Mastering Conditional Probability & Tree Diagrams | 掌握条件概率与树形图

📚 Mastering Conditional Probability & Tree Diagrams | 掌握条件概率与树形图

Conditional probability and tree diagrams are among the most frequently tested topics in IGCSE Mathematics. They appear in Paper 2 and Paper 4, often as multi-part questions worth 5 to 8 marks. Understanding the logic behind “given that” is the key to unlocking these questions.

条件概率与树形图是 IGCSE 数学中最常考查的知识点之一，常见于 Paper 2 和 Paper 4，通常作为 5 到 8 分的多小问综合题。理解”在……条件下”背后的逻辑，是攻克这类题目的关键。

1. What Is Conditional Probability? | 什么是条件概率？

Conditional probability asks: if we already know that event B has happened, what is the probability that event A also happens? We write this as P(A | B), read as “the probability of A given B”.

条件概率问的是：若我们已知事件 B 发生了，那么事件 A 也发生的概率是多少？我们将其记为 P(A | B)，读作”在 B 条件下 A 的概率”。

The formal definition connects conditional probability to the intersection of two events:

正式定义将条件概率与两个事件的交集联系起来：

P(A | B) = P(A ∩ B) ÷ P(B), where P(B) ≠ 0

This formula is not just a definition — it is also a powerful tool. If you know P(A ∩ B) and P(B), you can find the conditional probability directly.

这个公式不仅是定义，更是强有力的工具。若你知道 P(A ∩ B) 和 P(B)，就能直接求出条件概率。

2. Understanding “Given That” | 理解”在……条件下”

In exam questions, the phrase “given that” signals a conditional probability. It tells you that the sample space has already been reduced. You are no longer looking at all possible outcomes — only the outcomes that satisfy the given condition.

在考题中，”given that（在……条件下）”这一短语标志着条件概率。它告诉你样本空间已经缩小了。你不再考虑所有可能的结果，而只需关注满足给定条件的结果。

For example, if a student is chosen from a class and we are told the student is a girl, the probability that she wears glasses is conditional. The denominator is the number of girls, not the number of all students.

例如，若从班级中选一名学生，且已知该学生是女生，那么她戴眼镜的概率就是条件概率。分母是女生人数，而不是全班人数。

“Given that” reduces the sample space.
“在……条件下”缩小了样本空间。
The denominator changes to the size of the given condition.
分母变成给定条件对应的事件大小。
Do not use the total number of outcomes in a conditional probability.
在条件概率中，不能使用全部结果的总数。

3. Tree Diagrams: The Basics | 树形图的基础

A tree diagram is a visual way to list all outcomes of a multi-stage experiment. Each branch represents a possible outcome, and the probability written on each branch is the conditional probability of that outcome given the path so far.

树形图是列出多阶段试验所有结果的可视化方法。每条分支代表一种可能结果，分支上标注的概率是从起点到该分支为止的条件概率。

For two stages with two possible outcomes at each stage, the tree has four end branches. Multiply along the branches to find the probability of each combined outcome:

若每个阶段有两种可能结果，则两阶段的树形图有 4 条末端分支。沿分支相乘，即可求出每种组合结果的概率：

P(Path) = P(Stage 1 outcome) × P(Stage 2 outcome | Stage 1 outcome)

Each path probability must be written at the end of the branch. The sum of all path probabilities must equal 1, which is an excellent way to check your work.

每条路径的概率必须标在分支末端。所有路径概率之和必须等于 1，这是检验你是否算对的好方法。

4. Without Replacement: A Classic Example | 无放回：经典例题

Consider a bag containing 3 red balls and 2 blue balls. Two balls are drawn one after another without replacement. What is the probability that both balls are red?

考虑一个袋子中有 3 个红球和 2 个蓝球。不放回地连续抽取两个球。两个球都是红色的概率是多少？

On the first draw, P(Red) = 3/5. After removing one red ball, there are 2 red balls left out of 4 total balls. So the conditional probability P(Red on 2nd | Red on 1st) = 2/4 = 1/2.

第一次抽取时，P(红) = 3/5。拿走一个红球后，剩下 2 个红球，总球数为 4。因此条件概率 P(第二次红 | 第一次红) = 2/4 = 1/2。

P(Both red) = (3/5) × (2/4) = 6/20 = 3/10

Notice that the denominator decreased from 5 to 4 because one ball was removed and not replaced. Tree diagrams make this change visible.

注意分母从 5 变为 4，因为抽走的球没有放回。树形图能让这种变化一目了然。

5. With Replacement: The Comparison | 有放回：对比

If the ball is replaced after the first draw, the probabilities do not change between draws. The second draw is independent of the first. The same example with replacement gives:

若第一次抽取后将球放回，那么两次抽取的概率保持不变，第二次抽取与第一次相互独立。相同例题在有放回的情况下为：

P(Both red) = (3/5) × (3/5) = 9/25

Compare this with the 3/10 = 0.3 obtained without replacement. The without-replacement result is smaller because removing a red ball reduces the proportion of red balls remaining.

与无放回得到的 3/10 = 0.3 对比，无放回的结果更小，因为拿走一个红球后，剩余红球的比例变小了。

Scenario 情景	P(Both red) 两红概率
With replacement 有放回	9/25 = 0.36
Without replacement 无放回	3/10 = 0.3

6. The Multiplication Rule | 乘法法则

Tree diagrams rely on the multiplication rule for probabilities. For any two events A and B:

树形图依据的是概率的乘法法则。对于任意两个事件 A 和 B：

P(A ∩ B) = P(A) × P(B | A)

This rule always holds, whether the events are independent or not. If A and B are independent, then P(B | A) = P(B), and the formula simplifies to P(A ∩ B) = P(A) × P(B).

这一法则总是成立，无论事件是否独立。若 A 与 B 独立，则 P(B | A) = P(B)，公式简化为 P(A ∩ B) = P(A) × P(B)。

In a tree diagram, every path probability is obtained by multiplying the probabilities along the branches. This is why the multiplication rule is central to conditional probability problems.

在树形图中，每条路径的概率都是由沿途分支的概率相乘得到的。这就是乘法法则在条件概率问题中处于核心地位的原因。

7. Independent vs Dependent Events | 独立事件与非独立事件

Two events are independent if the occurrence of one does not affect the probability of the other. For example, tossing a coin twice: the result of the second toss is unaffected by the first.

若一个事件的发生不影响另一个事件发生的概率，则这两个事件是独立的。例如，掷两次硬币：第二次掷出的结果不受第一次影响。

Two events are dependent if the occurrence of one does change the probability of the other. Drawing balls without replacement is a classic dependent-event scenario.

若一个事件的发生会改变另一个事件发生的概率，则这两个事件是相关的。无放回抽取球就是典型的相关事件情景。

Independent: P(B | A) = P(B) — 独立：P(B | A) = P(B)
Dependent: P(B | A) ≠ P(B) — 相关：P(B | A) ≠ P(B)
Check for replacement in the question — 注意题目中是否有”放回”
Without replacement means dependent — 无放回意味着相关

8. Total Probability: Finding P(B) from a Tree | 全概率：从树形图中求 P(B)

A tree diagram can find the probability of a later-stage event by adding all paths that end with that event. For example, the probability that the second ball is red is the sum of the path where the first is red and the path where the first is blue.

树形图可以通过把所有以某事件结束的路径概率相加，来求该事件在较后阶段发生的概率。例如，第二个球为红色的概率，等于第一次为红的路径与第一次为蓝的路径的概率之和。

Using the 3 red, 2 blue bag without replacement:

以 3 红 2 蓝、无放回为例：

P(2nd red) = (3/5)(2/4) + (2/5)(3/4) = 6/20 + 6/20 = 12/20 = 3/5

Interestingly, the probability that the second ball is red equals the probability that the first ball is red. This symmetry holds when the draws follow the same rule.

有趣的是，第二个球为红色的概率等于第一个球为红色的概率。在两次抽取遵循相同规则时，这种对称性成立。

9. Conditional Probability from a Table | 用表格求条件概率

Two-way tables are often used to present data in conditional probability questions. Consider a survey of 50 students: 20 are girls who like maths, 10 are boys who like maths, 8 are girls who do not like maths, and 12 are boys who do not like maths.

双向表格常用于条件概率题中呈现数据。假设对 50 名学生进行调查：20 名喜欢数学的女生，10 名喜欢数学的男生，8 名不喜欢数学的女生，12 名不喜欢数学的男生。

	Like Maths 喜欢数学	Not Like Maths 不喜欢数学	Total 合计
Girls 女生	20	8	28
Boys 男生	10	12	22
Total 合计	30	20	50

To find P(Girl | Likes Maths), restrict attention to the 30 students who like maths. Among them, 20 are girls. So P(Girl | Likes Maths) = 20/30 = 2/3.

求 P(女生 | 喜欢数学) 时，只需关注喜欢数学的 30 名学生。其中有 20 名女生，因此 P(女生 | 喜欢数学) = 20/30 = 2/3。

Always ask yourself: what is the restricted group? The “given” phrase defines the denominator.

始终问自己：限制在哪个群体？”在……条件下”中的条件就定义了分母。

10. Common Traps and Exam Tips | 常见陷阱与考试技巧

Students often make predictable errors in conditional probability questions. One common trap is using the original sample space instead of the reduced one. Another is forgetting to check whether replacement occurs.

学生在条件概率题中常犯可预测的错误。一个常见陷阱是使用原始样本空间而非缩小后的样本空间。另一个是忘记检查题目中是否有放回。

Read the question carefully: with or without replacement? — 仔细读题：有放回还是无放回？
Write down the formula P(A | B) = P(A ∩ B) ÷ P(B) before substituting — 代入前先写出公式
Check that all path probabilities sum to 1 — 检查所有路径概率之和为 1
Cancel fractions early to avoid arithmetic errors — 尽早约分，避免计算错误
Use the table or tree diagram to identify the restricted group — 用表格或树形图确定限制群体

11. Worked Exam-Style Question | 真题风格演练

A bag contains 4 red sweets and 6 yellow sweets. Two sweets are taken at random without replacement. Find: (a) the probability that both sweets are red; (b) the probability that exactly one sweet is red; (c) given that the first sweet is red, the probability that the second is yellow.

一个袋子中有 4 颗红色糖果和 6 颗黄色糖果。不放回地随机取两颗。求：(a) 两颗糖果都为红色的概率；(b) 恰好一颗为红色的概率；(c) 已知第一颗为红色，第二颗为黄色的概率。

(a) P(both red) = (4/10) × (3/9) = 12/90 = 2/15.

(a) P(两红) = (4/10) × (3/9) = 12/90 = 2/15。

(b) P(exactly one red) = P(RY) + P(YR) = (4/10)(6/9) + (6/10)(4/9) = 24/90 + 24/90 = 48/90 = 8/15.

(b) P(恰一红) = P(红黄) + P(黄红) = (4/10)(6/9) + (6/10)(4/9) = 24/90 + 24/90 = 48/90 = 8/15。

(c) Given that the first sweet is red, there are 3 red and 6 yellow sweets remaining, 9 sweets in total. So P(second yellow | first red) = 6/9 = 2/3.

12. Summary and Final Advice | 总结与建议

Conditional probability and tree diagrams are not just about memorising formulas — they are about carefully reducing the sample space and tracking how probabilities change at each stage. With replacement keeps probabilities constant; without replacement changes them.

条件概率与树形图不只是记忆公式，而是要仔细缩小样本空间，并追踪每个阶段概率如何变化。有放回时概率保持不变；无放回时概率会改变。

When you see “given that” in an exam, your first task is to identify the condition, then count or calculate within that reduced group. Practise past-paper questions until this process becomes automatic.

在考试中看到”given that（在……条件下）”时，你的首要任务是确定条件，然后在该缩小后的群体内计数或计算。反复练习真题，直到这个流程变成直觉反应。

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Mastering Conditional Probability & Tree Diagrams | 掌握条件概率与树形图