Tree diagrams | 树状图

📚 Tree diagrams | 树状图

Tree diagrams provide a visual way to represent all possible outcomes of a sequence of events, making it easier to calculate probabilities of combined events. They are particularly useful for multi-stage experiments in A-Level probability and form a foundation for understanding conditional probability and Bayes’ theorem.

树状图提供了一种可视化的方法来表示一连串事件的所有可能结果，从而更容易计算复合事件的概率。它们在处理A-Level概率中的多阶段试验时尤其有用，并为理解条件概率和贝叶斯定理奠定了基础。

---

1. What is a Tree Diagram? | 什么是树状图？

A tree diagram is a branching structure where each branch represents a possible outcome at a certain stage of an experiment. The branches are labelled with probabilities, and the diagram expands to show further outcomes at subsequent stages. It helps systematically list all outcomes in the sample space for two or more trials, ensuring no outcome is missed.

树状图是一种分支结构，每个分支代表试验某个阶段的一个可能结果。分支上标注着概率，并且图会扩展以显示后续阶段的更多结果。它有助于系统地列出两次或多次试验中样本空间的所有结果，确保不会遗漏任何结果。

---

2. Constructing a Tree Diagram | 构建树状图

To construct a tree diagram, start with a single node representing the start of the experiment. For each possible outcome at the first stage, draw a branch and label it with its probability. At the end of each first-stage branch, draw further branches for the second stage, and label these with the conditional probabilities given that the first outcome has occurred. Repeat for further stages as needed.

要构建一个树状图，从一个代表试验开始的节点出发。对于第一阶段每个可能的结果，画出一条分支并标注其概率。在第一阶段各分支的末端，为第二阶段画出更多分支，并标注以第一阶段结果为条件的条件概率。根据需要为后续阶段重复这一过程。

---

3. Probabilities on Branches | 分支上的概率

The probabilities on the first set of branches are unconditional probabilities and must sum to 1. The probabilities on second-stage branches are conditional probabilities, such as P(B|A), and each set of branches coming from a single node must also sum to 1. This reflects the fact that, given the previous outcome, one of the subsequent outcomes must occur.

第一组分支上的概率是无条件概率，且总和必须为1。第二阶段分支上的概率是条件概率，例如P(B|A)，并且从同一节点出发的每一组分支的概率总和也必须为1。这反映了给定之前的结果后，后续结果中必有一个会发生这一事实。

---

4. Multiplication Rule for Probabilities | 概率的乘法法则

To find the probability of a sequence of outcomes along a single path, multiply the probabilities along that path. For example, the probability of outcome A followed by outcome B is P(A ∩ B) = P(A) × P(B|A). In a tree diagram, this corresponds to multiplying the branch probabilities from the start to the end of that path.

要找到沿某一条路径的一系列结果的概率，将该路径上的概率相乘。例如，结果A后再发生结果B的概率为P(A ∩ B) = P(A) × P(B|A)。在树状图中，这对应于将从起点到该路径末端的所有分支概率相乘。

---

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

If you need the probability of an event that can occur via multiple mutually exclusive paths on the tree, add the probabilities of all the relevant paths. For instance, P(B) can be found by adding the probabilities of all paths that end in B: P(B) = Σ P(Aᵢ) × P(B|Aᵢ) over all prior outcomes Aᵢ. This is the law of total probability.

如果你需要求一个可以通过树上多条互斥路径发生的事件的概率，将所有相关路径的概率相加。例如，P(B)可以通过将所有以B结尾的路径的概率相加得到：P(B) = Σ P(Aᵢ) × P(B|Aᵢ)，对之前的所有结果Aᵢ求和。这就是全概率公式。

---

6. Conditional Probability and Second Branches | 条件概率与第二分支

Second-stage branches are always conditional on the first-stage outcome. If you are asked to find a conditional probability such as P(A|B), you can use the tree diagram to identify both the joint probability P(A ∩ B) and the marginal probability P(B). Then apply P(A|B) = P(A ∩ B) / P(B). The tree makes these components explicit.

第二阶段分支总是以第一阶段结果为条件。如果要求计算P(A|B)这样的条件概率，可以利用树状图找出联合概率P(A ∩ B)和边缘概率P(B)。然后应用公式P(A|B) = P(A ∩ B) / P(B)。树状图使这些组成部分一目了然。

---

7. Dependent and Independent Events | 相依事件与独立事件

When events are independent, the probability on a second-stage branch does not depend on the outcome of the first stage; the same probabilities appear on branches from every first-stage outcome. When events are dependent, the second-stage probabilities change depending on the first outcome. A tree diagram reveals this dependence clearly.

当事件独立时，第二阶段分支上的概率不依赖于第一阶段的结果；从每一第一阶段结果出发的分支上的概率都相同。当事件相依时，第二阶段概率会根据第一阶段的结果而变化。树状图可以清楚地揭示这种相关性。

---

8. Tree Diagrams with Replacement vs Without Replacement | 有放回与无放回试验的树状图

In a ‘with replacement’ scenario, the probabilities on second-stage branches are identical to those on the first stage because the composition of the population remains unchanged. In a ‘without replacement’ scenario, the probabilities on second-stage branches are updated to reflect the reduced number of items. The tree diagram visually distinguishes these two cases.

在有放回的情况下，第二阶段分支上的概率与第一阶段相同，因为总体构成保持不变。在无放回的情况下，第二阶段分支上的概率会被更新，以反映物品数量的减少。树状图能从视觉上区分这两种情况。

Consider a bag with 5 red and 3 blue balls. Draw two balls without replacement:

考虑一个装有5个红球和3个蓝球的袋子。无放回地抽取两次：

• 1st draw: Red (5/8)
• 第一次抽取：红球 (5/8)
• 2nd if 1st Red: Red (4/7), Blue (3/7)
• 若第一次红，第二次：红球 (4/7)，蓝球 (3/7)
• 1st draw: Blue (3/8)
• 第一次抽取：蓝球 (3/8)
• 2nd if 1st Blue: Red (5/7), Blue (2/7)
• 若第一次蓝，第二次：红球 (5/7)，蓝球 (2/7)

---

9. Finding Probabilities of Combined Events | 求复合事件的概率

Tree diagrams are ideal for calculating probabilities like ‘exactly one red’ or ‘both the same colour’. You simply identify all paths that satisfy the condition, multiply along each path, and then add these path probabilities. This method reduces complex combinatorial problems to a simple systematic process.

树状图非常适合计算像“恰好一个红球”或“两个球同色”这样的概率。你只需确定所有满足条件的路径，沿每条路径相乘，然后将这些路径概率相加。这种方法将复杂的组合问题简化为了一个简单的系统性过程。

Using the bag above, P(exactly one red) = (5/8 × 3/7) + (3/8 × 5/7) = 15/56 + 15/56 = 30/56 = 15/28.

使用上述袋子，P(恰好一个红球) = (5/8 × 3/7) + (3/8 × 5/7) = 15/56 + 15/56 = 30/56 = 15/28。

---

10. Using Tree Diagrams to Solve ‘At Least One’ Problems | 用树状图解决“至少一次”问题

For problems involving ‘at least one’, it is often more efficient to calculate the probability of the complementary event ‘none’ and subtract from 1. A tree diagram quickly gives the probability of the path where the event does not occur in any trial, which can then be used to find the desired probability.

对于涉及“至少一次”的问题，计算互补事件“一个都没有”的概率并用1来减往往更高效。树状图能够快速给出任何一次试验中事件都不发生的路径的概率，然后可将其用于求得所需的概率。

---

11. Tree Diagrams and Bayes’ Theorem | 树状图与贝叶斯定理

Bayes’ theorem reverses the order of conditioning and can be easily applied using a tree diagram. To find P(A|B), you compute the joint probability P(A ∩ B) from the path containing both A and B, and divide it by the total probability of B, which is the sum of all paths ending in B. This is exactly the Bayes formula: P(A|B) = [P(B|A) × P(A)] / [ Σ P(B|Aᵢ) × P(Aᵢ) ].

贝叶斯定理颠倒了条件的顺序，并且可以借助树状图轻松应用。要计算P(A|B)，你从包含A和B的路径中计算出联合概率P(A ∩ B)，再除以B的全概率（即所有以B结尾的路径之和）。这正是贝叶斯公式：P(A|B) = [P(B|A) × P(A)] / [ Σ P(B|Aᵢ) × P(Aᵢ) ]。

---

12. Common Mistakes and Tips | 常见错误与技巧

Common errors include forgetting that the second-stage branch probabilities must sum to 1 for each node, mixing up ‘and’ and ‘or’, and treating dependent events as independent. Always label every branch with its probability, and double-check that probabilities on branches from a single node add up to 1. Use a highlighter to mark the paths relevant to the question.

常见错误包括忘记第二阶段各分支从一个节点出发的概率之和必须为1，混淆“与”和“或”，以及将相依事件误作独立事件处理。务必在每个分支上标注概率，并再次检查从同一节点出发的分支概率之和是否为1。用荧光笔标出与问题相关的路径也会有所帮助。

Also remember that tree diagrams can be extended to three or more stages, but the same multiplication and addition rules apply. Practice with a variety of contexts, such as dice, coins, cards, and real-world scenarios.

还要记住，树状图可以扩展到三个阶段或更多，但同样的乘法和加法法则依然适用。多在各种情境下练习，例如骰子、硬币、扑克牌以及现实世界的情景。

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）