📚 Conditional Probability using Venn Diagrams | 使用维恩图的条件概率
Conditional probability is one of the most important ideas in IGCSE Mathematics, and Venn diagrams provide one of the clearest visual tools for understanding it. In this revision guide, we will explore how to calculate P(A|B) — the probability of event A occurring given that event B has already occurred — using Venn diagrams, with worked examples and exam-style tips tailored to the Edexcel IGCSE syllabus.
条件概率是 IGCSE 数学中最重要的概念之一，而维恩图是理解条件概率最清晰的视觉工具之一。在本复习指南中，我们将探索如何利用维恩图计算 P(A|B)——即在事件 B 已经发生的前提下事件 A 发生的概率——并配有符合 Edexcel IGCSE 考纲的例题和考试技巧。
1. What is Conditional Probability? | 什么是条件概率？
Conditional probability answers the question: “If we know that one event has already happened, how does that change the probability of another event?” The notation P(A|B) is read as “the probability of A given B”. Crucially, the occurrence of B restricts the set of possible outcomes — we are no longer looking at the whole sample space, but only at the outcomes inside B.
条件概率回答的问题是：”如果我们已知某一事件已经发生，那么另一事件发生的概率会如何改变？”记号 P(A|B) 读作”在 B 发生的条件下 A 的概率”。关键在于，B 的发生缩小了可能结果的集合——我们不再观察整个样本空间，而只关注 B 内部的结果。
For example, if we know that a randomly chosen student plays football, the probability that they also play tennis is a conditional probability. The condition “plays football” restricts our focus to the football set only.
例如，如果我们知道随机选中的一名学生会踢足球，那么他同时也会打网球的概率就是一个条件概率。”会踢足球”这个条件将我们的关注范围限制在足球集合之内。
2. Venn Diagram Notation for Probability | 概率中的维恩图符号
A Venn diagram represents events as circles inside a rectangle. The rectangle is the sample space, with total probability 1. Each circle represents an event, and the overlapping region represents the intersection of two events. Key notation you must master includes P(A) for the probability of A, P(A∩B) for the probability of both A and B occurring, and P(A∪B) for the probability of A or B (or both) occurring.
维恩图用矩形内的圆圈表示事件。矩形是样本空间，总概率为 1。每个圆圈代表一个事件，重叠区域代表两个事件的交集。你必须掌握的关键记号包括：P(A) 表示 A 的概率，P(A∩B) 表示 A 和 B 同时发生的概率，P(A∪B) 表示 A 或 B（或两者）发生的概率。
Each region of the Venn diagram has a specific meaning. If we label the four regions of a two-circle diagram, we have: A only, B only, both A and B, and neither A nor B. These four values always sum to 1 (or to the total number of items if using counts).
维恩图的每个区域都有特定含义。如果我们将两圆的维恩图划分为四个区域，则有：仅 A、仅 B、A 和 B 都发生、A 和 B 都不发生。这四个值之和始终等于 1（若使用频数则表示它们之和等于总数）。
P(A only) + P(B only) + P(A∩B) + P(neither) = 1
仅A + 仅B + P(A∩B) + 两者都不 = 1
3. The Conditional Probability Formula | 条件概率公式
The single most important formula for conditional probability is given below. You must memorise it and understand why it works:
条件概率最重要的公式如下所示。你必须记住它并理解其原理：
P(A|B) = P(A∩B) ÷ P(B), provided P(B) ≠ 0
P(A|B) = P(A∩B) ÷ P(B)，前提是 P(B) ≠ 0
The denominator P(B) reflects the fact that we already know B has occurred, so B becomes our new sample space. The numerator P(A∩B) counts the outcomes where both A and B happen, which is the part of A that lies inside B. The formula rearranges neatly: P(A∩B) = P(A|B) × P(B), which is often used in tree diagrams.
分母 P(B) 反映了一个事实：我们已经知道 B 发生了，因此 B 成为新的样本空间。分子 P(A∩B) 计算 A 和 B 同时发生的结果，即 A 落在 B 内部的部分。该公式可以灵活变形：P(A∩B) = P(A|B) × P(B)，这一形式常用于树状图中。
4. Reading P(A|B) Directly from a Venn Diagram | 直接从维恩图中读出 P(A|B)
When a Venn diagram has probabilities written in each region, you can compute conditional probability simply by comparing two numbers. To find P(A|B), first identify the region inside B: this includes the B-only section plus the intersection. Then identify the portion of that region that also belongs to A — that is exactly the intersection A∩B. Divide the intersection probability by the total probability of B.
当维恩图的每个区域都标有概率时，你可以通过比较两个数字轻松计算条件概率。要求 P(A|B)，首先找出 B 内部的区域：这包括仅 B 部分加上交集部分。然后找出该区域中同时属于 A 的部分——这正是交集 A∩B。用交集的概率除以 B 的总概率即可。
Consider the following Venn diagram with probabilities: the region A only has 0.25, B only has 0.35, both A and B have 0.15, and neither region has 0.25. Here P(B) = 0.15 + 0.35 = 0.50, and P(A∩B) = 0.15. Therefore P(A|B) = 0.15 ÷ 0.50 = 0.30.
考虑如下概率维恩图：仅 A 区域为 0.25，仅 B 区域为 0.35，A 和 B 交集为 0.15，两者都不的区域为 0.25。这里 P(B) = 0.15 + 0.35 = 0.50，P(A∩B) = 0.15。因此 P(A|B) = 0.15 ÷ 0.50 = 0.30。
| B occurs B 发生 |
B does not occur B 不发生 |
Total 合计 |
|
| A occurs A 发生 |
0.15 | 0.25 | 0.40 |
| A does not occur A 不发生 |
0.35 | 0.25 | 0.60 |
| Total 合计 |
0.50 | 0.50 | 1.00 |
The table above shows the same information in a two-way table format. Notice that the row and column totals help you quickly read off P(B) = 0.50 and P(A∩B) = 0.15. In an exam, you may be given either representation, so practice converting between them.
上表以双向表的形式展示了同样的信息。注意行与列的合计可帮助你快速读出 P(B) = 0.50 和 P(A∩B) = 0.15。在考试中，你可能会遇到其中任何一种表示形式，因此要练习在两者之间转换。
5. Worked Example 1: Using Counts in a Venn Diagram | 实例 1：使用维恩图中的频数
A survey of 80 students asked whether they study chemistry and whether they study biology. The results are: 30 study chemistry only, 25 study biology only, 15 study both, and 10 study neither. Find the probability that a randomly chosen student studies chemistry, given that they study biology.
一项针对 80 名学生的调查询问他们是否学习化学以及是否学习生物。结果如下：30 人仅学习化学，25 人仅学习生物，15 人两科都学，10 人两科都不学。求随机选中的一名学生在已知其学习生物的条件下学习化学的概率。
Step 1: Write down the known probabilities by dividing by 80. The Venn diagram probabilities are: P(C only) = 30/80 = 0.375, P(B only) = 25/80 = 0.3125, P(C∩B) = 15/80 = 0.1875, P(neither) = 10/80 = 0.125.
第一步：将频数除以 80 写出已知概率。维恩图概率为：P(仅化学) = 30/80 = 0.375，P(仅生物) = 25/80 = 0.3125，P(化学∩生物) = 15/80 = 0.1875，P(两者都不) = 10/80 = 0.125。
Step 2: Identify the condition. Since we are given that the student studies biology, our new sample space is P(B) = 25/80 + 15/80 = 40/80 = 0.5.
第二步：确定条件。由于已知该学生学习生物，我们的新样本空间为 P(生物) = 25/80 + 15/80 = 40/80 = 0.5。
Step 3: Apply the formula: P(C|B) = P(C∩B) ÷ P(B) = 0.1875 ÷ 0.5 = 0.375 = 3/8.
第三步：套用公式：P(化学|生物) = P(化学∩生物) ÷ P(生物) = 0.1875 ÷ 0.5 = 0.375 = 3/8。
Notice that the answer 0.375 is actually the same as P(C only) in this case because the intersection equals the chemistry-only region — but do not assume this always happens. Always use the formula rather than guessing.
注意在本例中答案 0.375 恰好等于仅化学的概率，因为交集恰好等于仅化学区域——但不要假设这种情况总是成立。务必使用公式，而非凭猜测。
6. Worked Example 2: Interpreting “Given That” with a Completed Diagram | 实例 2：用完整维恩图理解”已知”条件
A Venn diagram shows two events X and Y with probabilities written in each region: P(X only) = 0.2, P(Y only) = 0.35, P(X∩Y) = 0.25, P(neither) = 0.2. Find P(X|Y) and P(Y|X).
一个维恩图显示两个事件 X 和 Y 各区域的概率为：P(仅X) = 0.2，P(仅Y) = 0.35，P(X∩Y) = 0.25，P(两者都不) = 0.2。求 P(X|Y) 和 P(Y|X)。
For P(X|Y), the denominator is P(Y) = 0.25 + 0.35 = 0.60. The numerator is P(X∩Y) = 0.25. Hence P(X|Y) = 0.25 ÷ 0.60 = 5/12 ≈ 0.4167.
对于 P(X|Y)，分母为 P(Y) = 0.25 + 0.35 = 0.60。分子为 P(X∩Y) = 0.25。因此 P(X|Y) = 0.25 ÷ 0.60 = 5/12 ≈ 0.4167。
For P(Y|X), the denominator is P(X) = 0.25 + 0.20 = 0.45. The numerator is still P(X∩Y) = 0.25. Hence P(Y|X) = 0.25 ÷ 0.45 = 5/9 ≈ 0.5556.
对于 P(Y|X)，分母为 P(X) = 0.25 + 0.20 = 0.45。分子仍然是 P(X∩Y) = 0.25。因此 P(Y|X) = 0.25 ÷ 0.45 = 5/9 ≈ 0.5556。
This example illustrates a vital point: P(X|Y) and P(Y|X) are different unless P(X) = P(Y). Always check which event comes after the vertical bar, because that event determines the denominator.
这个例子说明了一个关键点：除非 P(X) = P(Y)，否则 P(X|Y) 和 P(Y|X) 是不同的。始终检查竖线后面是哪个事件，因为该事件决定分母。
7. Using Complements in Conditional Probability | 条件概率中的补事件
Sometimes you need to find the probability of the complement of a condition or the event itself. The complement of A, written A′, represents “not A”. In a Venn diagram, P(A′) = 1 − P(A), which is the total probability outside circle A.
有时你需要求某个条件或事件的补事件的概率。A 的补事件写作 A′，代表”非 A”。在维恩图中，P(A′) = 1 − P(A)，即圆圈 A 之外的总概率。
For example, using the previous X and Y diagram, P(X′|Y) is the probability that X does not occur given that Y occurs. Since the only part of Y that is not in X is the Y-only region, P(X′∩Y) = 0.35. Therefore P(X′|Y) = 0.35 ÷ 0.60 = 7/12 ≈ 0.5833. Note that P(X|Y) + P(X′|Y) = 1, as expected.
例如，使用之前 X 和 Y 的维恩图，P(X′|Y) 表示在 Y 发生的条件下 X 不发生的概率。由于 Y 中不在 X 内的部分只有仅 Y 区域，P(X′∩Y) = 0.35。因此 P(X′|Y) = 0.35 ÷ 0.60 = 7/12 ≈ 0.5833。注意 P(X|Y) + P(X′|Y) = 1，这是符合预期的。
8. Common Mistakes and How to Avoid Them | 常见错误及其避免方法
One of the most frequent errors students make is using the wrong denominator. When asked for P(A|B), some students mistakenly divide P(A∩B) by P(A) instead of P(B). Always ask yourself: “Which event do I already know has happened?” That event’s total probability goes in the denominator.
学生最常犯的错误之一是使用错误的分母。当要求 P(A|B) 时，有些学生错误地将 P(A∩B) 除以 P(A) 而不是 P(B)。始终问自己：”我已经知道哪个事件发生了？”该事件的总概率应放在分母位置。
Another common mistake is forgetting to include the intersection when calculating the total probability of the condition. For example, P(B) is not simply the B-only region — it is B-only plus the intersection A∩B. This error can change your answer significantly.
另一个常见错误是在计算条件事件的总概率时忘记包含交集。例如，P(B) 不仅仅是仅 B 区域——它是仅 B 加上交集 A∩B。这个错误会显著改变你的答案。
A third error is treating “given that” as the same as “and”. The probability of A given B is not the same as P(A∩B). The latter is the probability that both occur without any prior condition, while the former is a relative probability within the restricted space of B.
第三个错误是将”已知”等同于”且”。已知 B 条件下 A 的概率并不等于 P(A∩B)。后者是无任何先决条件时两者同时发生的概率，而前者是在 B 的受限空间内的相对概率。
9. Exam-Style Question with Solution | 考试风格例题及解答
Question: In a school of 200 students, 90 play rugby, 120 play cricket, and 50 play both. A student is chosen at random. (a) Draw a Venn diagram to represent this information. (b) Find the probability that the student plays rugby given that they play cricket. (c) Find the probability that the student plays neither sport.
题目：某学校有 200 名学生，其中 90 人打橄榄球，120 人打板球，50 人两项都打。随机选择一名学生。(a) 用维恩图表示此信息。(b) 求已知该学生打板球的条件下他打橄榄球的概率。(c) 求该学生两项运动都不参加的概率。
Solution (a): Rugby only = 90 − 50 = 40. Cricket only = 120 − 50 = 70. Both = 50. Neither = 200 − 40 − 50 − 70 = 40. The Venn diagram has these four values: 40, 70, 50, 40.
解答 (a)：仅打橄榄球 = 90 − 50 = 40。仅打板球 = 120 − 50 = 70。两项都打 = 50。两项都不打 = 200 − 40 − 50 − 70 = 40。维恩图的四个值为：40、70、50、40。
Solution (b): P(rugby|cricket) = P(rugby∩cricket) ÷ P(cricket). The probability P(cricket) = 120/200 = 0.6, and P(rugby∩cricket) = 50/200 = 0.25. Therefore P(rugby|cricket) = 0.25 ÷ 0.6 = 5/12 ≈ 0.4167.
解答 (b)：P(橄榄球|板球) = P(橄榄球∩板球) ÷ P(板球)。P(板球) = 120/200 = 0.6，P(橄榄球∩板球) = 50/200 = 0.25。因此 P(橄榄球|板球) = 0.25 ÷ 0.6 = 5/12 ≈ 0.4167。
Solution (c): P(neither) = 40/200 = 0.2. This is found by dividing the “neither” region by the total number of students.
解答 (c)：P(两者都不) = 40/200 = 0.2。将”两者都不”区域的频数除以学生总数即可得到。
10. Quick Revision Checklist | 快速复习清单
Before your exam, make sure you can confidently do the following: correctly identify the region representing A∩B in a Venn diagram; write down the conditional probability formula P(A|B) = P(A∩B) ÷ P(B); convert between probability values and counts; determine which event is the condition and therefore the denominator; and calculate P(A′|B) using 1 − P(A|B).
在考试之前，请确保你能自信地完成以下内容：正确识别维恩图中代表 A∩B 的区域；写出条件概率公式 P(A|B) = P(A∩B) ÷ P(B)；在概率值与频数之间转换；确定哪个事件是条件、从而确定分母；以及利用 1 − P(A|B) 计算 P(A′|B)。
Practice drawing Venn diagrams from worded problems. In Edexcel IGCSE papers, the diagram itself is often provided, but you may need to complete it with given numbers or probabilities. Always check that the values in all regions sum to the total or to 1.
练习从文字题目中绘制维恩图。在 Edexcel IGCSE 试卷中，通常会给出图形，但你可能需要用给定的数字或概率完整填写。始终检查所有区域的值之和等于总数或 1。
Finally, remember that conditional probability is not just a formula — it is a logical process. Read “given that” as a clue to change your sample space. With clear diagrams and careful reading, conditional probability questions become straightforward marks.
最后，记住条件概率不仅仅是一个公式——它是一个逻辑过程。将”已知”视为改变样本空间的提示。借助清晰的图形与仔细审题，条件概率题目就能成为稳拿的分数。
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