/6 × 1/2 = 1/12.

示例：掷一枚均匀骰子并抛一枚均匀硬币，掷出 6 且正面朝上的概率为 1/6 × 1/2 = 1/12。

Two events are independent only if this equation holds. In exam questions, look for phrases like “with replacement” or “the dice is rolled again” — these signal independence.

只有当该等式成立时，两个事件才是独立的。在考试题目中，注意”放回”或”再次掷骰子”等表述——这些标志着独立性。

P(A ∩ B) = P(A) × P(B) (for independent events)

3. Tree Diagrams | 树形图

A tree diagram is a visual tool for tracking all possible outcomes of a compound experiment. Each branch represents a possible outcome, and probabilities are written along the branches. To find the probability of a particular sequence of outcomes, multiply along the branches; to find the probability of multiple sequences, add the resulting products.

树形图是追踪复合试验所有可能结果的直观工具。每条分支代表一个可能结果，概率标注在分支上。求某一特定结果序列的概率时，沿分支相乘；求多个序列的概率时，将所得乘积相加。

For example, tossing a coin twice creates a tree with four paths: HH, HT, TH, TT. Each path has probability 1/2 × 1/2 = 1/4.

例如，抛两次硬币可构建一棵树，共四条路径：HH、HT、TH、TT。每条路径的概率为 1/2 × 1/2 = 1/4。

When drawing tree diagrams in exams, always label each branch with its probability and each endpoint with the full outcome description. This prevents calculation errors and earns method marks.

在考试中绘制树形图时，务必在每条分支上标注概率，并在每个终点标注完整的结果描述。这样可以避免计算错误并获得方法分。

4. Dependent Events (Without Replacement) | 相依事件（不放回）

When sampling without replacement, the probability of the second event changes depending on the outcome of the first. For example, drawing two marbles from a bag without returning the first marble means the total number of marbles decreases.

不放回抽样时，第二个事件的概率取决于第一个事件的结果。例如，从袋子中连续取出两个弹珠且不将第一个放回，弹珠总数会减少。

The probability of both events A and B occurring is: P(A ∩ B) = P(A) × P(B given A). Here, P(B given A) is the probability of B after A has occurred.

事件 A 和 B 同时发生的概率为：P(A ∩ B) = P(A) × P(B | A)。其中 P(B | A) 表示在 A 发生后 B 发生的概率。

Example: A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. The probability of drawing two red marbles is 5/8 × 4/7 = 20/56 = 5/14. Notice the denominator changes from 8 to 7 after the first draw.

示例：袋子中有 5 个红弹珠和 3 个蓝弹珠，不放回地连续抽取两个。抽到两个红弹珠的概率为 5/8 × 4/7 = 20/56 = 5/14。注意第一次抽取后分母从 8 变为 7。

5. Conditional Probability | 条件概率

Conditional probability P(B | A) is read as “the probability of B given A”. It asks: knowing that event A has already occurred, what is the probability of B? The formula is: P(B | A) = P(A ∩ B) / P(A), provided P(A) ≠ 0.

条件概率 P(B | A) 读作”在 A 已发生的条件下 B 发生的概率”。公式为：P(B | A) = P(A ∩ B) / P(A)，前提是 P(A) ≠ 0。

In Edexcel IGCSE problems, conditional probability often appears in the context of two-way tables or without-replacement scenarios. For example, in a class of 30 students where 12 wear glasses and 18 do not, if 5 of the glasses-wearers are girls, the probability of selecting a girl given that the student wears glasses is 5/12.

在 Edexcel IGCSE 题目中，条件概率常出现在双向表格或不放回情境中。例如，一个班有 30 名学生，其中 12 人戴眼镜，18 人不戴；若戴眼镜者中有 5 名女生，则在已知该生戴眼镜的条件下选到女生的概率为 5/12。

A common misconception is to confuse P(B | A) with P(A | B). These are generally not equal. Always read the question carefully to identify which event is the condition.

一个常见的误解是将 P(B | A) 与 P(A | B) 混淆。两者通常不相等。务必仔细审题，明确哪个事件是条件。

P(B | A) = P(A ∩ B) ÷ P(A)

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

Mutually exclusive events cannot occur at the same time. For example, when rolling a single die, rolling a 2 and rolling a 5 are mutually exclusive. For mutually exclusive events, P(A ∩ B) = 0, so the addition rule simplifies to P(A ∪ B) = P(A) + P(B).

互斥事件不可能同时发生。例如，掷一颗骰子时，掷出 2 和掷出 5 是互斥的。对于互斥事件，P(A ∩ B) = 0，因此加法法则简化为 P(A ∪ B) = P(A) + P(B)。

Independent events can occur simultaneously, and one does not influence the other. Independence is about the relationship between probabilities, whereas mutual exclusivity is about whether outcomes can coexist.

独立事件可以同时发生，且一个事件不影响另一个事件。独立性描述的是概率之间的关系，而互斥性描述的是结果能否共存。

These two concepts are frequently confused. Remember: mutually exclusive events cannot both happen (P(A ∩ B) = 0), while independent events satisfy P(A ∩ B) = P(A) × P(B). If two events have non-zero probabilities, they cannot be both mutually exclusive and independent.

这两个概念经常被混淆。请记住：互斥事件不能同时发生（P(A ∩ B) = 0），而独立事件满足 P(A ∩ B) = P(A) × P(B)。若两个事件的概率均不为零，则它们不可能同时是互斥且独立的。

7. Worked Example: With Replacement | 实例：放回抽样

A bag contains 4 green balls and 6 yellow balls. A ball is drawn, its colour is noted, and then it is returned to the bag. This process is repeated twice. Find the probability that both balls drawn are green.

袋子中有 4 个绿球和 6 个黄球。抽取一个球并记录颜色，然后将其放回袋中。重复该过程两次。求两次都抽到绿球的概率。

Since the ball is replaced, the two draws are independent. The probability of drawing a green ball each time is 4/10 = 2/5. Therefore, P(both green) = 2/5 × 2/5 = 4/25.

由于球被放回，两次抽取是独立的。每次抽到绿球的概率均为 4/10 = 2/5。因此，P(两次均为绿色) = 2/5 × 2/5 = 4/25。

Now find the probability of at least one green ball. It is easier to use the complement: P(at least one green) = 1 − P(no green) = 1 − (3/5 × 3/5) = 1 − 9/25 = 16/25.

现在求至少一次抽到绿球的概率。使用互补事件更简单：P(至少一次绿色) = 1 − P(没有绿色) = 1 − (3/5 × 3/5) = 1 − 9/25 = 16/25。

This example demonstrates the two key strategies: multiplying along branches for “and” outcomes, and using complements (1 − P(not event)) when “at least one” appears in the question.

该实例展示了两种关键策略：在树上沿分支相乘求解”且”事件，以及当题目出现”至少一次”时使用互补事件（1 − P(非事件)）。

8. Worked Example: Without Replacement | 实例：不放回抽样

A box contains 3 red counters and 5 blue counters. Two counters are drawn at random without replacement. Find the probability that the two counters are of the same colour.

盒子中有 3 个红色筹码和 5 个蓝色筹码。不放回地随机抽取两个筹码。求两个筹码颜色相同的概率。

Let R₁ represent red on the first draw and B₁ represent blue on the first draw, with R₂ and B₂ for the second draw. The probability of two red counters is: P(R₁) × P(R₂ | R₁) = 3/8 × 2/7 = 6/56. The probability of two blue counters is: P(B₁) × P(B₂ | B₁) = 5/8 × 4/7 = 20/56.

设 R₁ 表示第一次抽到红色，B₁ 表示第一次抽到蓝色，R₂ 和 B₂ 对应第二次抽取。两个红色筹码的概率为：P(R₁) × P(R₂ | R₁) = 3/8 × 2/7 = 6/56。两个蓝色筹码的概率为：P(B₁) × P(B₂ | B₁) = 5/8 × 4/7 = 20/56。

Since “both red” and “both blue” are mutually exclusive events, we add them: P(same colour) = 6/56 + 20/56 = 26/56 = 13/28.

由于”两个红色”和”两个蓝色”是互斥事件，我们将它们相加：P(颜色相同) = 6/56 + 20/56 = 26/56 = 13/28。

Notice how the numerator and denominator both decrease on the second branch because no replacement occurs. Constructing a complete tree diagram with symmetric branches can help verify that all probabilities sum to 1.

注意在第二条分支上分子和分母均减少，因为未放回。构建包含对称分支的完整树形图有助于验证所有概率之和为 1。

9. Extending to Three Events | 扩展到三个事件

Tree diagrams extend naturally to three or more stages. For three independent events A, B, and C, the probability of all three occurring is P(A) × P(B) × P(C). For dependent events, each subsequent branch must be adjusted based on the outcomes that came before.

树形图自然可以扩展到三个或更多阶段。对于三个独立事件 A、B、C，三者同时发生的概率为 P(A) × P(B) × P(C)。对于相依事件，每条后续分支必须根据先前结果进行调整。

Example: A biased coin has P(H) = 0.6. It is tossed three times. The probability of getting exactly two heads is: P(HHT) + P(HTH) + P(THH) = (0.6 × 0.6 × 0.4) × 3 = 0.144 × 3 = 0.432.

示例：一枚偏倚硬币正面朝上的概率为 0.6，连续抛掷三次。恰好出现两次正面的概率为：P(HHT) + P(HTH) + P(THH) = (0.6 × 0.6 × 0.4) × 3 = 0.144 × 3 = 0.432。

This “exactly k successes” pattern involves multiplying the probability of each branch, then summing over all arrangements that meet the condition. Be systematic: list all relevant ordered outcomes first.

这种”恰好 k 次成功”的模式涉及计算各分支的概率，然后对所有满足条件的排列求和。要系统化：先列出所有相关的有序结果。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Common mistakes in compound probability include: adding probabilities when you should multiply (“and” vs “or”), forgetting to change the probability after drawing without replacement, and incorrectly identifying whether events are independent. Also, many students forget that P(A ∪ B) = P(A) + P(B) only works for mutually exclusive events — the general formula includes the subtraction of P(A ∩ B).

复合概率中的常见错误包括：该乘时却加（”且”与”或”混淆）、在不放回抽取后忘记调整概率，以及未能正确判断事件是否独立。此外，许多学生忘记 P(A ∪ B) = P(A) + P(B) 仅适用于互斥事件——一般公式需要减去 P(A ∩ B)。

Useful exam strategies:

有用的考试策略：

Read the question to identify “with replacement” (independent) or “without replacement” (dependent). 判断题目是”放回”（独立）还是”不放回”（相依）。
Draw a tree diagram whenever there are two or more stages — it structures your thinking. 当有两个或更多阶段时，绘制树形图——它使思考结构化。
For “at least one” questions, use the complement 1 − P(none). 对于”至少一次”的题目，使用互补事件 1 − P(无)。
Check that all terminal branch probabilities sum to 1. 检查所有末端分支概率之和是否为 1。
Write probabilities as simplified fractions for full marks. 将概率写为最简分数以获得满分。

Practising past paper questions is essential, as Edexcel often sets compound probability problems in the context of everyday situations such as games, surveys, and quality control.

练习历年真题至关重要，因为 Edexcel 经常将复合概率问题置于日常生活情境中，如游戏、调查和质量检测。

By mastering the addition rule, multiplication rule, conditional probability, and tree diagrams, you will be fully equipped to handle any compound probability question on the IGCSE Edexcel Mathematics paper.

通过掌握加法法则、乘法法则、条件概率和树形图，你将完全有能力应对 IGCSE Edexcel 数学试卷中的任何复合概率问题。

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More Compound Probability | 复合概率进阶