Tree Diagrams in Edexcel A-Level Maths | Edexcel A-Level 数学中的树状图

📚 Tree Diagrams in Edexcel A-Level Maths | Edexcel A-Level 数学中的树状图

Tree diagrams are one of the most reliable tools in Edexcel A-Level probability. They turn multi-stage experiments into a clear branching structure, showing every possible path and making it easier to apply the multiplication and addition rules correctly.

树状图是 Edexcel A-Level 概率部分最可靠的工具之一。它把多阶段试验转化为清晰的分支结构，展示每一条可能路径，从而更容易正确应用乘法法则和加法法则。

1. What Is a Tree Diagram? | 什么是树状图？

A tree diagram is a branching diagram used to list all possible outcomes of a sequence of events. Each branch represents one possible outcome at a given stage, and the probability of that outcome is written on the branch. The diagram grows from left to right, with the first event’s outcomes shown first, then the second event’s outcomes branching from each first-stage outcome.

树状图是一种分支图，用来列出一系列事件的所有可能结果。每个分支代表某一阶段的一个可能结果，该结果的概率写在分支上。图从左向右展开，首先显示第一个事件的结果，然后从每个第一阶段结果延伸出第二个事件的结果。

At A-Level, tree diagrams are especially useful for two or three stages where outcomes are not equally likely. A complete path from the start to a terminal node represents one combined outcome, such as “red then blue”. The probability of that path is found by multiplying the probabilities along its branches.

在 A-Level 中，当结果不是等可能时，树状图对两个或三个阶段的试验特别有用。从起点到终结点的一条完整路径代表一个复合结果，例如 “先红后蓝”。该路径的概率通过将沿路径各分支上的概率相乘得到。

2. The Multiplication Rule along Branches | 沿分支的乘法法则

For any path through the diagram, the probability of that specific sequence is the product of the probabilities on the branches that make up the path. This is the multiplication rule: P(A ∩ B) = P(A) × P(B|A), where P(B|A) is the probability of B given that A has already occurred.

对于图中的任意一条路径，该特定事件序列的概率等于构成该路径的各分支概率的乘积。这就是乘法法则：P(A ∩ B) = P(A) × P(B|A)，其中 P(B|A) 表示在 A 已经发生的条件下 B 发生的概率。

For example, a bag contains 4 red and 5 blue counters. If two counters are drawn without replacement, the probability of drawing red first and blue second is:

例如，一个袋子里有 4 个红色计数器和 5 个蓝色计数器。如果不放回地抽取两次，先抽到红色再抽到蓝色的概率为：

P(Red then Blue) = 4/9 × 5/8 = 20/72 = 5/18

The denominator changes from 9 to 8 because the first counter is not replaced, so only 8 counters remain for the second draw.

分母从 9 变为 8，因为第一个计数器没有被放回，所以第二次抽取时只剩下 8 个计数器。

3. The Addition Rule across Branches | 跨分支的加法法则

When an event can occur through more than one path, the probabilities of those mutually exclusive paths must be added. For example, the event “one red and one blue” can occur as red then blue or as blue then red. The total probability is the sum of the two path probabilities.

当一个事件可以通过多条路径发生时，这些互斥路径的概率必须相加。例如，事件 “一红一蓝” 可以通过先红后蓝或先蓝后红发生。总概率是这两条路径概率之和。

Using the same bag with 4 red and 5 blue counters drawn without replacement:

使用同样的装有 4 个红色和 5 个蓝色计数器的袋子，不放回抽取：

P(one red and one blue) = P(R then B) + P(B then R) = (4/9 × 5/8) + (5/9 × 4/8) = 20/72 + 20/72 = 40/72 = 5/9

The two paths are mutually exclusive, so they cannot both happen in the same two-draw experiment. Adding their probabilities is therefore valid and avoids double counting.

这两条路径是互斥的，它们不会在同一次两次抽取试验中同时发生。因此将它们的概率相加是有效的，并且可以避免重复计算。

4. With Replacement and Without Replacement | 有放回与无放回

With replacement, the item is returned to the bag before the next draw, so the probabilities at each stage remain exactly the same. This makes the successive events independent. Without replacement, the probabilities change because the total number of items decreases and the composition of the bag changes.

有放回时，物品在下一次抽取之前被放回袋中，因此每个阶段的概率保持完全相同。这使得连续事件相互独立。无放回时，概率会发生改变，因为物品总数减少，袋中的组成也发生了变化。

Feature	With replacement	Without replacement
Second-stage probability	Unchanged	Changed
Events	Independent	Dependent
Denominator	Stays the same	Decreases by 1

For example, a bag has 3 red and 2 blue counters. With replacement, P(R then R) = 3/5 × 3/5 = 9/25. Without replacement, P(R then R) = 3/5 × 2/4 = 6/20 = 3/10. Always check whether the question says “replaced” or “not replaced”.

例如，一个袋子有 3 个红色和 2 个蓝色计数器。有放回时，P(红再红) = 3/5 × 3/5 = 9/25。无放回时，P(红再红) = 3/5 × 2/4 = 6/20 = 3/10。始终检查题目说的是 “放回” 还是 “不放回”。

5. Independent Events on Tree Diagrams | 树状图中的独立事件

Two events A and B are independent if P(A ∩ B) = P(A) × P(B), or equivalently P(B|A) = P(B). On a tree diagram, independence means that the second-stage probabilities are identical no matter which first-stage outcome occurred.

两个事件 A 和 B 如果满足 P(A ∩ B) = P(A) × P(B)，或者等价地 P(B|A) = P(B)，则它们是独立的。在树状图中，独立意味着无论第一阶段发生哪个结果，第二阶段的概率都完全相同。

In Edexcel questions, you may be given P(A) and P(B) and asked to construct a tree diagram for independent events. In this case, the second-stage branch probabilities are not affected by the first-stage result. This is also the reason why with replacement produces independent events.

在 Edexcel 的题目中，你可能会被给到 P(A) 和 P(B)，并被要求为独立事件构建树状图。在这种情况下，第二阶段的分支概率不受第一阶段结果的影响。这也是为什么有放回抽取会产生独立事件。

If a question says events are independent, you can multiply their probabilities directly without needing to adjust any branch. If you are testing for independence, compare P(A ∩ B) with P(A) × P(B) or compare P(B|A) with P(B).

如果题目说明事件是独立的，你可以直接将它们的概率相乘，而无需调整任何分支。如果你在检验独立性，可以比较 P(A ∩ B) 与 P(A) × P(B)，或比较 P(B|A) 与 P(B)。

6. Conditional Probability and Reversing Order | 条件概率与逆序推理

Tree diagrams naturally give conditional probabilities in the forward direction, such as P(B|A). However, exam questions often require the reverse conditional probability P(A|B). This cannot be read directly from a forward tree; it must be calculated using the formula:

树状图自然地给出正向的条件概率，例如 P(B|A)。然而，考试题经常要求反向条件概率 P(A|B)。这不能直接从正向树状图中读出，必须使用公式计算：

P(A|B) = P(A ∩ B) / P(B)

A classic example is a disease test. Let A be the event “has the disease” and B be the event “tests positive”. The tree gives P(Positive|Disease) and P(Positive|No disease), as well as the prevalence P(Disease). Suppose the prevalence is 1%, the sensitivity is 95%, and the false positive rate is 4%.

一个经典例子是疾病检测。设 A 表示 “患病”，B 表示 “检测呈阳性”。树状图给出 P(阳性|患病) 和 P(阳性|未患病)，以及患病率 P(患病)。假设患病率为 1%，灵敏度为 95%，假阳性率为 4%。

Then:

那么：

P(Disease ∩ Positive) = 0.01 × 0.95 = 0.0095

P(No disease ∩ Positive) = 0.99 × 0.04 = 0.0396

Total P(Positive) = 0.0095 + 0.0396 = 0.0491

P(Disease|Positive) = 0.0095 / 0.0491 ≈ 0.1935

This shows that even with a positive test result, the probability of actually having the disease is only about 19.35% when the disease is rare. The tree diagram makes this counter-intuitive result easier to see because it separates the two paths that lead to a positive test.

这表明即使检测结果呈阳性，当疾病很罕见时，真正患病的概率也只有约 19.35%。树状图使这一违反直觉的结果更容易被理解，因为它分离了导致检测阳性的两条路径。

7. Bayes’ Theorem via Tree Diagrams | 通过树状图理解贝叶斯定理

Bayes’ theorem formalises the process of reversing a conditional probability. When a tree diagram is used, the numerator is one joint probability from a specific path, and the denominator is the total probability of the conditioning event, found by adding all paths that lead to that event.

贝叶斯定理将逆转条件概率的过程形式化。使用树状图时，分子是来自某一特定路径的联合概率，分母是条件事件的总概率，通过将所有通向该事件的路径相加得到。

P(A|B) = [P(B|A) × P(A)] / [P(B|A) × P(A) + P(B|A’) × P(A’)]

Here, A’ denotes the complement of A and represents “not A”. The denominator is simply the sum of all path probabilities that produce B. For Edexcel students, using a tree diagram is often clearer than memorising the abstract formula because it visualises exactly which paths contribute to the total.

这里 A’ 表示 A 的补集，即 “非 A”。分母就是所有产生 B 的路径概率之和。对 Edexcel 学生来说，使用树状图通常比记忆抽象公式更清晰，因为它直观地显示了哪些路径贡献了总概率。

8. Common Mistakes and How to Avoid Them | 常见错误与避免方法

A common mistake is adding probabilities along a path instead of multiplying them. Along a path you are finding the probability of a sequence of events all happening together, so the correct operation is multiplication.

一个常见错误是沿路径相加概率而不是相乘。沿路径你是在求一系列事件同时发生的概率，因此正确的运算是乘法。

Another mistake is forgetting to adjust the denominators when there is no replacement. If you draw without replacement, the second draw has one fewer item, and the probabilities must reflect that change.

另一个常见错误是在无放回时忘记调整分母。如果你不放回抽取，第二次抽取的物品少了一个，概率必须反映这一变化。

Students also sometimes ignore alternative orderings. If the question asks for “exactly one red” in two draws, both red-then-blue and blue-then-red paths must be included and their probabilities added.

学生有时也会忽略不同的顺序。如果题目要求在两次抽取中 “恰好一个红色”，则必须先红后蓝和先蓝后红两条路径都包括在内，并且它们的概率要相加。

Finally, be careful with percentages and decimals. Convert percentages to decimals before multiplying, so 4% becomes 0.04 and 95% becomes 0.95. Final answers can be given as exact fractions unless the question asks for decimals.

最后，要注意百分数和小数。在相乘之前将百分数转换为小数，所以 4% 变成 0.04，95% 变成 0.95。除非题目要求小数，最终答案可以用精确分数表示。

9. Exam-Style Worked Example | 考试风格例题解析

Worked example: A bag contains 5 blue counters and 3 green counters. David draws two counters at random without replacement. Find:

例题：一个袋子装有 5 个蓝色计数器和 3 个绿色计数器。David 无放回地随机抽取两个计数器。求：

(a) the probability that both counters are blue;

(a) 两个计数器都是蓝色的概率；

(b) the probability that exactly one counter is green;

(b) 恰好一个计数器是绿色的概率；

Step 1: draw the tree diagram. First draw: P(Blue) = 5/8 and P(Green) = 3/8. Second draw without replacement: P(Blue after Blue) = 4/7, P(Green after Blue) = 3/7, P(Blue after Green) = 5/7, P(Green after Green) = 2/7.

第一步：画树状图。第一次抽取：P(蓝色) = 5/8，P(绿色) = 3/8。第二次无放回抽取：P(蓝色后蓝色) = 4/7，P(绿色后蓝色) = 3/7，P(蓝色后绿色) = 5/7，P(绿色后绿色) = 2/7。

For part (a), multiply along the blue-blue path:

对于 (a) 部分，沿蓝-蓝路径相乘：

P(Both Blue) = 5/8 × 4/7 = 20/56 = 5/14

For part (b), exactly one green can occur in two ways: blue then green, or green then blue. Add these two path probabilities:

对于 (b) 部分，恰好一个绿色可以通过两种方式发生：先蓝后绿，或先绿后蓝。将这两条路径概率相加：

P(Exactly one green) = (5/8 × 3/7) + (3/8 × 5/7) = 15/56 + 15/56 = 30/56 = 15/28

For part (c), use the conditional probability formula. First find P(at least one blue):

对于 (c) 部分，使用条件概率公式。首先求 P(至少一个蓝色)：

P(At least one blue) = 1 − P(Green then Green) = 1 − (3/8 × 2/7) = 1 − 6/56 = 50/56 = 25/28

Then:

然后：

P(Both Blue | At least one blue) = P(Both Blue) / P(At least one blue) = (5/14) / (25/28) = (10/28) / (25/28) = 10/25 = 2/5

Always show these steps clearly in an exam. The tree diagram is used for the joint probabilities, while the conditional probability formula handles the “given that” part.

在考试中要清楚地展示这些步骤。树状图用于求联合概率，而条件概率公式用于处理 “已知” 的部分。

10. Summary and Exam Tips | 总结与考试技巧

Tree diagrams are a systematic way to handle multi-stage probability problems. Label every branch with its probability, check that the probabilities at each branch point sum to 1, multiply along paths, and add between paths. Always identify whether events are with or without replacement before assigning second-stage probabilities.

树状图是处理多阶段概率问题的一种系统方法。给每个分支标上概率，检查每个分支点处的概率之和为 1，沿路径相乘，在路径之间相加。在分配第二阶段概率之前，始终先确定事件是有放回还是无放回。

For reverse conditional probability questions, write the formula P(A|B) = P(A ∩ B) / P(B) and use the tree diagram to find the numerator and denominator. In Edexcel exams, showing explicit products like 4/9 × 5/8 is better than just writing a final fraction, because method marks are available even if a small arithmetic error occurs.

对于反向条件概率问题，写出公式 P(A|B) = P(A ∩ B) / P(B)，并使用树状图求出分子和分母。在 Edexcel 考试中，展示像 4/9 × 5/8 这样的明确乘积比只写最终分数更好，因为即使出现小的算术错误，也可以获得方法分。

Use exact fractions unless the question asks for a decimal or percentage. If a probability is given as a percentage, convert it to a decimal first. Practise drawing tree diagrams for two-stage and three-stage experiments until the process becomes automatic.

除非题目要求小数或百分数，否则使用精确分数。如果概率以百分数给出，先将其转换为小数。练习两阶段和三阶段试验的树状图绘制，直到这个过程变得自动化

Published by TutorHao | A-Level Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）

Tree Diagrams in Edexcel A-Level Maths | Edexcel A-Level 数学中的树状图