Bayes' Theorem: Derivation and Applications — 贝叶斯定理：公式推导与应用

📚 Bayes’ Theorem: Derivation and Applications | 贝叶斯定理：公式推导与应用

中文：在 IB 数学（IB Mathematics AA 与 AI 均覆盖）的概率与统计部分中，条件概率是核心概念之一，而贝叶斯定理（Bayes’ theorem）则是建立在条件概率之上的最重要结论。它回答了一个极具实用价值的问题：当我们观察到某个结果已经发生时，如何反过来更新我们对”原因”概率的判断？医学检测、垃圾邮件过滤、法庭证据评估、金融风险控制等领域都离不开它。本文将从条件概率的定义出发，一步步完成贝叶斯定理的公式推导，讲解每个符号的含义，并结合 IB 考试的典型题型与真实应用场景，帮助你彻底掌握这个考点。

English: In IB Mathematics (covered by both AA and AI), conditional probability is one of the core ideas in the Probability and Statistics strand, and Bayes’ theorem is the most important result built on top of it. It answers a highly practical question: once we observe that an outcome has occurred, how do we update our judgement about the probability of its cause? Medical testing, spam filtering, courtroom evidence evaluation, and financial risk control all rely on it. In this article we start from the definition of conditional probability, derive Bayes’ theorem step by step, explain what every symbol means, and work through typical IB exam questions and real-world applications.

1. 条件概率的定义与乘法法则 | The Definition of Conditional Probability and the Multiplication Rule

English: The conditional probability P(A|B) reads “the probability of A given B”: the probability that event A occurs under the condition that event B has already occurred. It is defined as P(A|B) = P(A∩B) / P(B), provided P(B) > 0. The vertical bar is read as “given” — it shrinks the sample space from the whole Ω down to the subset B. Multiplying both sides of the definition by P(B) gives the Multiplication Rule immediately: P(A∩B) = P(A) × P(B|A) = P(B) × P(A|B). This rule is the only tool we need to derive Bayes’ theorem.

中文：举一个 IB 例题中常见的场景：某班有 30 名学生，其中 18 人选了数学 AA，12 人选了数学 AI；在选了 AA 的学生中，有 6 人同时选了物理 HL。设 A = “选数学 AA”，B = “选物理 HL”。那么 P(A) = 18/30 = 0.6，P(B|A) = 6/18 = 1/3，于是 P(A∩B) = 0.6 × 1/3 = 0.2，即全班 20% 的学生同时选了 AA 和物理 HL。这个例子展示了乘法法则如何把”先发生的事件概率”与”条件下一步概率”相乘，得到”两者同时发生”的概率。

English: Consider a typical IB-style example: a class of 30 students, 18 taking Mathematics AA and 12 taking Mathematics AI; among the AA students, 6 also take Physics HL. Let A = “takes Mathematics AA” and B = “takes Physics HL”. Then P(A) = 18/30 = 0.6 and P(B|A) = 6/18 = 1/3, so P(A∩B) = 0.6 × 1/3 = 0.2 — that is, 20% of the class take both AA and Physics HL. This shows how the multiplication rule combines the probability of the earlier event with the conditional probability of the next step to give the probability that both happen.

2. 从乘法法则出发：贝叶斯定理的推导 | Deriving Bayes’ Theorem from the Multiplication Rule

English: The problem Bayes’ theorem solves is reversing a conditional probability: given P(B|A), how do we find P(A|B)? The derivation takes only three lines. First, write down the definition of P(A|B): P(A|B) = P(A∩B) / P(B). Second, replace the numerator using the multiplication rule: P(A∩B) = P(B|A) × P(A). Third, substitute and rearrange to obtain Bayes’ theorem: P(A|B) = P(B|A) × P(A) / P(B). The core idea of the whole derivation is one sentence: the two conditional probabilities are connected through the bridge of the intersection probability, and the multiplication rule is that bridge.

中文：从树状图的角度理解会更直观。一棵两步概率树的第一层分支是 A 和 A’（A 不发生），第二层分支是 B 和 B’。沿”先 A 再 B”的路径走，概率是 P(A) × P(B|A)，这正是 P(A∩B)。而贝叶斯定理求的 P(A|B)，本质上是在问：在所有到达 B 的路径中（P(B)），经过 A 的那条路径（P(A∩B)）占了多少比例？所以贝叶斯定理也可以写成”占比”的形式：P(A|B) = 经过 A 到达 B 的路径概率 / 所有到达 B 的路径概率之和。这个”路径占比”的视角对解决 IB 题目非常有用。

English: A tree diagram makes this intuitive. The first layer of a two-step probability tree branches into A and A’ (not A), and the second layer branches into B and B’. Walking along the path “A first, then B” gives probability P(A) × P(B|A), which is exactly P(A∩B). What Bayes’ theorem asks for, P(A|B), is essentially: among all paths that reach B, what fraction go through A? So Bayes’ theorem can be rewritten as a fraction of path weights: P(A|B) = (probability of paths reaching B via A) / (sum of probabilities of all paths reaching B). This “fraction of paths” viewpoint is extremely useful for solving IB questions.

3. 贝叶斯定理的标准形式与各部分含义 | The Standard Form of Bayes’ Theorem and the Meaning of Each Term

中文：IB 公式手册（Formula Booklet）中给出的标准形式是：P(A|B) = P(B|A) × P(A) / P(B)。公式里的每个部分都有专门的名字，理解这些名字有助于记忆和应用。P(A) 称为先验概率（Prior Probability），是观察到任何新证据之前，我们对事件 A 原本的判断；P(B|A) 称为似然（Likelihood），表示在 A 为真的条件下观察到 B 的可能性；P(B) 称为证据（Evidence），是观察到 B 的总概率；P(A|B) 称为后验概率（Posterior Probability），是结合了证据 B 之后修正过的判断。贝叶斯定理的哲学含义是：后验 ∝ 似然 × 先验。

English: The standard form given in the IB Formula Booklet is: P(A|B) = P(B|A) × P(A) / P(B). Each part of the formula has a dedicated name, and knowing these names helps with memorising and applying it. P(A) is the prior probability: our original judgement about A before seeing any new evidence. P(B|A) is the likelihood: the chance of observing B given that A is true. P(B) is the evidence: the total probability of observing B. P(A|B) is the posterior probability: the updated judgement after incorporating the evidence B. The philosophical meaning of Bayes’ theorem is: posterior ∝ likelihood × prior.

中文：下表总结了四个术语及其对应符号，方便考前快速复习：

English: The table below summarises the four terms and their symbols for quick revision before exams:

术语 Term	符号 Symbol	含义 Meaning
先验概率 Prior	P(A)	证据前的判断 Judgement before evidence
似然 Likelihood	P(B\|A)	A 为真时观察到 B 的概率 Chance of B if A is true
证据 Evidence	P(B)	观察到 B 的总概率 Total probability of B
后验概率 Posterior	P(A\|B)	证据后的修正判断 Updated judgement

4. 全概率公式：分母的秘密 | The Law of Total Probability: The Secret of the Denominator

中文：在实际计算中，P(B) 往往不会直接给出，需要用全概率公式（Law of Total Probability）来求。当 A 与 A’ 构成样本空间的一个完备划分时（A ∪ A’ = Ω 且 A ∩ A’ = ∅），有 P(B) = P(B|A) × P(A) + P(B|A’) × P(A’)。直观理解：B 发生的总概率，等于”经过 A 到达 B”与”经过 A’ 到达 B”两条路径概率之和。若划分成 n 个互斥事件 A₁, A₂, …, Aₙ，一般形式为 P(B) = Σ P(B|Aᵢ) × P(Aᵢ)。把全概率公式代入贝叶斯定理，就得到完整形式：P(A|B) = P(B|A) × P(A) / [P(B|A) × P(A) + P(B|A’) × P(A’)]。

English: In real calculations P(B) is usually not given directly — it must be found using the Law of Total Probability. When A and A’ form a complete partition of the sample space (A ∪ A’ = Ω and A ∩ A’ = ∅), we have P(B) = P(B|A) × P(A) + P(B|A’) × P(A’). Intuitively: the total probability of B equals the sum of the probabilities of the two paths “to B via A” and “to B via A’”. If the partition has n mutually exclusive events A₁, A₂, …, Aₙ, the general form is P(B) = Σ P(B|Aᵢ) × P(Aᵢ). Substituting the law of total probability into Bayes’ theorem gives the complete form: P(A|B) = P(B|A) × P(A) / [P(B|A) × P(A) + P(B|A’) × P(A’)].

中文：IB 考试中，最常见的出题方式就是”套完整形式”：题目给出两个互斥原因（如两台机器、两个工厂、两种人群），再给出各自的生产占比与缺陷率，最后问”已知一件产品有缺陷，它来自机器甲的概率”。这种题目本质上就是让你填完整形式中的每一项。建议先画树状图把四个概率 P(A)、P(A’)、P(B|A)、P(B|A’) 标出来，再代入公式，可以显著降低出错率。

English: In IB exams, the most common setup is “apply the complete form”: the question gives two mutually exclusive causes (two machines, two factories, two population groups), together with their production shares and defect rates, and then asks “given that a product is defective, what is the probability it came from machine A?”. Such questions are essentially asking you to fill in every term of the complete form. A recommended strategy is to draw a tree diagram, label the four probabilities P(A), P(A’), P(B|A), P(B|A’), and only then substitute into the formula — this dramatically reduces errors.

5. IB 考试中的三大典型题型 | Three Typical Question Types in IB Exams

中文：根据近年 IB 真题（Paper 1 与 Paper 2 均出现过），贝叶斯定理相关题目主要有三种考法。题型一：树状图法。题目直接给出两步概率树，要求先补全树枝上的概率，再求 P(A|B)，此时只需计算”经过 A 到达 B 的路径权重 ÷ 到达 B 的总权重”。题型二：表格法。题目给出一张 2×2 联列表（例如按”是否生病”与”检测是否阳性”分类的频数表），要求先算边际概率再套公式。题型三：纯公式应用。不给背景，直接给 P(A)、P(B|A)、P(B|A’)，要求用完整形式求 P(A|B)，主要考查对公式结构的记忆。

English: According to recent IB papers (both Paper 1 and Paper 2), questions on Bayes’ theorem come in three main styles. Type 1: tree diagram. The question provides a two-step probability tree and asks you to complete the probabilities on the branches, then find P(A|B) by computing (path weight reaching B via A) ÷ (total weight reaching B). Type 2: table. The question gives a 2×2 contingency table (for example, counts classified by “disease status” and “test result”), and you compute marginal probabilities first and then apply the formula. Type 3: direct formula. No story context — you are given P(A), P(B|A) and P(B|A’) and must use the complete form to find P(A|B); this mainly tests memory of the formula structure.

中文：下面用一个完整的例题演示题型三的完整解法。设工厂有甲、乙两台机器，产量分别占 60% 和 40%；甲机器生产的产品次品率为 2%，乙机器为 5%。随机抽取一件产品，发现它是次品，求它来自甲机器的概率。设 A = “来自甲机器”，B = “是次品”。则 P(A) = 0.6，P(A’) = 0.4，P(B|A) = 0.02，P(B|A’) = 0.05。代入完整形式：P(A|B) = (0.02 × 0.6) / (0.02 × 0.6 + 0.05 × 0.4) = 0.012 / 0.032 = 0.375。即：虽然甲机器产量占 60%，但已知次品来自甲的概率只有 37.5%，因为甲的次品率更低。这个”反直觉”的结论正是贝叶斯定理的价值所在。

English: Here is a complete example demonstrating Type 3. A factory has two machines, A and B (call them Jia and Yi). Jia produces 60% of the items with a 2% defect rate; Yi produces 40% with a 5% defect rate. A randomly selected item is found defective: what is the probability it came from Jia? Let A = “came from Jia” and D = “is defective”. Then P(A) = 0.6, P(A’) = 0.4, P(D|A) = 0.02, P(D|A’) = 0.05. Substituting into the complete form: P(A|D) = (0.02 × 0.6) / (0.02 × 0.6 + 0.05 × 0.4) = 0.012 / 0.032 = 0.375. So even though Jia produces 60% of all items, given a defective item the probability it came from Jia is only 37.5%, because Jia’s defect rate is lower. This counter-intuitive conclusion is exactly the value of Bayes’ theorem.

6. 应用一：医学检测中的假阳性 | Application 1: False Positives in Medical Testing

中文：贝叶斯定理最经典的案例是医学检测。设某种罕见病在人群中的患病率为 1%（即先验 P(D) = 0.01），某检测方法的灵敏度（Sensitivity，患病者检出阳性的概率 P(+|D)）为 95%，特异度（Specificity，健康者检出阴性的概率 P(−|H)）为 90%，即误报率 P(+|H) = 10%。现在一个人检测结果为阳性，他真正患病的概率 P(D|+) 是多少？代入完整形式：P(D|+) = (0.95 × 0.01) / (0.95 × 0.01 + 0.10 × 0.99) = 0.0095 / 0.1085 ≈ 0.0876，约为 8.8%。

English: The classic case study for Bayes’ theorem is medical testing. Suppose a rare disease has a prevalence of 1% in the population (prior P(D) = 0.01). A test has sensitivity 95% (the probability of a positive result given disease, P(+|D)) and specificity 90% (the probability of a negative result given healthy, P(−|H)), so the false positive rate is P(+|H) = 10%. If a person tests positive, what is the probability they actually have the disease, P(D|+)? Substituting into the complete form: P(D|+) = (0.95 × 0.01) / (0.95 × 0.01 + 0.10 × 0.99) = 0.0095 / 0.1085 ≈ 0.0876, about 8.8%.

中文：这个结果非常反直觉：一个灵敏度高达 95% 的检测，阳性结果对应的真实患病率居然不到 9%。原因是患病率太低（1%），健康人群基数太大，即使误报率只有 10%，也会产生大量假阳性，把真阳性”稀释”了。这也解释了为什么临床上对阳性结果通常还要做二次确认（如更精确的复查），以及为什么医生强调”检测阳性 ≠ 确诊”。IB 考试中常把这类场景改编为选择题或简答题，考察你是否会用完整形式而不是只看灵敏度。

English: This result is highly counter-intuitive: a test with 95% sensitivity yields a true disease probability of under 9% for a positive result. The reason is that the prevalence is very low (1%), so the healthy population is huge; even a 10% false positive rate produces a large number of false positives that “dilute” the true positives. This explains why in clinical practice positive results are usually confirmed by a second, more accurate test, and why doctors emphasise that “a positive test does not mean a diagnosis”. IB exams often adapt this scenario into multiple-choice or short-answer questions to check whether you use the complete form rather than just quoting the sensitivity.

7. 应用二：垃圾邮件过滤与朴素贝叶斯 | Application 2: Spam Filtering and Naive Bayes

中文：另一个重要的应用是垃圾邮件过滤。设某邮箱收到的邮件中 20% 是垃圾邮件，即 P(S) = 0.2。统计发现，垃圾邮件中有 8% 包含单词”free”，正常邮件中只有 1% 包含该词，即 P(free|S) = 0.08，P(free|H) = 0.01。那么一封包含”free”的邮件是垃圾邮件的概率 P(S|free) 为：P(S|free) = (0.08 × 0.2) / (0.08 × 0.2 + 0.01 × 0.8) = 0.016 / 0.024 ≈ 0.667，约 2/3。也就是说，看到”free”这个词后，该邮件是垃圾邮件的概率从先验的 20% 跃升到约 67%。

English: Another important application is spam filtering. Suppose 20% of the emails received by an account are spam, so P(S) = 0.2. Statistics show that 8% of spam emails contain the word “free”, while only 1% of normal (ham) emails contain it: P(free|S) = 0.08 and P(free|H) = 0.01. Then the probability that an email containing “free” is spam is P(S|free) = (0.08 × 0.2) / (0.08 × 0.2 + 0.01 × 0.8) = 0.016 / 0.024 ≈ 0.667, about 2/3. In other words, after seeing the word “free”, the probability that the email is spam jumps from the prior of 20% to roughly 67%.

中文：真实的垃圾邮件过滤器会把成千上万个词的概率相乘（假设各词条件独立，即”朴素”贝叶斯 Naive Bayes 的含义），综合判断一封邮件是垃圾邮件的概率，超过阈值就拦截。虽然”词之间条件独立”的假设在严格意义上并不成立，但朴素贝叶斯在实践中依然表现得非常好，是机器学习中最简单有效的分类器之一。IB 数学 AI（Applications and Interpretation）课程会涉及这类”从数据估计概率并用贝叶斯思想更新判断”的建模思维，理解这个例子对 Paper 2 的应用题很有帮助。

English: Real spam filters multiply the probabilities of thousands of words (assuming the words are conditionally independent — that is what “Naive” Bayes means) to decide whether an email is spam, blocking it if the probability exceeds a threshold. Although the conditional independence assumption is not strictly true, naive Bayes still performs remarkably well in practice and is one of the simplest and most effective classifiers in machine learning. IB Mathematics AI (Applications and Interpretation) touches on this kind of modelling thinking — estimating probabilities from data and updating judgements with Bayesian reasoning — so understanding this example helps with Paper 2 applied questions.

中文：说到贝叶斯定理的经典应用，不得不提蒙提霍尔问题（Monty Hall Problem）。三扇门中只有一扇后面有奖品，你先选了一扇（假设 1 号门），主持人知道奖品位置并打开另一扇没有奖品的门（假设 3 号门），然后问你是否换到剩下那扇门（2 号门）。直觉上换不换似乎一样，都是 1/2，但贝叶斯分析表明换门的中奖概率是 2/3。设 A₁ = “奖品在 1 号门”、A₂ = “奖品在 2 号门”、A₃ = “奖品在 3 号门”，B = “主持人打开 3 号门”。先验 P(A₁) = P(A₂) = P(A₃) = 1/3。关键在于似然：若奖品在 1 号门，主持人可开 2 号或 3 号门，P(B|A₁) = 1/2；若奖品在 2 号门，主持人只能开 3 号门，P(B|A₂) = 1；若奖品在 3 号门，主持人不能打开它，P(B|A₃) = 0。代入公式：P(A₂|B) = (1 × 1/3) / (1/2 × 1/3 + 1 × 1/3 + 0) = 2/3。这正是”换门更优”的数学解释，也是 IB 拓展讨论（AA 的证明题与 AI 的建模题）中常见的素材。

English: When it comes to classic applications of Bayes’ theorem, the Monty Hall Problem must be mentioned. Behind one of three doors is a prize. You pick a door (say door 1); the host, who knows where the prize is, opens another door with no prize (say door 3) and asks whether you want to switch to the remaining door (door 2). Intuitively switching seems to make no difference — 1/2 either way — but a Bayesian analysis shows that switching wins with probability 2/3. Let A₁ = “prize behind door 1”, A₂ = “prize behind door 2”, A₃ = “prize behind door 3”, and B = “host opens door 3”. The priors are P(A₁) = P(A₂) = P(A₃) = 1/3. The key lies in the likelihoods: if the prize is behind door 1, the host may open door 2 or door 3, so P(B|A₁) = 1/2; if it is behind door 2, the host must open door 3, so P(B|A₂) = 1; if it is behind door 3, the host cannot open it, so P(B|A₃) = 0. Substituting: P(A₂|B) = (1 × 1/3) / (1/2 × 1/3 + 1 × 1/3 + 0) = 2/3. This is the mathematical explanation of why switching is better, and a common discussion topic for IB extension work (proof questions in AA and modelling questions in AI).

8. 常见错误与避坑指南 | Common Mistakes and How to Avoid Them

中文：错误一：混淆方向。把 P(A|B) 与 P(B|A) 混为一谈，这是最典型的错误，在法庭证据领域被称为”检方谬误”（Prosecutor’s Fallacy）。例如”检测阳性的人中有 8.8% 患病”与”患病的人中有 95% 检测阳性”是两个完全不同的命题，二者不能互换。错误二：分母算错。忘记用全概率公式展开 P(B)，只把 P(B|A) × P(A) 当作分母，导致后验概率大于 1 这种明显荒谬的结果。错误三：忽略先验。在题目明确给出 P(A) 的情况下仍把它当作 1/2，或把两个互斥原因的概率当成相等，导致答案偏离正确值。

English: Mistake 1: confusing the direction. Treating P(A|B) as the same as P(B|A) is the most typical error; in the field of legal evidence it is known as the prosecutor’s fallacy. “8.8% of people who test positive actually have the disease” and “95% of people with the disease test positive” are two completely different statements and cannot be interchanged. Mistake 2: getting the denominator wrong. Forgetting to expand P(B) using the law of total probability and using only P(B|A) × P(A) as the denominator can produce absurd results such as a posterior probability greater than 1. Mistake 3: ignoring the prior. When the question explicitly gives P(A), treating it as 1/2, or assuming the two mutually exclusive causes are equally likely, will push your answer away from the correct value.

中文：避坑建议：第一，动笔前先明确”A 是什么、B 是什么”，把事件用中文或符号写出来，防止方向混淆。第二，凡是用到贝叶斯定理，就检查分母是否等于”所有到达 B 的路径之和”；写完整形式时把分母中每一项与树状图的树枝一一对应。第三，算完后做合理性检验：后验概率必须介于 0 和 1 之间，而且应当介于先验与 1 之间（新证据只会把你从先验推向更确定的结论，不会反向）。这三条检查可以在 30 秒内完成，却能挡住绝大部分失分。

English: Advice for avoiding these traps: first, before writing anything, clarify what A and B are — write the events in symbols or words so that you never confuse the direction. Second, whenever you use Bayes’ theorem, check that the denominator equals “the sum of all paths reaching B”; when writing the complete form, match each term of the denominator to a branch of the tree diagram. Third, after computing, do a sanity check: the posterior must lie between 0 and 1, and it should lie between the prior and 1 (new evidence only pushes you from the prior towards a more certain conclusion, never in the opposite direction). These three checks take 30 seconds and prevent most lost marks.

9. 巩固练习：三道真题风格题目 | Practice Problems: Three Exam-Style Questions

中文：练习一（题型二/表格法）：某学校调查 200 名学生，其中 120 人参加课外补习，80 人不参加；参加补习的学生中 90 人在期末数学考试中取得 7 分及以上，不参加的学生中 30 人取得 7 分及以上。随机选一名取得 7 分及以上的学生，求他参加过补习的概率。提示：先补全 2×2 表格，再套完整形式，答案应为 0.75。

English: Problem 1 (Type 2, table method): A school surveys 200 students; 120 attend after-school tuition and 80 do not. Among those who attend, 90 scored 7 or above in the final maths exam; among those who do not attend, 30 scored 7 or above. A student who scored 7 or above is chosen at random: find the probability that they attended tuition. Hint: complete the 2×2 table first, then apply the complete form — the answer is 0.75.

中文：练习二（题型三/纯公式）：某地区有 1% 的人携带某种基因，检测该基因的准确率为：携带者检出阳性概率 98%，非携带者误报阳性概率 5%。求检测为阳性的人真正携带该基因的概率。答案约为 16.5%。这个结果再次提醒我们：低患病率场景下，阳性结果需要谨慎解读。

English: Problem 2 (Type 3, direct formula): In a certain region 1% of people carry a particular gene. The test detects carriers with 98% sensitivity, and the false positive rate for non-carriers is 5%. Find the probability that a person who tests positive actually carries the gene. The answer is about 16.5%. This again reminds us that in low-prevalence settings, positive results must be interpreted with care.

中文：练习三（题型一/树状图）：甲、乙两家供应商向某餐厅供货，甲供应 70% 的食材且变质率为 1%，乙供应 30% 且变质率为 3%。某日发现一批食材变质，求它来自甲的概率。提示：先画两步树，标出四条路径的概率，再算”经甲变质路径 ÷ 变质总路径”。答案约为 0.7 ÷ (0.7 × 0.01 + 0.3 × 0.03) 对应的比例，即约 43.8%。

English: Problem 3 (Type 1, tree diagram): Two suppliers, Jia and Yi, supply a restaurant: Jia provides 70% of the ingredients with a spoilage rate of 1%, and Yi provides 30% with a spoilage rate of 3%. One day a batch of ingredients is found spoiled: find the probability it came from Jia. Hint: draw the two-step tree, label the four path probabilities, then compute (path via Jia and spoiled) ÷ (total spoiled paths). The answer is about 43.8%.

10. 考前复习要点 | Key Points for Exam Revision

中文：最后总结贝叶斯定理考点的复习要点。第一，必须能在 1 分钟内默写贝叶斯定理及其完整形式（含全概率公式展开），这是计算题的保底分。第二，掌握三种题型对应的三种工具：树状图（路径占比）、2×2 表格（边际概率）、完整公式（直接代入）；看到题目先判断题型再选择工具。第三，牢记”后验 ∝ 似然 × 先验”这一核心思想，遇到新情境（检测、筛查、分类、预测）时能识别出贝叶斯结构。第四，做完务必检查分母是否是”所有到达 B 的路径之和”，并验证后验落在 0 到 1 之间。把这四点落实，贝叶斯定理相关的题目就可以稳定拿分。

English: Finally, here are the key points for revising Bayes’ theorem. First, you must be able to write Bayes’ theorem and its complete form (with the law of total probability expansion) from memory within one minute — this is guaranteed easy marks in calculation questions. Second, master the three tools for the three question types: the tree diagram (path fractions), the 2×2 table (marginal probabilities) and the complete formula (direct substitution); identify the type before choosing the tool. Third, keep the core idea “posterior ∝ likelihood × prior” in mind so that you can recognise the Bayesian structure in new contexts (screening, classification, prediction). Fourth, always check that the denominator is the sum of all paths reaching B, and verify that the posterior lies between 0 and 1. If you implement these four points, questions on Bayes’ theorem become reliable marks.

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Bayes’ Theorem: Derivation and Applications — 贝叶斯定理：公式推导与应用