Tree Diagrams in Probability | 概率中的树状图

📚 Tree Diagrams in Probability | 概率中的树状图

Tree diagrams are a systematic visual tool used to enumerate all possible outcomes of multi-stage experiments. They clearly display how probabilities branch out at each step, making it much easier to handle conditional probability, independent and dependent events, and the core rules of probability.

树状图是一种系统化的可视化工具，用于枚举多阶段试验的所有可能结果。它们清晰展示了概率在每一步如何分支扩展，大大方便了处理条件概率、独立与相关事件以及概率的核心法则。

In IB Mathematics, both Analysis and Approaches (AA) and Applications and Interpretation (AI) rely on tree diagrams to build intuition for compound events. This article covers the essential theory, worked examples, common pitfalls, and exam tips, ensuring you can confidently use tree diagrams in any problem.

在IB数学中，无论是分析与方法（AA）还是应用与解释（AI），都依赖树状图来培养对复合事件的直觉。本文涵盖核心理论、典型例题、常见错误和应考建议，确保你能在任何问题中自信地运用树状图。

---

1. What is a Tree Diagram? | 什么是树状图？

A tree diagram is a graphical model that starts from a single root node and splits into branches at each stage to represent all possible outcomes of a sequential experiment. Every branch corresponds to a specific event at that stage, labelled with its probability.

树状图是一种图形模型，从一个根节点开始，在每个阶段分叉为多个分支，以表示连续试验的所有可能结果。每一分支对应那个阶段的一个具体事件，并标有其概率。

The power of a tree diagram lies in its ability to make the sample space visible and structured. Instead of listing outcomes in a table, you follow paths from root to leaf, each path representing a unique compound outcome.

树状图的强大之处在于它能让样本空间变得可见且有结构。你无需用表格列出结果，而是沿着从根到叶的路径，每条路径代表唯一的复合结果。

---

2. Basic Structure and Components | 基本结构与组成部分

Every tree diagram has three key elements: the root node (start of the experiment), intermediate nodes (representing stages), and terminal leaves (final outcomes). Branches originating from the same node must have probabilities that sum to exactly 1, because they cover all mutually exclusive possibilities at that decision point.

每个树状图有三个关键元素：根节点（实验起点）、中间节点（代表各阶段）和终端叶节点（最终结果）。从同一节点出发的分支，其概率之和必须恰好为1，因为它们覆盖了该决策点的所有互斥可能性。

The number of branches at each node equals the number of possible outcomes at that stage. For example, if you flip a coin, two branches emerge from that node: Heads (0.5) and Tails (0.5). If you then roll a die, six branches grow from each of those, forming a comprehensive tree.

每个节点处的分支数量等于该阶段可能的结果数。例如，抛一枚硬币时，从该节点伸出两条分支：正面（0.5）和反面（0.5）。如果你接着掷一个骰子，每条分支又会长出六条分支，形成一棵完整的树。

---

3. Multiplication Rule on Tree Diagrams | 树状图中的乘法法则

To find the probability of a specific sequence of events, you multiply the probabilities along the corresponding path. This is the fundamental multiplication rule for 'and' events, which works because each branch probability beyond the first is a conditional probability.

要找出某一特定事件序列的概率，你需要沿着相应路径将概率相乘。这是“与”事件的基本乘法法则，之所以成立，是因为第一级之后每个分支的概率都是条件概率。

P(A ∩ B) = P(A) × P(B | A)

makes this multiplication intuitive by following the rain branch then the traffic branch.

例如，今天下雨的概率是0.4，而给定下雨的情况下堵车的概率是0.8，则既下雨又堵车的概率为 0.4 × 0.8 = 0.32。树状图通过沿着“下雨”分支再走“堵车”分支，使这种乘法变得直观。

---

4. Addition Rule for Mutually Exclusive Outcomes | 互斥结果的加法法则

When several final outcomes are mutually exclusive — meaning they cannot happen at the same time — the total probability of any one of them occurring is the sum of their individual path probabilities. On a tree diagram, you identify all paths that satisfy your condition and add the probabilities you already obtained through multiplication.

当多个最终结果互斥时——即它们不可能同时发生——其中任何一个发生的总概率就是它们各自路径概率的总和。在树状图上，你找出所有符合条件的路径，然后把已经通过乘法得到的概率相加。

P(A or B) = P(A) + P(B)

rams naturally separate mutually exclusive paths, making the addition step straightforward.

例如，要计算抛两次硬币恰好得到一次正面的概率，你会分别计算HT和TH路径（各为0.25），然后相加：0.25 + 0.25 = 0.5。树状图自然地将互斥路径分开，使加法步骤清晰明了。

---

5. Conditional Probability and Tree Diagrams | 条件概率与树状图

Tree diagrams are inherently built on conditional probabilities. Every branch after the first stage is labelled with a probability that assumes the previous outcomes have occurred. Thus, the second-level branches directly represent P(second event | first event).

树状图本质上是建立在条件概率之上的。第一阶段之后的每条分支，其标志的概率都假定前面结果已经发生。因此，第二级分支直接表示 P（第二个事件 | 第一个事件）。

Reversing conditional probabilities — finding P(A|B) when you know P(B|A) — is also simplified with a tree diagram. You compute the joint probability P(A ∩ B) from the path, and divide by the total probability of B, which is the sum of all paths that lead to B. This is an application of Bayes' theorem, fully visualised.

对于反向条件概率——即在已知P(B|A)的情况下求P(A|B)——树状图也能简化。你从路径中计算出联合概率P(A ∩ B)，再除以B的总概率，即所有导向B的路径概率之和。这是贝叶斯定理的应用，完全可视化。

P(A | B) = P(A ∩ B) / P(B)

---

6. Independent Events: Simplifying the Branches | 独立事件：简化分支

If two events are independent, the occurrence of the first does not affect the probability of the second. In a tree diagram, this means that the second-stage probability for a given event is the same regardless of which first-stage branch was taken. All corresponding branches carry identical probabilities.

如果两个事件独立，第一个事件的发生不会影响第二个事件发生的概率。在树状图中，这意味着无论沿哪条第一阶段分支前进，第二阶段某特定事件的概率都相同。所有对应分支的概率完全一致。

This symmetry allows you to simplify notation: you do not need to recalculate or relabel those branches. For example, when rolling a fair die twice, the probability of rolling a 6 on the second roll is always 1/6, no matter the first outcome. The tree diagram thus explicitly reflects independence.

这种对称性使你可以简化标记：无需重新计算或重新标注这些分支。例如，掷两次均匀的骰子，第二次掷出6的概率始终是1/6，无论第一次结果如何。树状图因此明确体现了独立性。

---

广告 / Sponsored

7. Dependent Events and Without Replacement | 相关事件与不放回抽样

In sampling without replacement, outcomes become dependent because the sample space changes after each draw. The probabilities on the second-stage branches must be recalculated based on what has been removed. Tree diagrams manage this perfectly by showing distinct probability sets at different second-stage nodes.

在不放回抽样中，由于样本空间在每次抽取后改变，结果变得相关。第二阶段分支的概率必须根据已被取走的物体重新计算。树状图通过在不同的第二阶段节点显示不同的概率集合，完美地处理了这一点。

For instance, drawing balls from a bag without replacement: if you first draw a red ball, the remaining composition shifts, so the chance of drawing another red decreases. Each first-stage outcome leads to a unique subtree with updated conditional probabilities.

例如，从袋中不放回地取球：如果你先抽到一个红球，剩余球的结构会改变，因此再抽到红球的概率下降。每一个第一阶段结果都会导向一个具有更新条件概率的独特子结构。

---

8. Using Tree Diagrams for Multiple Trials | 多次试验的树状图

Tree diagrams can be extended to three or more trials, although the number of final branches grows exponentially (2ⁿ for binary outcomes, mⁿ for m-ary). This is especially useful for modelling binomial experiments like repeated coin tosses or success/failure tests.

树状图可以扩展到三次或更多次试验，尽管最终分支数呈指数增长（二元结果为2ⁿ，m元结果为mⁿ）。这对于模拟二项试验（如反复抛硬币或成功/失败测试）尤其有用。

t the full tree for small n (n ≤ 4) builds a deep understanding of how combinations arise.

在这种图上，每条路径代表一个特定的有序序列。然后你可以根据相同的成功次数将路径分组，以验证二项概率公式。对于较小的n（n ≤ 4），画出完整的树状图可以深入理解组合数是如何产生的。

P(X = k) = C(n, k) × p^k × (1 − p)^{n − k}

---

9. Probability Distribution from Tree Diagrams | 从树状图获得概率分布

By aggregating the path probabilities of all final outcomes that share a numerical value (say, the count of successes), you can construct the probability mass function of a discrete random variable. The tree diagram acts as a visual bridge between raw outcomes and the probability distribution table.

通过汇总共享同一个数值（如成功次数）的所有最终结果的路径概率，你可以构建离散随机变量的概率质量函数。树状图充当了原始结果与概率分布表之间的可视化桥梁。

This approach is particularly helpful in questions that ask you to "find the probability distribution of X", where X is the number of red balls drawn in two or three picks. You simply list the final leaves, group by X, and sum the grouped probabilities.

这种方法在要求“求X的概率分布”的题目中特别有用，其中X是两到三次抽取中红球的数量。你只需列出最终叶子结果，根据X分组，并将分组概率相加即可。

X (Number of successes)	Paths	P(X = x)
0	FFF	(1−p)³
1	SFF, FSF, FFS	3p(1−p)²
2	SSF, SFS, FSS	3p²(1−p)
3	SSS	p³

(S = success, F = failure / S = 成功, F = 失败)

---

10. Common Mistakes and Tips | 常见错误与提示

A frequent mistake occurs with without-replacement problems: students forget to update the denominator and the number of favourable outcomes at the second stage. Always clearly write the new composition of items after each draw before assigning second-stage probabilities.

不放回问题中一个常见错误是：学生忘记在第二阶段更新分母和有利结果的数量。务必在给出第二阶段概率前，清楚地写下每次抽取后物品的新构成。

Another typical error is confusing the multiplication and addition rules. Remember the mantra: multiply along a path to get the probability of a specific sequence, then add probabilities of distinct paths that meet the event condition. Never multiply across paths.

另一个典型错误是混淆乘法和加法法则。记住口诀：沿一条路径相乘，得到特定序列的概率；然后把符合事件条件的不同路径的概率相加。切不可跨路径相乘。

Always check that probabilities from a single node sum to 1. If they do not, you have missed an outcome or miscounted possibilities. Verify independence claims by comparing branch probabilities at different nodes; if they differ, the events are dependent.

务必检查单个节点出发的概率之和是否为1。如果不是，说明你漏掉了某个结果或算错了可能性。通过比较不同节点处的分支概率来验证独立性声明；若概率不同，则事件相关。

---

11. Worked Example: Drawing Balls from a Bag | 典型例题：从袋中取球

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

一个袋子里有4个红球和6个蓝球。从中随机不放回地抽取两个球。求两个都是红球的概率，以及恰好有一个是红球的概率。

pend on the first outcome. If first is red, remaining bag contains 3 red, 6 blue; so P(Red | Red) = 3/9 = 1/3, P(Blue | Red) = 6/9 = 2/3. If first is blue, remaining: 4 red, 5 blue; so P(Red | Blue) = 4/9, P(Blue | Blue) = 5/9.

第一次抽取：P(红) = 4/10 = 0.4，P(蓝) = 6/10 = 0.6。第二次抽取概率依赖于第一次结果。若第一次是红，袋中剩3红6蓝，故P(红|红) = 3/9 = 1/3，P(蓝|红) = 6/9 = 2/3。若第一次是蓝，剩下4红5蓝，故P(红|蓝) = 4/9，P(蓝|蓝) = 5/9。

Published by TutorHao | IB Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）