📚 Multiple Events (Tree Diagrams) | 多事件（树状图）
Tree diagrams offer a systematic and visual method for mapping out all possible outcomes of multiple events, making complex probability calculations more manageable. They are particularly powerful when events occur in sequence or when each outcome influences the next. This article explores how to construct and interpret tree diagrams, apply the multiplication and addition rules, and handle both independent and conditional probabilities in IB Mathematics.
树状图为我们提供了一种系统化、可视化的方法，用以梳理多个事件的所有可能结果，使复杂的概率计算变得更容易掌握。当事件顺序发生，或者每个结果会影响后续事件时，树状图尤为强大。本文探讨如何构建和解读树状图，应用乘法和加法法则，并在 IB 数学中处理独立概率和条件概率。
1. What is a Tree Diagram? | 什么是树状图？
A tree diagram is a branching structure that displays all possible outcomes of a multi‑step random experiment. Each branch represents a single outcome of one trial, labelled with its probability. Starting from a single point, the diagram branches outward for each successive event, creating a complete map of the sample space.
树状图是一种分支结构，展示多步骤随机试验的所有可能结果。每个分支代表一次试验的单一结果，并标有其概率。从一个点开始，树状图为每一个后续事件向外延伸分支，生成完整的样本空间地图。
In IB Mathematics, tree diagrams are used extensively for problems involving coin tosses, dice rolls, drawing cards or balls from bags, and any scenario with conditional or sequential probabilities.
在 IB 数学中，树状图广泛用于投掷硬币、掷骰子、从袋中抽牌或抽球，以及任何涉及条件概率或顺序概率的情境。
2. Constructing a Tree Diagram | 构建树状图
To construct a tree diagram, begin with a starting node on the left. For the first event, draw a branch for each possible outcome and write its probability along the branch. From the end of each first‑event branch, draw further branches for the second event, labelling them with the probabilities that apply given the preceding outcome. Continue for subsequent events. The final nodes at the rightmost ends represent complete sequences of outcomes.
构建树状图时，从左侧的起始节点开始。为第一个事件的每一个可能结果画一条分支，并沿着分支写上其概率。从每个第一事件分支的末端，再为第二个事件画出更多分支，标注在给定前一个结果下适用的概率。对后续事件重复这一过程。最右端的终结点代表完整的结果序列。
A well‑labelled tree diagram has two key features: each set of branches that emerge from a single node must have probabilities that sum to 1, and every pathway from start to a terminal node represents a unique compound outcome.
一张标注清晰的树状图有两个关键特征：从同一节点延伸出的每一组分支，其概率之和必须等于 1；从起点到终结点之间的每一条路径都代表一个唯一的复合结果。
3. Independent Events on a Tree | 树状图中的独立事件
When events are independent, the probability of an outcome at a later stage does not change regardless of what happened earlier. On a tree diagram, this means the second‑stage probabilities are identical on every branch that represents the same type of outcome. For example, tossing a fair coin twice: the probability of heads on the second toss is always ½, whether the first toss was heads or tails.
当事件相互独立时，后续阶段结果的概率不会因前期发生的情况而改变。在树状图上，这意味着代表同一类型结果的第二阶段分支的概率完全相同。例如，抛掷一枚公平硬币两次：无论第一次是正面还是反面，第二次得到正面的概率始终是 ½。
The branches for independent events use the same unchanging probabilities stage after stage. The multiplication rule along any path simply multiplies the branch probabilities to obtain the compound event’s probability.
独立事件的分支在各阶段使用相同且不变的概率。沿任意路径应用乘法法则，只需将各分支概率相乘，即可得到该复合事件的概率。
4. Dependent Events & Conditional Probability | 非独立事件与条件概率
When events are dependent, the probability of a later event changes based on earlier outcomes. Tree diagrams handle this naturally by writing different probabilities on branches that stem from different preceding outcomes. These are conditional probabilities, often written as P(B|A), meaning “the probability of event B given that event A has occurred”.
当事件不独立时，后续事件的概率会根据之前的结果改变。树状图对此能够自然地处理：从不同前驱结果延伸出来的分支上标注不同的概率。这些就是条件概率，通常写作 P(B|A)，意为“在事件 A 已发生的条件下，事件 B 发生的概率”。
A classic example is drawing two balls from a bag without replacement. Suppose a bag contains 3 red and 2 blue balls. The probability that the second ball is red depends on the colour of the first ball drawn. The second‑stage branches will show the updated probabilities based on the remaining balls.
一个经典例子是从袋中无放回地抽取两个球。假设袋中有 3 个红球和 2 个蓝球。第二个球是红色的概率取决于第一个球抽到的颜色。第二阶段的分支将根据剩余球的数量显示更新后的概率。
5. Multiplication Rule Along a Path | 路径乘法法则
To find the probability of a specific sequence of outcomes—an intersection of events—multiply the probabilities along the branches of the corresponding path. For a two‑stage tree, if the first event is A and the second is B, then P(A ∩ B) = P(A) × P(B|A). This rule extends directly to three or more stages: multiply all branch probabilities along the path from start to finish.
要计算某个特定结果序列（即事件的交集）的概率，只需将对应路径上各分支的概率相乘。对于两阶段树状图，若第一事件为 A，第二事件为 B，则 P(A ∩ B) = P(A) × P(B|A)。该法则可直接扩展到三个阶段或更多：将从起点到终点的路径上所有分支概率相乘。
This multiplication rule is the core computation on a tree diagram. When events are independent, P(B|A) simplifies to P(B), and the expression becomes P(A) × P(B). The visual nature of the tree helps avoid forgetting to condition on earlier outcomes.
这条乘法法则是树状图上的核心计算。当事件独立时，P(B|A) 简化为 P(B)，表达式变为 P(A) × P(B)。树状图可视化的特性有助于避免忘记基于先前结果的条件设定。
6. Addition Rule for Multiple Paths | 多路径的加法法则
Often we are interested in an event that can occur through several different compound paths—for instance, “exactly one head in two coin tosses”. This event consists of the sequences (H, T) and (T, H). The probability of the overall event is the sum of the probabilities of each mutually exclusive path that satisfies the condition.
我们通常关注那些可通过多个不同复合路径发生的事件，例如，“两枚硬币投掷中恰好出现一次正面”。这一事件包含序列 (H, T) 和 (T, H)。该整体事件的概率等于所有满足条件且互斥的各路径概率之和。
Symbolically, if an event E can be realised by k disjoint sequences, then P(E) = P(path₁) + P(path₂) + … + P(pathₖ). Once the tree is fully drawn and terminal probabilities computed, simply add the relevant terminal values.
用符号表示：若事件 E 可通过 k 个互不相交的序列实现，则 P(E) = P(路径₁) + P(路径₂) + … + P(路径ₖ)。一旦完整绘制出树状图并求出终点概率，只需将相关终点的值相加即可。
7. Finding Conditional Probabilities from a Tree | 由树状图求条件概率
Tree diagrams also make it possible to work backwards and compute conditional probabilities such as P(A|B) when P(B|A) is known. Using the formula P(A|B) = P(A ∩ B) / P(B), the numerator comes directly from a branch product, while the denominator is the sum of all paths that end in event B.
树状图还使我们能够反向求解，在已知 P(B|A) 的情况下计算 P(A|B) 这样的条件概率。运用公式 P(A|B) = P(A ∩ B) / P(B)，分子直接来自分支的乘积，分母则是所有以事件 B 结尾的路径概率之和。
This technique is essentially Bayes’ theorem visualised. By identifying the “reverse” condition, students can avoid memorising the formula and instead rely on the tree to extract the needed numerator and denominator.
这种方法本质上就是可视化的贝叶斯定理。通过识别“反向”条件，学生可以不必死记公式，而是依靠树状图提取所需的分子和分母。
8. With Replacement vs Without Replacement | 有放回 vs 无放回
One of the most important distinctions in tree diagram problems is whether selections are made with or without replacement. With replacement, the probabilities on the second and subsequent selections remain unchanged, so the events are independent. Without replacement, the probabilities update, and the events are dependent.
树状图问题中最重要的区别之一是选择是否放回。在有放回的情况下，第二次及之后的选择概率保持不变，因此事件相互独立。在无放回的情况下，概率会更新，事件不独立。
For example, drawing two cards from a deck of 52: with replacement, P(ace on second draw) = 4/52 regardless of the first draw. Without replacement, if the first was an ace, P(ace on second draw) = 3/51. The tree branches must reflect the changing denominators and numerators.
例如，一副 52 张的扑克牌中抽两张：有放回时，无论第一张抽到什么，P(第二次抽到 A) = 4/52。无放回时，若第一张是 A，则 P(第二次抽到 A) = 3/51。树状图的分支必须反映变化的分母和分子。
9. Tree Diagrams with More Than Two Stages | 超过两个阶段的树状图
Tree diagrams are not limited to two events. For three or more sequential events, simply add extra layers of branches. The same principles apply: each branching set sums to 1, path probabilities multiply, and the probability of a compound event that occurs via multiple routes is the sum of those route probabilities.
树状图并不局限于两个事件。对于三个或更多顺序事件，只需增加额外的分支层级。相同的原理适用：每一组分支总和为 1，路径概率相乘，通过多条路径实现的复合事件的概率为这些路径概率之和。
For instance, a student taking three multiple‑choice questions with four options each has a tree of depth 3. Each question’s branch splits into 4. The total number of terminal nodes is 4³ = 64, but the tree helps focus on specific sequences like “two correct and one wrong” by adding the suitable path probabilities.
例如，一位学生做三道四选一的选择题，树状图深度为 3。每道题分出 4 条分支。终节点总数为 4³ = 64，但树状图有助于聚焦特定序列，如“答对两题答错一题”，只需将相应路径概率相加。
10. Common Mistakes to Avoid | 常见错误与避免方法
A frequent mistake is failing to ensure that the probabilities on all branches coming from a single node sum to exactly 1. If the sum is less than or greater than 1, the diagram does not model a complete sample space and will produce incorrect results.
一个常见错误是未能确保从同一节点伸出的所有分支的概率之和恰好为 1。如果总和小于或大于 1，树状图就没有模拟完整的样本空间，会产生错误的结果。
Another pitfall is mixing up the multiplication and addition rules: multiplying when paths should be added, or adding branch probabilities on the same path instead of multiplying. Carefully identifying whether you are tracking a single sequence (multiply) or combining multiple sequences (add) prevents this.
另一个易犯错误是混淆乘法法则和加法法则：在该相加路径时却相乘，或把同一路径上的分支概率相加而非相乘。仔细辨别你是要追踪单个序列（相乘）还是要合并多个序列（相加），就能避免这个问题。
Students also sometimes label a branch with the probability of the intersection instead of the conditional probability. Remember that each branch shows P(this outcome | previous outcomes), not P(previous outcome ∩ this outcome).
学生有时还会将分支标成交集概率，而不是条件概率。请记住，每个分支展示的是 P(该结果 | 先前结果)，而不是 P(先前结果 ∩ 该结果)。
11. Worked Example: Without Replacement | 完整例题：无放回抽样
A bag contains 5 green sweets and 3 yellow sweets. Two sweets are drawn at random without replacement. Construct a tree diagram and find the probability that the two sweets drawn are of different colours.
一个袋子装有 5 颗绿色糖果和 3 颗黄色糖果。随机无放回地抽取两颗。构建树状图，并求抽到两个不同颜色糖果的概率。
First draw: P(G) = 5/8, P(Y) = 3/8. Second draw branches: if first was G, then P(G|G) = 4/7 and P(Y|G) = 3/7; if first was Y, then P(G|Y) = 5/7 and P(Y|Y) = 2/7. The event “different colours” occurs on the paths (G, Y) and (Y, G).
第一次抽取：P(G) = 5/8，P(Y) = 3/8。第二阶段分支：若第一次为 G，则 P(G|G) = 4/7，P(Y|G) = 3/7；若第一次为 Y，则 P(G|Y) = 5/7，P(Y|Y) = 2/7。事件“不同颜色”发生在路径 (G, Y) 和 (Y, G) 上。
Path probability for (G, Y) = (5/8) × (3/7) = 15/56. Path for (Y, G) = (3/8) × (5/7) = 15/56. Total probability = 15/56 + 15/56 = 30/56 = 15/28.
路径 (G, Y) 的概率 = (5/8) × (3/7) = 15/56。路径 (Y, G) 的概率 = (3/8) × (5/7) = 15/56。总概率 = 15/56 + 15/56 = 30/56 = 15/28。
This example highlights how the tree structure makes it straightforward to capture the subtle change in probability due to the lack of replacement, and how the addition rule combines the two favourable sequences.
这个例子突显了树状结构如何直接捕捉到因无放回而导致的概率细微变化，以及加法规则如何将两个有利序列组合起来。
12. Summary and Key Tips | 总结与核心技巧
Tree diagrams are an indispensable tool for handling multiple events in probability. Start every branch probability from the correct conditioning point, check that sibling branches sum to 1, multiply along paths, and add between paths. Always pause to decide whether events are independent or dependent—this governs whether second‑stage probabilities stay constant or vary.
树状图是处理概率中多事件问题不可或缺的工具。从正确的条件点开始标注每个分支的概率，检查同级分支概率之和是否为 1，沿路径相乘，并在不同路径间相加。始终先停下来判断事件是独立还是非独立——这决定了第二阶段概率是保持不变还是随之变化。
When faced with complex multi‑stage problems, drawing a clear, well‑spaced tree diagram transforms abstract conditions into a concrete visual, reduces mistakes, and reveals all necessary paths. Practice building trees for both with‑replacement and without‑replacement scenarios until the process becomes automatic.
面对复杂的多阶段问题时，绘制一张清晰、布局良好的树状图能将抽象的条件转化为具体的可视形式，减少错误，并揭示所有必要的路径。请反复练习有放回和无放回两种情境下的树状图构建，直到整个过程变得流畅自如。
Published by TutorHao | Mathematics Revision Series | aleveler.com
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