📚 More Tree Diagrams | 进阶树形图
Tree diagrams are an essential visual tool in probability, helping you to break down multi-stage events into clear branches. In IGCSE Edexcel Mathematics, "More Tree Diagrams" goes beyond simple coin tosses to cover conditional probability, dependent events without replacement, reverse working, and "at least one" scenarios. This article will guide you through constructing, interpreting, and solving advanced tree diagram problems with confidence.
树形图是概率中必不可少的可视化工具，它能将多阶段事件分解为清晰的分支。在 IGCSE Edexcel 数学中，“进阶树形图”的考察范围超出了简单的抛硬币，涵盖了条件概率、不放回的相依事件、逆向求解以及“至少一次”等典型场景。本文将带你全面掌握高级树形图的构建、解读与解题技巧。
1. What Are Tree Diagrams? | 什么是树形图？
A tree diagram is a branching structure that shows every possible outcome of an experiment that consists of two or more stages. Each branch represents a possible result at that stage, and the probability of that result is written along the branch. The outcomes at the end of the branches are combined outcomes, and their probabilities are found by multiplying along the branches. Tree diagrams are especially powerful when events are not equally likely or when probabilities change from stage to stage.
树形图是一种分支结构，用来展示包含两个或更多阶段的试验中所有可能的结果。每个分支代表该阶段的一个可能结果，概率标注在分支上。分支末端的结局是组合结果，其概率可通过沿着分支相乘得到。当事件并非等可能，或者概率随时间改变时，树形图的威力就尤为突出。
2. Independent Events and the Multiplication Rule | 独立事件与乘法法则
e probability of rolling a 6 is 1/6 each time. The probability of rolling two sixes is found by multiplying along the branches: P(6 then 6) = (1/6) × (1/6) = 1/36. The same principle applies to any sequence of independent events.
当事件相互独立时，第一次事件的结果不会影响第二次事件的结果。在描绘独立事件的树形图中，第二组分支上的概率与第一组完全相同。例如，投掷一枚均匀的六面骰子两次，每次掷出 6 的概率都是 1/6。两次都掷出 6 的概率可通过沿分支相乘求得：P(先6,后6) = (1/6) × (1/6) = 1/36。同样的乘法法则适用于任何独立事件序列。
3. Dependent Events (Without Replacement) | 相依事件（不放回）
g without replacing them. Suppose a bag contains 3 red balls and 5 blue balls. If you draw two balls without replacement, the probability of drawing a red ball second changes depending on what was drawn first. In the tree diagram, the second-stage probabilities must be updated: after drawing a red first, there are 2 red left out of 7, so P(red second | red first) = 2/7. The branches no longer carry identical numbers.
相依事件发生在第一次事件的结果改变了第二次事件的概率时。经典的例子是从袋子中不放回地抽取物品。假设一个袋子里有 3 个红球和 5 个蓝球。如果不放回地抽取两个球，第二次抽到红球的概率会因第一次抽到的颜色而改变。在树形图中，第二阶段的概率必须更新：第一次抽到红球后，剩下 2 个红球和总共 7 个球，故 P(第二次红 | 第一次红) = 2/7。此时各分支上的数值不再相同。
4. Constructing a Probability Tree | 构建概率树形图
Follow these steps to build an accurate tree diagram: (1) Identify the stages and list all possible outcomes for the first stage. (2) Draw a starting point and branch out the first-stage outcomes with their probabilities; these must sum to 1. (3) For each first-stage outcome, draw a second set of branches for the second stage, again ensuring probabilities sum to 1. (4) Label each branch clearly with the probability in fraction or decimal form. (5) At the end of each final branch, write the combined outcome and multiply the probabilities along the path to find its probability. (6) Check that all final probabilities sum to 1.
按照以下步骤构建精确的树形图：(1) 明确阶段，列出第一阶段的所有可能结果。(2) 画一个起点，用分支表示第一阶段结果并标出概率；这些概率之和必须为 1。(3) 对每一个第一阶段的结果，再画出第二组分支代表第二阶段，同样确保概率之和为 1。(4) 在每个分支上清晰地标明分数或小数形式的概率。(5) 在每条末端分支上写出组合结果，沿路径相乘得到该结局的概率。(6) 检查所有末端概率之和是否为 1。
5. Using the Addition Rule for Combined Outcomes | 使用加法法则求组合结果
Once you have found the probabilities of the final outcomes, the addition rule allows you to find the probability of one event or another occurring, provided the outcomes are mutually exclusive. For example, if you flip two coins, you can calculate the probability of getting exactly one head by adding P(HT) and P(TH). On a tree diagram, simply identify the relevant end points and add their probabilities. The addition rule is essential for solving "at least one" problems and for summing probabilities across different branches.
一旦求出各最终结局的概率，就可以利用加法法则来计算某一事件或另一事件发生的概率，前提是这些结局互斥。例如，抛掷两枚硬币时，正好出现一个正面的概率就是 P(正反) 与 P(反正) 的和。在树形图上，只需找出相关的末端点，将其概率相加。加法法则对于解决“至少一次”问题以及汇总不同分支上的概率至关重要。
6. Conditional Probability on Tree Diagrams | 树形图中的条件概率
ability of an event occurring given that another event has already happened. On a tree diagram, P(A| ion to the branch where B has occurred. For example, a bag has 3 red and 2 blue balls; two balls are drawn without replacement. To find P(first red | ue) by adding the probabilities of (red, blue) and (blue, blue). Then use the formula P(first red | blue) / P(second blue). Tree diagrams make this partitioned calculation visual and systematic.
条件概率指的是在已知另一事件已经发生的情况下，某事件发生的概率。在树形图中，可将注意力限定在 B 已发生的分支上，进而求出 P(A|B)。例如，袋中有 3 红 2 蓝，不放回地抽两次。要求 P(第一次红 | 第二次蓝)，首先通过 (红,蓝) 与 (蓝,蓝) 的概率之和求出 P(第二次蓝)。然后代入公式：P(第一次红 | 第二次蓝) = P(第一次红且第二次蓝) / P(第二次蓝)。树形图使这种分区计算变得直观而有序。
7. Reverse Tree Diagrams: Finding Prior Probabilities | 逆树形图：求先验概率
In some exam questions you are given a conditional probability at the end and must work backwards to find an unknown probability on an earlier branch. This is often done by introducing a variable, say x, for the unknown branch probability, calculating the expressions for final outcomes, and then setting up an equation using the given conditional probability. Solving for x gives the required prior probability. Reverse tree diagram problems test your ability to handle algebraic fractions and proportional reasoning under probability constraints.
有些考题会给出末端的条件概率，要求你逆向推导出较早分支上的未知概率。常用的方法是设未知分支概率为 x，写出各末端概率的表达式，再利用已知的条件概率建立方程。解出 x 便得到所求的先验概率。逆树形图题目考验你在概率约束下用代数分数和比例推理的能力。
8. "At Least One" Problems | “至少一次”问题
Questions that ask for the probability of "at least one" one 6 is 1 - (5/6 × 5/6 × 5/6) = 1 - 125/216 = 91/216. The tree diagram confirms that the "no sixes" path is just one branch out of many.
对于求解“至少一次”成功的问题，用补集法则往往更快捷：P(至少一次) = 1 - P(全部失败)。在树形图中，你可以找出那条唯一的零成功路径，计算它的概率，再从 1 中减去。例如，掷骰子三次，至少出现一次 6 的概率为 1 - (5/6 × 5/6 × 5/6) = 1 - 125/216 = 91/216。树形图可以直观地显示“完全没有 6”只是众多分支中的一条。
9. Tree Diagrams with Three or More Stages | 三阶段或多阶段树形图
. Each branch at each stage still requires probabilities that sum to 1. More complex scenarios, like drawing three cards without replacement or repeating a medical test, can be handled by extending the tree further. To keep diagrams tidy, use clear labelling and consider folding symmetrical branches when outcomes are identical.
当试验包含三个或更多阶段时，树形图会变得更大，但原理完全一致。比如掷三次硬币会产生 2³ = 8 个最终结局。每个阶段的各分支仍需要概率之和为 1。更复杂的情形，如不放回地抽三张牌或重复进行某项医学测试，可以通过进一步延伸树形图来处理。为了保持图示整洁，应当使用清晰的标注，并考虑在结果对称时合并分支。
10. Common Mistakes and Exam Tips | 常见错误与考试技巧
One common mistake is forgetting to update probabilities for dependent events, e.g. using 3/8 for a second draw when a ball has already been removed. Another is misapplying the addition rule by adding probabilities of non-mutually exclusive outcomes without adjusting for overlap. Students sometimes write probabilities greater than 1 or fail to check that branch probabilities at each node sum to 1. In exams, always read carefully whether replacement occurs, label branches with fractions rather than decimals for accuracy, and leave your final answer as an unsimplified fraction if the question allows, showing all working on the tree.
常见的错误包括忘记为相依事件更新概率，例如在已拿走一个球后仍用 3/8；或是错误地使用加法法则，把非互斥事件的概率直接相加而未调整重叠部分。有些学生会写出大于 1 的概率，或者忘记检查每个节点上分支概率之和是否为 1。考试时，一定要仔细读题，看清是否放回；为了提高准确度，用分数而不是小数标示概率；如果题目允许，最终答案可以保留未化简的分数，并在树形图上展示全部计算过程。
11. Practice Problem Walkthrough | 例题讲解
Let's work through a typical IGCSE "More Tree Diagrams" problem. A bag contains 4 red pencils and 5 green pencils. Two pencils are taken at random without replacement. (a) Draw and label a tree diagram. (b) Find the probability that both pencils are red. (c) Find the probability that exactly one pencil is red. (d) Given that the second pencil is green, find the probability that the first pencil was red.
我们来演练一道典型的 IGCSE “进阶树形图”问题。一个袋子里有 4 支红铅笔和 5 支绿铅笔。随机抽取两支，且不放回。 (a) 画出并标注树形图。(b) 求两支都是红色的概率。(c) 求恰好一支为红色的概率。(d) 已知第二支是绿色，求第一支是红色的概率。
Solution: 9, P(Green) = 5/9. Second draw: after Red, remaining: 3 Red, 5 Green -> totals 8, so P(Red|Red)=3/8, P(Green|Red)=5/8. After Green first, remaining: 4 Red, 4 Green -> totals 8, so P(Red|Green)=4/8=1/2, P(Green| (5/8)=20/72; P(Green, Red) = (5/9)×(4/8)=20/72; Sum = 40/72 = 5/9. (d) We need P(first Red | (5/9)×(4/8)=20/72 + 20/72 = 40/72. P(first Red and second Green) = 20/72. Thus P(first Red | second Green) = (20/72) / (40/72) = 1/2.
解答：(a) 第一次抽取概率：P(红) = 4/9，P(绿) = 5/9。第二次抽取：若第一次是红，剩下 3 红、5 绿，共 8 支，则 P(红|红)=3/8，P(绿|红)=5/8。若第一次是绿，剩下 4 红、4 绿，共 8 支，则 P(红|绿)=4/8=1/2，P(绿|绿)=4/8=1/2。(b) P(两支全红) = (4/9)×(3/8) = 12/72 = 1/6。(c) 恰好一支红：(红,绿) 或 (绿,红)。P(红,绿) = (4/9)×(5/8)=20/72；P(绿,红) = (5/9)×(4/8)=20/72；总和 = 40/72 = 5/9。(d) 要求 P(第一次红 | 第二次绿)。P(第二次绿) = P(红,绿) + P(绿,绿) = 20/72 + 20/72 = 40/72。P(第一次红且第二次绿) = 20/72。因此 P(第一次红 | 第二次绿) = (20/72) / (40/72) = 1/2。
12. Summary and Key Takeaways | 总结与要点
More Tree Diagrams in IGCSE Mathematics teach you to model complex, multi-stage probability problems with accuracy. Always note whether events are independent or dependent, update branch probabilities correctly, multiply along branches for combined outcomes, and add across branches for mutually exclusive events. For conditional probability and reverse problems, let the tree structure guide your partitioning. With systematic drawing and clear fractions, you can confidently tackle even the trickiest Edexcel probability questions.
IGCSE 数学中的进阶树形图教会你精确地建模复杂的多阶段概率问题。务必留意事件是独立还是相依，正确更新分支概率，沿分支相乘以获得组合结果，并在互斥事件间横向相加。对于条件概率和逆向问题，让树形结构的天然分区为你导航。凭借系统的作图方式和清晰的分数量值，再棘手的 Edexcel 概率题你都能从容应对。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图