📚 Conditional Probability | 条件概率
Conditional probability is a fundamental concept in probability theory that deals with the likelihood of an event occurring given that another event has already taken place. This idea is essential for modelling real-world situations where outcomes depend on prior information, such as medical diagnoses, quality control, and strategic decision-making. In the Edexcel A-Level Mathematics syllabus, conditional probability appears across Statistics topics, requiring students to compute probabilities from tree diagrams, Venn diagrams, two-way tables, and to understand the relationships between dependent events.
条件概率是概率论中的一个基本概念，它研究在已知另一事件已经发生的前提下，某事件发生的可能性。这一概念对于建模现实世界中依赖于先验信息的结果至关重要，例如医学诊断、质量控制和策略决策。在Edexcel A-Level数学大纲中，条件概率分布于统计学的多个主题，要求学生能够利用树形图、维恩图、双向表计算概率，并理解相依事件之间的关系。
1. Definition and Notation of Conditional Probability | 条件概率的定义与记号
Conditional probability is written as P(A|B) and read as "the probability of A given B". It represents the chance that event A occurs, knowing that event B has already occurred. The vertical bar ‘|’ indicates that the occurrence of B is treated as a condition, effectively reducing the sample space to only those outcomes in which B happens.
条件概率记作 P(A|B)，读作“在B发生的条件下A发生的概率”。它表示在已知事件B已经发生的情况下，事件A发生的可能性。中间的竖线‘|’表示B的发生被看作一个条件，从而将样本空间有效地缩小到仅包含那些B发生的结果。
For example, if we roll a fair six-sided die, let A be the event “rolling a 4” and B be the event “rolling an even number”. The unconditional probability P(A) is 1/6, but if we already know that B has occurred (the outcome is even), the conditional probability P(A|B) becomes 1/3, because among the three even outcomes {2,4,6}, only one is a 4.
例如，抛掷一枚均匀的六面骰子，设A为“掷出4”的事件，B为“掷出偶数”的事件。无条件概率 P(A) 为1/6，但若我们已知B已经发生（结果为偶数），则条件概率 P(A|B) 变为 1/3，因为在三个偶数结果{2,4,6}中，只有一个是4。
2. The Formula for Conditional Probability | 条件概率公式
The formal definition of conditional probability is given by the formula
P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0.
条件概率的正式定义由以下公式给出
P(A|B) = P(A ∩ B) / P(B), 其中P(B) > 0。
Here P(A ∩ onal probability is simply the proportion of the probability of B that is also shared with A. When probabilities are expressed as frequencies, you can also write P(A|B) = n(A ∩ B) / n(B), where n denotes the number of equally likely outcomes.
这里 P(A ∩ B) 是A和B同时发生的概率。分母 P(B) 是条件事件的概率。这个公式表明，条件概率不过是B的概率中与A共享的那一部分所占的比例。当概率以频率表示时，也可写成 P(A|B) = n(A ∩ B) / n(B)，其中 n 表示等可能结果的个数。
It is essential to ensure that P(B) is not zero; otherwise, the conditional probability is undefined. In exam questions, the event B is often clearly given, but you should always check that a non-zero probability is provided or implied.
必须确保 P(B) 不等于零；否则条件概率无定义。在考试题目中，事件B通常会明确给出，但你应该始终检查是否提供了或隐含了非零概率。
3. The Multiplication Law | 乘法法则
Rearranging the conditional probability formula gives the multiplication law, which is used to find the probability that two events both occur:
P(A ∩ B) = P(B) × P(A|B) or equivalently P(A ∩ B) = P(A) × P(B|A).
将条件概率公式变形即得到乘法法则，用于求两个事件同时发生的概率：
P(A ∩ B) = P(B) × P(A|B) 或等价地 P(A ∩ B) = P(A) × P(B|A)。
multiplication law also makes clear that if A and B are independent, it simplifies to P(A ∩ B) = P(A) × P(B).
当概率沿着树形图的分支给出时，这一定律特别有用。例如，若事件A的概率为0.4，且在A发生的条件下B的概率为0.7，则A和B同时发生的概率为 0.4 × 0.7 = 0.28。乘法法则也清楚地表明，如果A与B独立，公式简化为 P(A ∩ B) = P(A) × P(B)。
4. Independent Events and Conditional Probability | 独立事件与条件概率
Two events A and B are independent if the occurrence of one does not affect the probability of the other. In terms of conditional probability, this means
P(A|B) = P(A) and P(B|A) = P(B).
如果两个事件A与B互不影响，即一个事件的发生不影响另一个事件的概率，则称它们为独立事件。用条件概率表达，即为
P(A|B) = P(A) 以及 P(B|A) = P(B)。
When either of these equalities holds, the multiplication rule P(A ∩ Edexcel exam questions, you may be asked to test for independence by checking whether P(A ∩ rresponding unconditional probability. Always state your conclusion clearly, for instance, “Since P(A|B) = P(A), A and B are independent.”
当这些等式之一成立时，乘法规则 P(A ∩ B) = P(A) × P(B) 即成立。在Edexcel的考试题中，可能会要求你通过检查 P(A ∩ B) 是否等于 P(A) × P(B)，或通过验证条件概率是否等于相应的无条件概率，来检验独立性。务必清晰陈述结论，例如：“因为 P(A|B) = P(A)，所以A与B独立。”
5. Conditional Probability on Tree Diagrams | 树形图与条件概率
nd 5 blue marbles. Two marbles are drawn without replacement. The first branch shows P(Red₁) = 3/8 and P(Blue₁) = 5/8. The second-level branch for “2nd is red” given “1st is red” is P(Red₂|Red₁) = 2/7, because there are now only 2 reds left out of 7 marbles.
树形图是A-Level统计学中处理事件序列的标准工具。每组分支表示条件概率：第二层分支上的概率是给定第一阶段结果的条件概率。例如，一个袋子里有3颗红色和5颗蓝色弹珠，不放回地抽取两次。第一分支显示 P(Red₁) = 3/8 和 P(Blue₁) = 5/8。第二层“第二次为红色”在“第一次为红色”条件下的分支概率为 P(Red₂|Red₁) = 2/7，因为此时7颗弹珠中只剩2颗红色。
To find the conditional probability of an earlier event given a later one, say P(Red₁| se the multiplication law together with the law of total probability. First compute P(Red₁ ∩ Blue₂) = P(Red₁) × P(Blue₂| 15/56) and from Blue₁ then Blue₂ (5/8 × 4/7 = 20/56). So P(Blue₂) = 35/56. Finally, P(Red₁|Blue₂) = (15/56) / (35/56) = 15/35 = 3/7. This technique is tested regularly.
要求由后一事件推前一事件的条件概率，例如 P(Red₁|Blue₂)，可以结合使用乘法法则和全概率公式。首先计算 P(Red₁ ∩ Blue₂) = P(Red₁) × P(Blue₂|Red₁) = 3/8 × 5/7 = 15/56。然后通过求所有以 Blue₂ 结尾的路径概率之和得到 P(Blue₂)：从 Red₁ 再到 Blue₂ (15/56) 加上从 Blue₁ 再到 Blue₂ (5/8 × 4/7 = 20/56)。因此 P(Blue₂) = 35/56。最后 P(Red₁|Blue₂) = (15/56) / (35/56) = 15/35 = 3/7。这种技巧在考试中经常出现。
6. Conditional Probability on Venn Diagrams | 维恩图与条件概率
bility. Suppose a Venn diagram shows that for events A and B within a sample space of 50 equally likely outcomes, n(A ∩ B) = 8, n(B) = 20. Then
P(A|B) = n(A ∩ B) / n(B) = 8/20 = 0.4.
维恩图以图形化方式表示集合及其交集，便于提取计算条件概率所需的频数。假设一个维恩图显示，在包含50个等可能结果的样本空间中，事件A与B满足 n(A ∩ B) = 8, n(B) = 20。则
P(A|B) = n(A ∩ B) / n(B) = 8/20 = 0.4。
t by the sum of all probabilities that make up B. For example, if a Venn diagram shows P(A ∩ B) = 0.15 and P(B) = 0.45, then P(A|B) = 0.15/0.45 = 1/3. Always identify the conditioning event’s total probability first – this denominator includes all parts of B, even those not in A.
当概率直接标在区域上时，你只需取交集区域的概率，并除以构成B的所有概率之和。例如，若维恩图显示 P(A ∩ B) = 0.15, P(B) = 0.45，则 P(A|B) = 0.15/0.45 = 1/3。务必先确定条件事件的总概率——这个分母包含B的所有部分，即使是那些不在A中的部分。
7. Two-Way Tables and Conditional Probability | 双向表与条件概率
Two-way tables (contingency tables) display joint frequencies for two categorical variables. They are extremely useful for calculating conditional probabilities directly. Consider the following table showing the number of students who study Mathematics and Physics:
双向表（列联表）展示两个分类变量的联合频数。它们对于直接计算条件概率极为有用。考虑下表展示学习数学与物理的学生人数：
| Physics Yes | Physics No | Total | |
| Maths Yes | 30 | 15 | 45 |
| Maths No | 10 | 25 | 35 |
| Total | 40 | 40 | 80 |
To find P(Physics Yes | Maths Yes), restrict your attention to the ‘ take Physics is 30, so the conditional probability is 30/45 = 2/3. Similarly, P(Maths Yes | Physics No) looks at the ‘ udents, giving 15/40 = 3/8. Always use the correct row or column total as the denominator.
求 P(Physics Yes | Maths Yes) 时，只需关注‘Maths Yes’这一行（45名学生）。这些学生中同时选修物理的人数为30，因此条件概率为 30/45 = 2/3。类似地，P(Maths Yes | Physics No) 则看‘Physics No’列（40名学生）中有15名数学学生，得 15/40 = 3/8。务必以正确的行合计或列合计作为分母。
8. The Law of Total Probability | 全概率公式
When the sample space can be partitioned into a set of mutually exclusive and exhaustive events B₁, B₂, …, Bₙ, the probability of any event A can be expressed as a weighted sum of conditional probabilities:
P(A) = Σ P(Bᵢ) × P(A|Bᵢ).
当样本空间可被划分成一组互斥且完备的事件 B₁, B₂, …, Bₙ 时，任何事件A的概率都可以表示为条件概率的加权和：
P(A) = Σ P(Bᵢ) × P(A|Bᵢ)。
inator for reverse conditional probabilities. In a typical problem, the events Bᵢ might be “having a disease” and “not having the disease”, while A could be “testing positive”. You compute P(A) by considering all paths that lead to A: P(Disease) × P(Positive|Disease) + P(No Disease) × P(Positive| The result then feeds into Bayes’ theorem or direct conditional probability calculations.
在求取反向条件概率的分母时，该公式不可或缺。典型问题中，事件 Bᵢ 可能是“患病”和“未患病”，而A可能是“检测呈阳性”。通过考虑所有通往A的路径来计算 P(A)：P(患病) × P(阳性|患病) + P(未患病) × P(阳性|未患病)。所得结果随后用于贝叶斯定理或直接的条件概率计算。
9. Bayes’ Theorem – Reversing the Condition | 贝叶斯定理——条件的逆转
Bayes’ theorem provides a systematic way to “reverse” conditional probabilities. It states
P(B|A) = [ P(A|B) × P(B) ] / P(A),
r example, in a disease testing scenario with prevalence P(D) = 0.01, test sensitivity P(+|D) = 0.95, and specificity P(-|D' find the probability that a person actually has the disease given a positive test result.
贝叶斯定理提供了系统“逆转”条件概率的方法。它表示为
P(B|A) = [ P(A|B) × P(B) ] / P(A)，
其中 P(A) 可运用全概率公式展开。虽然Edexcel A-Level不总要求该定理的名称，但逆转条件的方法却经常考查。例如，在疾病检测情景中，患病率 P(D) = 0.01，检测灵敏度 P(+|D) = 0.95，特异度 P(-|D') = 0.90，你可以求出一个已知检测呈阳性的人实际患病的概率。
First, find the probability of a positive result P(+) = P(D) × P(+|D) + P(D') × P(+|D') = 0.01×0.95 + 0.99×0.10 = 0.0095 + 0.099 = 0.1085. Then, by Bayes, P(D|+) = (0.0095) / 0.1085 ≈ 0.0876. Surprisingly, only about 8.8% of those testing positive truly have the disease – -intuitive result that highlights the importance of understanding conditional probability.
首先，求阳性结果概率 P(+) = P(D) × P(+|D) + P(D') × P(+|D') = 0.01×0.95 + 0.99×0.10 = 0.0095 + 0.099 = 0.1085。接着，根据贝叶斯定理，P(D|+) = (0.0095) / 0.1085 ≈ 0.0876。令人惊讶的是，检测呈阳性的人中仅有约8.8%真正患病——这个反直觉的结果凸显了理解条件概率的重要性。
10. Worked Example – Exam-Style Question | 例题解析——考试风格问题
Problem: A factory has two machines, X and Y. Machine X produces 60% of the items, and machine Y produces 40%. Of the items produced by X, 2% are defective; of those produced by Y, 5% are defective. An item is selected at random and found to be defective. What is the probability that it was produced by machine X?
问题：一家工厂有两台机器X和Y。机器X生产60%的产品，机器Y生产40%。X生产的产品中有2%为次品；Y生产的产品中有5%为次品。随机抽取一件产品，发现是次品。求它是由机器
Published by TutorHao | A-Level 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