A-Level Math: Using Tree Diagrams in Probability Calculations | A-Level数学：树形图在概率计算中的应用

📚 A-Level Math: Using Tree Diagrams in Probability Calculations | A-Level数学：树形图在概率计算中的应用

Tree diagrams are powerful visual tools for solving probability problems, especially those involving sequential events or conditional probabilities. They help break complex situations into clear stages, making it easier to apply the multiplication and addition rules. This article covers the structure of tree diagrams, key rules, and worked examples aligned with the A-Level mathematics syllabus.

树形图是解决概率问题的强大可视化工具，特别适用于涉及顺序事件或条件概率的题目。它能把复杂情形拆解成清晰的阶段，让我们更容易应用乘法法则与加法法则。本文将介绍树形图的结构、关键规则以及贴合 A-Level 数学考纲的实例。

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1. What Is a Tree Diagram? | 什么是树形图

A tree diagram is a systematic way to list all possible outcomes of one or more random events. It uses branches to connect each event stage, and each branch carries a probability. By following a path from the root to the end, you can identify a single combined outcome.

树形图是一种系统化列举一个或多个随机事件所有可能结果的方法。它用分支连接每个事件阶段，每条分支上都标有概率。从根节点沿着路径走到末端，就可以确定一个组合结果。

• Tree diagrams are especially useful when there are two or three stages.

  树形图尤其适用于两到三个阶段的问题。

• They work naturally with both independent and dependent events.

  无论是独立事件还是依赖事件，树形图都能自然表示。

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2. The Structure: Branches and Nodes | 结构：分支与节点

Every tree begins at a root. The root is where the first event starts. Each branch ends at a node, which represents the outcome of that event. From every node, new branches can grow for the next event. The probabilities on all branches leaving the same node must add up to 1.

每棵树都始于根节点。根节点是第一个事件开始的地方。每条分支终止于一个节点，代表该事件的一个结果。从每个节点出发，可以长出下一个事件的新分支。从同一节点出发的所有分支概率之和必须等于 1。

For example, if a bag contains 3 red and 2 blue balls, the first draw has two branches: red with probability 3/5 and blue with probability 2/5. These two probabilities add to 1.

例如，若袋中有 3 个红球和 2 个蓝球，第一次抽取会有两条分支：红球的概率为 3/5，蓝球的概率为 2/5。这两个概率之和等于 1。

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3. The Multiplication Rule Along Branches | 乘法法则：沿分支相乘

To find the probability of following a particular path, multiply the probabilities written on each branch of that path. For events A and B, the probability that both A and B occur is P(A) multiplied by P(B) given A.

要计算沿某条路径发生的概率，将该路径上每条分支标注的概率相乘。对于事件 A 和 B，A 与 B 同时发生的概率等于 P(A) 乘以“在 A 发生的条件下 B 发生”的概率。

P(A ∩ B) = P(A) × P(B|A)

When A and B are independent, the conditional probability P(B|A) is simply P(B), so the rule becomes:

当 A 与 B 独立时，条件概率 P(B|A) 就等于 P(B)，因此法则变为：

P(A ∩ B) = P(A) × P(B)

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4. The Addition Rule Across Outcomes | 加法法则：不同结果相加

If two outcomes cannot happen at the same time, their probabilities can be added. In a tree diagram, this means summing the probabilities of the relevant paths. Each complete path represents a mutually exclusive outcome, so paths may be added without double-counting.

若两个结果不可能同时发生，则其概率可以直接相加。在树形图中，这意味着将所有相关路径的概率求和。每一条完整路径都代表一个互斥结果，因此路径之间可以相加而不会重复计算。

For example, the probability of getting exactly one red ball from two draws is the sum of the red-then-blue path and the blue-then-red path.

例如，两次抽取中恰好得到一个红球的概率，等于“红后蓝”路径与“蓝后红”路径的概率之和。

P(RB or BR) = P(RB) + P(BR)

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5. Independent vs Dependent Events | 独立事件与依赖事件

In a tree diagram, independent events have branch probabilities that do not change after the first event. Dependent events require conditional probabilities on the second set of branches. Always check whether the question says “with replacement” or “without replacement”.

在树形图中，独立事件的分支概率不会因第一次事件的结果而改变；依赖事件则需要在第二组分支上使用条件概率。解题时务必先看题目说的是“有放回”还是“无放回”。

• With replacement: the second branch probabilities are the same as the first.

  有放回：第二次分支的概率与第一次相同。

• Without replacement: the second branch probabilities change according to the first outcome.

  无放回：第二次分支的概率会根据第一次结果而改变。

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6. Conditional Probability on the Tree | 树形图中的条件概率

The tree diagram naturally shows conditional probabilities. For example, after selecting a red ball first, the probability of selecting a blue ball second is written on the branch from the “first red” node. This is P(B|R), not P(R|B).

树形图能自然地呈现条件概率。例如，在第一次取出红球后，第二次取出蓝球的概率就标注在从“第一次红球”节点出发的分支上。这就是 P(B|R)，而不是 P(R|B)。

If you need the reverse conditional probability, such as P(D|Positive), first find the probability of the relevant path and the total probability of all paths that end with the condition.

如果需要求反向条件概率，例如 P(患病 | 检测阳性)，可以先找出相关路径的概率，以及所有以该条件为末端的路径的总概率。

P(A|B) = P(A ∩ B) ÷ P(B)

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7. Worked Example 1: Drawing Balls without Replacement | 实例1：无放回取球

A bag contains 3 red and 2 blue balls. Two balls are drawn at random without replacement. Find:

袋中有 3 个红球、2 个蓝球。随机无放回抽取两次。求：

• (a) P(both red)

  (a) 两次都红

• (b) P(one red and one blue)

  (b) 一红一蓝

• (c) P(second ball is red)

  (c) 第二次为红

First draw probabilities:

第一次抽取的概率：

P(R) = 3/5, P(B) = 2/5

Second draw probabilities depend on the first draw. If the first ball is red, there are now 2 red and 2 blue balls left. If the first ball is blue, there are 3 red and 1 blue ball left.

第二次抽取的概率取决于第一次的结果。若第一次为红球，则袋中剩 2 红 2 蓝；若第一次为蓝球，则剩 3 红 1 蓝。

First draw 第一次抽取	Second draw 第二次抽取	Path probability 路径概率
R (3/5)	R (2/4)	3/5 × 2/4 = 3/10
R (3/5)	B (2/4)	3/5 × 2/4 = 3/10
B (2/5)	R (3/4)	2/5 × 3/4 = 3/10
B (2/5)	B (1/4)	2/5 × 1/4 = 1/10

The four path probabilities add to 1, which confirms the tree is correct.

四条路径的概率之和等于 1，说明树形图正确。

(a) P(both red) = 3/10

(a) P(两次都红) = 3/10

(b) P(one red and one blue) = P(RB) + P(BR) = 3/

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