📚 Probability Tree Diagrams and Conditional Probability | 概率树图与条件概率
Probability tree diagrams are among the most versatile tools in IGCSE Edexcel Mathematics. They allow you to visualise multi-stage experiments, clearly separate independent from dependent events, and compute the probabilities of combined outcomes without losing track of the conditions. Mastering tree diagrams gives you a reliable method for answering questions that range from simple coin tosses to tricky 'given that' problems involving conditional probability.
概率树图是 IGCSE Edexcel 数学中最通用的工具之一。它能让你将多阶段实验可视化，清楚地区分独立事件与依赖事件，并在不丢失条件的前提下计算组合结果的概率。掌握树图能让你获得一种可靠的方法，解答从简单的抛硬币到涉及条件概率的棘手“given that”问题。
1. What Is a Probability Tree Diagram? | 什么是概率树图
A probability tree diagram is a branching picture that illustrates every possible outcome of an experiment across one or more stages. The diagram starts from a single point on the left and extends rightwards, with each branch representing a possible event and its associated probability written alongside. When an experiment has two or more stages, additional branches are drawn from the ends of the previous ones, allowing you to follow every sequence of outcomes.
概率树图是一幅分支图，展示了一个或多个阶段实验中所有可能的结果。图表从左侧的一个点开始，向右延伸，每个分支代表一个可能的事件，并在旁边写上其概率。当实验有两个或更多阶段时，会从前一个分支的末端继续画出新的分支，让你可以追踪每一条结果序列。
Tree diagrams are particularly effective when outcomes have different probabilities at later stages depending on what happened earlier. Instead of trying to keep all possibilities in your head, you simply read the diagram. This makes tree diagrams an essential tool for both revision and exams.
当后续阶段的结果概率依赖于前面发生的事情时，树图尤为有效。你无需在脑海中记住所有可能性，只需阅读图表。这使得树图成为复习和考试中必不可少的工具。
2. Key Probability Rules You Need to Know | 你需要知道的关键概率规则
equal 1. If an event has only two possible outcomes A and B, and P(A) = 0.4, then P(B) = 1 – ular sequence of outcomes, you multiply the probabilities along the corresponding branches – this is often called the 'and' rule or product rule.
在构建树图之前，请确保你熟悉两个基本规则。第一，从同一点出发的所有分支上的概率之和必须始终等于 1。如果一个事件只有两个可能的结果 A 和 B，且 P(A) = 0.4，则 P(B) = 1 – 0.4 = 0.6。第二，要找到某一特定结果序列的概率，你将对应分支上的概率相乘——这通常称为“与”规则或乘积法则。
When two different sequences both satisfy a given condition, you add their probabilities together. This is the 'or' rule or addition rule. These two operations – multiplying along branches and adding between them – form the backbone of every tree-diagram calculation you will perform at IGCSE.
当两条不同的序列都满足某一给定条件时，你将它们的概率相加。这就是“或”规则或加法法则。这两个操作——沿分支相乘和在分支之间相加——构成了你在 IGCSE 中要进行的每一项树图计算的核心。
3. How to Draw a Tree Diagram Step by Step | 如何逐步绘制树图
Start by placing a dot on the left and deciding the outcomes of the first event. Draw one branch for each possible outcome, labelled with the outcome name and its probability written as a fraction, decimal or percentage. For the second event, imagine you are standing at the end of each first-stage branch. From that point, draw branches for all possible second outcomes, labelling them with the probabilities that apply given the first result.
首先在左侧画一个点，并确定第一个事件的结果。为每一个可能的结果画一条分支，标注结果名称及其概率，可以用分数、小数或百分比表示。对于第二个事件，想象你站在每个第一级分支的末端。从该点出发，为所有可能的第二个结果画出分支，并标注在第一个结果给定的情况下的概率。
Always check that the probabilities branching from any single node sum to 1. If they do not, you may have misread the question or forgotten to account for a third outcome. For a standard IGCSE question with a bag of counters and two picks, you will have two layers of branches and four final outcomes, as shown in the description below.
务必检查从任意单一节点出发的分支概率之和是否为 1。如果不是，你可能误读了题目，或者遗漏了第三个结果。对于标准的 IGCSE 题目，如一个装有计数器的袋子和两次抽取，你将拥有两层分支和四个最终结果，如下面的描述所示。
Example: A bag contains 3 red and 5 blue counters. First pick: red branch marked 3/8, blue branch 5/8. After one red is taken without replacement, second pick from the red endpoint: red branch 2/7, blue branch 5/7. After one blue is taken: red branch 3/7, blue branch 4/7.
示例：一个袋子装有 3 个红色和 5 个蓝色计数器。第一次抽取：红色分支标为 3/8，蓝色分支 5/8。当无放回地拿走一个红色后，从红色端点开始的第二次抽取：红色分支 2/7，蓝色分支 5/7。当拿走一个蓝色后：红色分支 3/7，蓝色分支 4/7。
4. Calculating Probabilities from Tree Diagrams | 从树图中计算概率
= 3/28. The probability of drawing one red followed by one blue is (3/8) × (5/7) = 15/56.
一旦树图完成，每个最终结果对应一条唯一路径。该结果的概率是该路径上所有分支概率的乘积。对于袋子的例子，抽到两个红色的概率是 (3/8) × (2/7) = 6/56 = 3/28。抽到先红后蓝的概率是 (3/8) × (5/7) = 15/56。
If a question asks for the probability of 'at least one red' /56 = 36/56 = 9/14. This systematic approach is much safer than trying to guess an answer.
如果题目要求“至少一个红”的概率，你可以将包含至少一个红的路径概率相加：(红, 红)、(红, 蓝) 和 (蓝, 红)。将 6/56 + 15/56 + 15/56 = 36/56 = 9/14 相加。这种系统的方法比试图猜测答案要安全得多。
5. Independent Events on a Tree Diagram | 树图中的独立事件
e, tossing a fair coin twice: from both the Head branch and the Tail branch, the second toss still has P(Head) = 0.5 and P(Tail) = 0.5.
如果第一个事件的结果不影响第二个事件的概率，则事件是独立的。在树图中，独立性表现为第二级分支上的概率完全相同，无论你来自哪条一级分支。例如，抛一枚公平硬币两次：无论是从正面分支还是反面分支出发，第二次抛掷仍然是 P(正面) = 0.5 和 P(反面) = 0.5。
, but the tree still offers a visual check. IGCSE questions frequently mix independent and dependent examples, so always read the wording carefully to decide whether probabilities change after the first selection.
对于独立事件，你通常可以直接使用简单的乘积公式 P(A and B) = P(A) × P(B) 而无需绘制树图，但树图仍然可以提供直观的检查。IGCSE 题目常常混合独立和依赖的例子，因此务必仔细阅读题目文字，判断概率在第一次选择后是否发生变化。
6. Dependent Events and the Role of Conditional Probability | 依赖事件与条件概率的作用
rst. The branches on the second level then carry conditional probabilities, written as P(B | A), meaning 'the probability of B given that A has occurred'. For the bag of counters without replacement, P(2nd red | 1st red) = 2/7 is a conditional probability.
当事件不是独立的，第二个事件的概率取决于首先发生了什么。此时第二级分支上携带的就是条件概率，写作 P(B | A)，意思是“在 A 已经发生的情况下 B 的概率”。对于无放回的计数器袋，P(第2个红 | 第1个红) = 2/7 就是一个条件概率。
m. You simply multiply along the path to get the joint probability P(A and B) = P(A) × P(B | There is no need to rearrange formulas mid-question if you follow the branches carefully.
树图让依赖问题变得简单，因为条件概率已经标注在图中。你只需沿路径相乘即可得到联合概率 P(A and B) = P(A) × P(B | A)。如果你仔细沿分支计算，就无需在解题中途重新调整公式。
7. The Conditional Probability Formula and Its Use | 条件概率公式及其应用
The formal definition of conditional probability is:
条件概率的正式定义为：
P(B | A) = P(A ∩ B) ÷ P(A)
where P(A ∩ B) is the probability that both A and B occur. In words, you take the probability of the combined event and divide by the probability of the condition that is 'given'.
其中 P(A ∩ B) 是 A 和 B 同时发生的概率。换句话说，你取组合事件的概率，然后除以所“给定”条件的概率。
In a tree diagram, P(A ∩ B) is simply the product along the branches that give both A and B, and P(A) is the sum of probabilities of all paths that include A. This relationship is often tested in IGCSE when a question says 'Given that..., find the probability...'. You can then use the tree to extract the numerator and denominator directly.
在树图中，P(A ∩ B) 就是沿产生 A 和 B 的分支的乘积，而 P(A) 是所有包含 A 的路径的概率之和。当 IGCSE 题目中说“已知……求概率……”时，常常会考查这一关系。此时你可以直接利用树图提取分子和分母。
8. Without Replacement: A Classic IGCSE Scenario | 无放回：经典的 IGCSE 情境
'Without replacement' is the context in which tree diagrams truly shine. When an item is removed and not replaced, the probabilities on the second draw change because the total number of items and the composition both change. IGCSE papers love this scenario because it tests your ability to adjust probabilities dynamically.
“无放回”是树图真正大放异彩的情境。当物品被取走且不放回时，第二次抽取的概率会发生变化，因为物品的总数和构成都改变了。IGCSE 试卷钟爱这种情境，因为它能考查你动态调整概率的能力。
) = 4/10 = 2/5. If a green marble is taken, 3 green and 6 yellow remain, so P(Green second | first green) = 3/9 = 1/3. If a yellow was taken first, P(Green second | . This change in denominators is a key sign of a dependent, without-replacement situation.
考虑一个装有 4 颗绿色弹珠和 6 颗黄色弹珠
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