Probability of Compound Events | 复合事件的概率计算

📚 Probability of Compound Events | 复合事件的概率计算

In IGCSE mathematics, a compound event combines two or more simple events. Understanding how to calculate the probability of compound events is essential for solving real-world problems and for achieving top marks in the Edexcel examination.

在IGCSE数学中，复合事件由两个或更多简单事件组合而成。掌握复合事件概率的计算方法，对于解决实际问题以及在Edexcel考试中取得高分都至关重要。

1. What is a Compound Event? | 什么是复合事件？

A compound event is an event that consists of two or more simple events. For example, rolling a die and getting an even number, then flipping a coin and getting heads, is a compound event.

复合事件是由两个或更多简单事件组成的事件。例如，掷骰子得到偶数，然后抛硬币得到正面，就是一个复合事件。

Simple events are individual outcomes, such as “drawing a red card” from a deck. When we combine these, we are interested in the probability of both happening, or at least one happening, depending on the situation.

简单事件是单个结果，例如从一副牌中“抽到一张红牌”。当我们把这些事件组合起来时，我们关注的是它们同时发生或至少其中一个发生的概率，具体取决于情境。

2. Sample Space and Events | 样本空间与事件

The sample space is the set of all possible outcomes of an experiment. For a single die, the sample space is {1, 2, 3, 4, 5, 6}. For two dice, the sample space consists of all ordered pairs (a, b) where a and b are each from 1 to 6.

样本空间是实验中所有可能结果组成的集合。对于单个骰子，样本空间是{1, 2, 3, 4, 5, 6}。对于两个骰子，样本空间由所有有序对(a, b)组成，其中a和b各自从1到6取值。

When calculating compound probabilities, always first identify the sample space. This helps you determine whether outcomes are equally likely and how many total outcomes exist.

计算复合概率时，首先要确定样本空间。这有助于你判断结果是否等可能，以及总共有多少个结果。

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. For example, rolling a 3 and rolling a 5 on a single die are mutually exclusive.

如果两个事件不能同时发生，则它们互斥。例如，掷一次骰子得到3和得到5是互斥的。

For mutually exclusive events A and B, the probability of A or B occurring is the sum of their individual probabilities:

对于互斥事件A和B，A或B发生的概率等于它们各自概率之和：

P(A ∪ B) = P(A) + P(B)

This is known as the addition rule for mutually exclusive events. For example, the probability of drawing a king or a queen from a standard pack is 4/52 + 4/52 = 8/52.

这就是互斥事件的加法法则。例如，从一副标准扑克牌中抽到K或Q的概率是4/52 + 4/52 = 8/52。

If events are not mutually exclusive, the formula becomes P(A ∪ B) = P(A) + P(B) − P(A ∩ B), where you subtract the overlapping probability to avoid double counting.

如果事件不是互斥的，公式变为P(A ∪ B) = P(A) + P(B) − P(A ∩ B)，减去重叠部分以避免重复计算。

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

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

如果一个事件的发生不影响另一个事件的概率，则这两个事件相互独立。例如，抛硬币和掷骰子是独立事件。

For independent events A and B, the probability of both occurring is the product of their probabilities:

对于独立事件A和B，两者同时发生的概率等于它们各自概率的乘积：

P(A ∩ B) = P(A) × P(B)

Example: Find the probability of getting a 6 on a die and heads on a coin. P(6) = 1/6, P(H) = 1/2, so P(6 and H) = 1/6 × 1/2 = 1/12.

例如：求掷骰子得到6且抛硬币得到正面的概率。P(6) = 1/6，P(H) = 1/2，所以P(6且正面) = 1/6 × 1/2 = 1/12。

Be careful: the multiplication rule only applies to independent events. If events are dependent, you must use conditional probability.

注意：乘法法则只适用于独立事件。如果事件是相互依赖的，你必须使用条件概率。

5. Conditional Probability | 条件概率

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

条件概率是指在事件B已经发生的条件下，事件A发生的概率。记作P(A|B)，读作“在B发生的条件下A的概率”。

The formula is:

公式为：

P(A|B) = P(A ∩ B) / P(B)

Rearranging this gives the general multiplication rule:

变形后得到通用的乘法法则：

P(A ∩ B) = P(A) × P(B|A)

For example, if you draw two cards without replacement, the probability that both are aces is P(first ace) = 4/52, then P(second ace | first ace) = 3/51, so P(both aces) = 4/52 × 3/51 = 12/2652.

例如，如果不放回地抽取两张牌，两张都是A的概率是P(第一张A) = 4/52，然后P(第二张A | 第一张A) = 3/51，所以P(两张都是A) = 4/52 × 3/51 = 12/2652。

6. Tree Diagrams | 树形图

Tree diagrams are a powerful visual tool for calculating probabilities of compound events. Each branch represents a possible outcome, and probabilities are written along the branches.

树形图是计算复合事件概率的强有力可视化工具。每条分支代表一个可能结果，概率写在分支旁边。

To find the probability of a particular combination, multiply the probabilities along the path. To find the probability of two or more different combinations, add the probabilities of the paths.

要找到某一特定组合的概率，沿着路径将各分支概率相乘。要找到两个或多个不同组合的概率，将各路径的概率相加。

For example, consider spinning a spinner with 3 equal sections (red, blue, green) twice. The probability of getting red first and blue second is 1/3 × 1/3 = 1/9. The probability of getting exactly one red in two spins is P(R then not R) + P(not R then R) = (1/3 × 2/3) + (2/3 × 1/3) = 4/9.

例如，考虑一个三等分转盘（红、蓝、绿），旋转两次。第一次红色且第二次蓝色的概率是1/3 × 1/3 = 1/9。两次旋转中恰好一次红色的概率是P(红→非红) + P(非红→红) = (1/3 × 2/3) + (2/3 × 1/3) = 4/9。

Tree diagrams are especially useful for problems involving “with replacement” and “without replacement”.

树形图特别适用于涉及“有放回”和“不放回”的问题。

7. Venn Diagrams | 维恩图

Venn diagrams help visualise the relationships between events. Circles represent events, and the rectangle (or universal set) represents the sample space.

维恩图有助于可视化事件之间的关系。圆圈代表事件，矩形（或全集）代表样本空间。

Key notation in Venn diagrams:

维恩图中的关键记号：

A ∩ B represents the intersection, where both events occur.
A ∪ B represents the union, where at least one event occurs.
A’ (or Aᶜ) represents the complement, meaning not A.

A ∩ B 表示交集，即两个事件同时发生。
A ∪ B 表示并集，即至少有一个事件发生。
A’（或Aᶜ）表示补集，即A不发生。

From a Venn diagram, you can directly read probabilities such as P(A), P(B), P(A ∩ B), and P(A ∪ B). This helps when solving problems with given information about overlapping groups.

从维恩图中，你可以直接读取概率，如P(A)、P(B)、P(A ∩ B)和P(A ∪ B)。这在解决给定重叠组信息的问题时很有帮助。

8. Complementary Events | 补事件

The complement of an event A is “not A”, written as A’. The probabilities of an event and its complement always sum to 1:

事件A的补事件是“非A”，记作A’。一个事件与其补事件的概率之和总为1：

P(A) + P(A’) = 1

Using complements is often the quickest way to solve probability problems. For example, “at least one” type questions are easier by finding the complement of “none”.

使用补事件往往是解决概率问题最快的方法。例如，“至少一个”类型的问题，通过求“一个都没有”的补事件更简单。

Example: In two independent trials, the probability of success is 0.4. Find the probability of at least one success. P(at least one) = 1 − P(none) = 1 − (0.6 × 0.6) = 1 − 0.36 = 0.64.

例如：在两次独立试验中，成功概率为0.4。求至少一次成功的概率。P(至少一次) = 1 − P(没有成功) = 1 − (0.6 × 0.6) = 1 − 0.36 = 0.64。

9. With Replacement vs Without Replacement | 有放回与不放回

When drawing objects from a set, the situation is different depending on whether the object is put back after each draw.

从集合中抽取物体时，根据每次抽取后是否放回，情况会有所不同。

With replacement: the events are independent, because the sample space stays the same. For example, drawing a red ball from a bag of 3 red and 2 blue balls, replacing it, then drawing again, P(R, R) = 3/5 × 3/5 = 9/25.

有放回：事件相互独立，因为样本空间保持不变。例如，从装有3个红球和2个蓝球的袋子中抽取一球，放回后再抽取一次，P(红, 红) = 3/5 × 3/5 = 9/25。

Without replacement: the events are dependent, because the composition changes after each draw. P(R, R) without replacement = 3/5 × 2/4 = 6/20 = 3/10.

不放回：事件相互依赖，因为每次抽取后成分发生变化。不放回时P(红, 红) = 3/5 × 2/4 = 6/20 = 3/10。

Always read the question carefully to determine whether replacement is allowed. This is one of the most common exam traps.

务必仔细阅读题目，判断是否允许放回。这是最常见的考试陷阱之一。

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

Many students lose marks on compound probability questions due to avoidable mistakes. Here are some key points to remember:

许多学生在复合概率问题上丢分，是因为一些本可避免的错误。以下是需要记住的关键点：

Do not use the addition rule when events are not mutually exclusive.
Do not use the multiplication rule for independent events when events are actually dependent.
Always simplify fractions where possible.
Check whether “or” means inclusive or exclusive.
Convert percentages and fractions consistently.

当事件不互斥时，不要使用加法法则。
当事件实际依赖时，不要使用独立事件的乘法法则。
尽量化简分数。
检查“或”是包含还是排除含义。
保持百分数和分数换算一致。

A good habit is to write down the type of events you are dealing with before applying any formula.

一个好习惯是在应用任何公式前，先写下你处理的是哪类事件。

11. Worked Example | 例题演练

Let us solve a typical Edexcel IGCSE style question step by step.

下面我们逐步解答一道典型的Edexcel IGCSE风格题目。

A bag contains 5 green marbles and 3 yellow marbles. Two marbles are drawn at random without replacement. Find the probability that the two marbles are the same colour.

一个袋子装有5颗绿色弹珠和3颗黄色弹珠。随机不放回地抽取两颗弹珠。求两颗弹珠颜色相同的概率。

Solution: The marbles can be the same colour in two ways: both green or both yellow. These are mutually exclusive outcomes.

解答：两颗颜色相同有两种情况：两绿或两黄。这两个结果是互斥的。

P(both green) = 5/8 × 4/7 = 20/56

P(both yellow) = 3/8 × 2/7 = 6/56

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

P(两绿) = 5/8 × 4/7 = 20/56

P(两黄) = 3/8 × 2/7 = 6/56

P(同色) = 20/56 + 6/56 = 26/56 = 13/28

Notice that we first identified the two cases, applied the multiplication rule for dependent events within each case, and then applied the addition rule for mutually exclusive cases.

注意，我们首先确定了两种情况，然后在每种情况下针对依赖事件应用了乘法法则，接着对互斥情况应用了加法法则。

12. Summary | 总结

Compound probability questions require you to identify the structure of the problem: are events independent or dependent? Mutually exclusive or not? Use the addition rule for “or” and the multiplication rule for “and”. Tree diagrams and Venn diagrams are powerful tools to organise information. Always compute with care, and check your answer for reasonableness.

复合概率问题要求你识别问题的结构：事件是独立还是依赖？互斥还是不互斥？对于“或”使用加法法则，对于“且”使用乘法法则。树形图和维恩图是组织信息的有力工具。计算时要细心，并检查答案是否合理。

With regular practice, you will master the techniques needed to excel in this topic on the Edexcel IGCSE mathematics exam.

通过定期练习，你将掌握在Edexcel IGCSE数学考试中熟练解答本主题所需的方法。

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Probability of Compound Events | 复合事件的概率计算