Probability of Combined Events | 复合事件的概率

📚 Probability of Combined Events | 复合事件的概率

Welcome to Lesson 16-4 of our IGCSE Mathematics series. In this lesson, we will explore how to calculate probabilities when two or more events occur together. We will focus on independent and dependent events, tree diagrams, and conditional probability — all essential tools for your exams.

欢迎来到 IGCSE 数学系列的第 16 课第 4 节。在本课中，我们将探讨如何计算两个或多个事件同时发生的概率。我们重点学习独立事件、非独立事件、树状图和条件概率——这些都是考试中至关重要的工具。

1. Review of Basic Probability | 基础概率复习

Before we combine events, recall the fundamental definition: for a single event A, the probability is the number of favourable outcomes divided by the total number of equally likely outcomes.

在组合事件之前，我们先回顾基本定义：对于单个事件 A，其概率等于有利结果的数量除以所有等可能结果的总数。

P(A) = (number of favourable outcomes) ÷ (total number of possible outcomes)

P(A) = 有利结果数 ÷ 可能结果总数

All probabilities lie between 0 and 1, inclusive. A probability of 0 means the event is impossible; a probability of 1 means the event is certain.

所有概率都介于 0 和 1 之间（含端点）。概率为 0 表示事件不可能发生；概率为 1 表示事件必然发生。

2. Independent Events | 独立事件

Two events A and B are independent if the occurrence of one does not affect the probability of the other. For example, tossing a coin and rolling a dice are independent events.

如果事件 A 的发生不影响事件 B 发生的概率，则称这两个事件是独立的。例如，抛硬币和掷骰子是独立事件。

For independent events, the multiplication rule applies:

对于独立事件，适用乘法法则：

P(A and B) = P(A) × P(B)

P(A 且 B) = P(A) × P(B)

For example, the probability of getting a head on a coin and a 6 on a dice is ½ × 1⁄6 = 1⁄12.

例如，抛硬币得到正面且掷骰子得到 6 点的概率为 ½ × 1⁄6 = 1⁄12。

3. Dependent Events | 非独立事件（依赖事件）

Two events are dependent if the outcome of one changes the probability of the other. This commonly occurs when items are drawn without replacement.

如果一个事件的结果改变了另一个事件发生的概率，则这两个事件是非独立的。这通常发生在不放回抽取物品的情况下。

When drawing two marbles from a bag without replacement, the first draw changes the contents of the bag, so the second probability must be recalculated.

当从袋子中不放回地抽取两颗弹珠时，第一次抽取会改变袋中的弹珠数量，因此第二次的概率必须重新计算。

For dependent events, we use conditional probability notation:

对于非独立事件，我们使用条件概率符号：

P(A then B) = P(A) × P(B given A)

P(先 A 后 B) = P(A) × P(在 A 条件下 B)

The term “P(B given A)” is usually written as P(B|A).

“在 A 条件下 B 的概率”通常写作 P(B|A)。

4. The Multiplication Law for Combined Events | 组合事件的乘法法则

The multiplication law unifies both cases. For any two events A and B:

乘法法则统一了这两种情况。对于任意两个事件 A 和 B：

P(A and B) = P(A) × P(B|A)

P(A 且 B) = P(A) × P(B|A)

If A and B are independent, then P(B|A) = P(B), so the formula simplifies to P(A) × P(B).

如果 A 和 B 独立，则 P(B|A) = P(B)，因此公式简化为 P(A) × P(B)。

This law also extends to three or more events:

该法则还可以扩展到三个或更多事件：

P(A and B and C) = P(A) × P(B|A) × P(C|A and B)

P(A 且 B 且 C) = P(A) × P(B|A) × P(C|A 且 B)

5. Drawing Tree Diagrams | 绘制树状图

A tree diagram is a visual way to represent combined events. Each branch shows an outcome and its probability. Multiply along branches to find the probability of a sequence of events.

树状图是一种直观表示复合事件的方法。每条分支显示一个结果及其概率。沿分支相乘即可求出事件序列的概率。

When drawing a tree diagram, remember these rules:

绘制树状图时，请记住以下规则：

Each branch must show the probability as a fraction or decimal.
Each branch set must sum to 1 (since outcomes are exhaustive).
Multiply along the branches for “and” situations.
Add across final branches for “or” situations.

每条分支必须用分数或小数标出概率。
每一组分支的概率之和必须为 1（因为结果穷尽所有可能）。
遇到“且”的情况时，沿分支相乘。
遇到“或”的情况时，将最终分支的概率相加。

For dependent events, the second set of branch probabilities must be adjusted based on what happened in the first draw.

对于非独立事件，第二组分支的概率必须根据第一次抽取的结果进行调整。

6. Conditional Probability Notation | 条件概率的符号

Conditional probability answers questions like “What is the probability that B occurs, given that A has already occurred?”

条件概率回答这样的问题：“在 A 已经发生的前提下，B 发生的概率是多少？”

P(B|A) = P(A and B) ÷ P(A)

P(B|A) = P(A 且 B) ÷ P(A)

This formula is extremely useful when you know the combined probability and one individual probability, and need to find the conditional probability.

当你已知组合概率和其中一个单独概率，需要求条件概率时，这个公式非常有用。

Be careful: P(B|A) is generally not equal to P(A|B). The order of conditioning matters.

请注意：P(B|A) 通常不等于 P(A|B)。条件的方向很重要。

7. Worked Example 1: Independent Events | 例题 1：独立事件

A spinner has 3 equal sections labelled 1, 2 and 3. It is spun twice. Find the probability of getting a 2 followed by a 3.

一个转盘有 3 个等分区域，分别标有 1、2、3。该转盘旋转两次。求先得到 2 再得到 3 的概率。

Since the spinner is the same each time, the events are independent. The probability of any specific number on one spin is 1⁄3.

由于每次旋转的是同一个转盘，因此事件是独立的。单次旋转得到任何特定数字的概率都是 1⁄3。

P(2 then 3) = 1⁄3 × 1⁄3 = 1⁄9

P(先 2 后 3) = 1⁄3 × 1⁄3 = 1⁄9

Now find the probability of getting two numbers that sum to 4. The favourable outcomes are (1,3), (2,2) and (3,1).

现在求两次数之和为 4 的概率。有利结果为 (1,3)、(2,2) 和 (3,1)。

P(sum = 4) = 1⁄9 + 1⁄9 + 1⁄9 = 3⁄9 = 1⁄3

P(和为 4) = 1⁄9 + 1⁄9 + 1⁄9 = 3⁄9 = 1⁄3

Here we multiplied for “and” and added for “or”.

这里我们在“且”时相乘，在“或”时相加。

8. Worked Example 2: Dependent Events | 例题 2：非独立事件

A bag contains 4 red and 6 blue marbles. Two marbles are drawn at random without replacement. Find the probability that both marbles are red.

一个袋子中有 4 颗红色弹珠和 6 颗蓝色弹珠。不放回地随机抽取两颗弹珠。求两颗弹珠都是红色的概率。

On the first draw, P(red) = 4⁄10. After one red is removed, only 3 red remain out of 9 total marbles.

第一次抽取时，P(红色) = 4⁄10。移除一颗红色后，9 颗弹珠中只剩下 3 颗红色。

P(both red) = 4⁄10 × 3⁄9 = 12⁄90 = 2⁄15

P(两红) = 4⁄10 × 3⁄9 = 12⁄90 = 2⁄15

Now find the probability that the two marbles are of the same colour. This includes both-red or both-blue.

现在求两颗弹珠颜色相同的概率。这包括“两红”或“两蓝”两种情况。

P(both blue) = 6⁄10 × 5⁄9 = 30⁄90 = 1⁄3

P(两蓝) = 6⁄10 × 5⁄9 = 30⁄90 = 1⁄3

P(same colour) = 2⁄15 + 1⁄3 = 2⁄15 + 5⁄15 = 7⁄15

P(颜色相同) = 2⁄15 + 1⁄3 = 2⁄15 + 5⁄15 = 7⁄15

9. Worked Example 3: Conditional Probability | 例题 3：条件概率

In a class of 20 students, 12 study physics and 8 study chemistry. 5 students study both subjects. A student is chosen at random. Given that the student studies physics, find the probability they also study chemistry.

一个班级有 20 名学生，12 人学习物理，8 人学习化学。有 5 人两科都学。随机选择一名学生。已知该学生学习物理，求他也学习化学的概率。

Using the conditional probability formula:

使用条件概率公式：

P(chemistry | physics) = P(physics and chemistry) ÷ P(physics)

P(化学 | 物理) = P(物理且化学) ÷ P(物理)

= (5⁄20) ÷ (12⁄20) = 5⁄12

So about 5 out of every 12 physics students also take chemistry.

因此，大约每 12 个物理学生中就有 5 人同时也学习化学。

10. Common Mistakes | 常见错误

Students often make the same errors when working with combined probabilities. Watch out for these:

学生在处理复合概率时经常犯同样的错误。请注意以下几点：

Mistake \| 错误	Correct approach \| 正确做法
Using the same probability for the second draw without replacement	Adjust the numerator and denominator after each draw
Adding probabilities for “and” situations	Multiply probabilities for “and” situations
Forgetting to sum all relevant branches for “or” situations	Identify every favourable branch and add them
Confusing P(A\|B) with P(B\|A)	Always check the condition in the question

错误	正确做法
不放回抽取时第二次仍使用相同概率	每次抽取后调整分子和分母
在“且”的情况下把概率相加	在“且”的情况下把概率相乘
在“或”的情况下忘记把所有相关分支相加	找出所有有利分支并将其相加
混淆 P(A\|B) 和 P(B\|A)	始终检查题目中的条件对象

Always read the question carefully to determine whether replacement is involved.

请务必仔细阅读题目，判断是否涉及放回抽取。

11. Exam Tips and Summary | 考试技巧与总结

In IGCSE exams, combined probability questions often appear as multi-part questions. The first part usually tests simple probability, while later parts require tree diagrams or conditional probability.

在 IGCSE 考试中，复合概率问题常以多小问的形式出现。第一问通常考查简单概率，后面几问则要求使用树状图或条件概率。

Write down each step of the calculation to gain method marks.
Check that all branch probabilities at each level sum to 1.
Leave answers as simplified fractions unless told otherwise.
Draw a tree diagram when there are two or more stages.

写出每一步计算过程以获得方法分。
检查每一层级上所有分支概率之和为 1。
除非另有说明，否则将答案化简为最简分数。
当有两个或更多阶段时，画出树状图。

Remember the golden rules: multiply along branches, add across branches, and adjust probabilities when drawing without replacement.

请记住黄金法则：沿分支相乘，跨分支相加，不放回抽取时调整概率。

With regular practice, combined probability becomes a reliable source of marks in your IGCSE mathematics paper.

通过定期练习，复合概率将稳拿分。

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Probability of Combined Events | 复合事件的概率