📚 Law of Total Probability with Worked Examples | 全概率公式与例题解析
The Law of Total Probability is a cornerstone of IB Mathematics Analysis and Approaches (AA) and Applications and Interpretation (AI) at Higher Level. It allows us to compute the probability of an event by considering all possible mutually exclusive scenarios that lead to it.
全概率公式是IB数学分析与方法（AA）以及应用与解释（AI）高级水平中的核心内容。它通过考虑导致某一事件发生的所有互斥情形，帮助我们计算该事件的概率。
1. What Is the Law of Total Probability? | 什么是全概率公式？
Suppose a sample space is partitioned into mutually exclusive events B₁, B₂, …, Bₙ that cover every possible outcome. Then for any event A, the law states that the total probability of A is the weighted sum of the conditional probabilities of A given each partition event.
假设样本空间被划分为互斥事件 B₁, B₂, …, Bₙ，这些事件覆盖了所有可能的结果。那么对于任意事件 A，全概率公式指出 A 的总概率等于在给定每个划分事件的条件下 A 的条件概率的加权和。
P(A) = P(A|B₁)P(B₁) + P(A|B₂)P(B₂) + … + P(A|Bₙ)P(Bₙ)
This is often written in compact summation notation as P(A) = Σ P(A|Bᵢ)P(Bᵢ), for i = 1 to n.
通常可以缩写为 P(A) = Σ P(A|Bᵢ)P(Bᵢ)，其中 i 从 1 到 n。
2. The Partition Condition | 划分条件
For the formula to be valid, the events B₁, B₂, …, Bₙ must form a partition of the sample space. This means two things: first, the events must be pairwise mutually exclusive, so Bᵢ ∩ Bⱼ = ∅ for i ≠ j; second, their union must equal the entire sample space, so B₁ ∪ B₂ ∪ … ∪ Bₙ = S.
要使公式成立，事件 B₁, B₂, …, Bₙ 必须构成样本空间的一个划分。这包含两个条件：第一，事件两两互斥，即当 i ≠ j 时，Bᵢ ∩ Bⱼ = ∅；第二，它们的并集必须等于整个样本空间，即 B₁ ∪ B₂ ∪ … ∪ Bₙ = S。
The most common partition in practice is a binary split: an event B and its complement B’. Since B ∪ B’ = S and B ∩ B’ = ∅, the law simplifies to P(A) = P(A|B)P(B) + P(A|B’)P(B’).
实际中最常见的划分是二元划分：事件 B 及其补事件 B’。由于 B ∪ B’ = S 且 B ∩ B’ = ∅，公式简化为 P(A) = P(A|B)P(B) + P(A|B’)P(B’)。
3. Derivation from Conditional Probability | 从条件概率推导
The law can be derived directly from the definition of conditional probability. Recall that P(A|B) = P(A ∩ B) / P(B), so P(A ∩ B) = P(A|B)P(B).
该公式可以直接从条件概率的定义推导出来。回顾 P(A|B) = P(A ∩ B) / P(B)，因此 P(A ∩ B) = P(A|B)P(B)。
Because the Bᵢ events form a partition, the event A can be written as the disjoint union of the intersections A ∩ Bᵢ. Therefore P(A) = P(A ∩ B₁) + P(A ∩ B₂) + … + P(A ∩ Bₙ), and substituting the conditional probability identity gives the law.
由于 Bᵢ 事件构成划分，事件 A 可以写成交集 A ∩ Bᵢ 的不相交并集。因此 P(A) = P(A ∩ B₁) + P(A ∩ B₂) + … + P(A ∩ Bₙ)，代入条件概率恒等式即可得到全概率公式。
P(A) = Σ P(A ∩ Bᵢ) = Σ P(A|Bᵢ)P(Bᵢ)
4. Standard Solution Procedure | 标准解题步骤
When tackling an IB exam question on the law of total probability, a systematic approach is essential. First, identify the event whose total probability you need to find.
在解答IB考试中涉及全概率公式的题目时，系统性方法至关重要。首先，确定需要计算总概率的事件。
Second, choose the partition. Look for a scenario where the sample space can naturally be divided into different cases or groups. These cases are usually introduced in the question as percentages or proportions of a population.
其次，选择合适的划分。寻找样本空间可以自然分为不同情形或群体的场景。这些情形通常在题目中以人口的比例或百分数的形式给出。
Third, write down P(Bᵢ) for each partition event and P(A|Bᵢ) for each conditional probability. Finally, multiply and sum all the terms to obtain P(A).
第三，写出每个划分事件的概率 P(Bᵢ) 以及每个条件概率 P(A|Bᵢ)。最后，将各项相乘并求和，得到 P(A)。
- Step 1: Define events clearly | 步骤一：清晰定义事件
- Step 2: Identify the partition | 步骤二：确定样本空间的划分
- Step 3: List P(Bᵢ) and P(A|Bᵢ) | 步骤三：列出各概率值
- Step 4: Apply the formula and sum | 步骤四：套用公式求和
5. Worked Example 1: Factory Production | 例题一：工厂生产
Consider a factory with three machines, M₁, M₂ and M₃, that produce 50%, 30% and 20% of the total output respectively. The defective rates are 2%, 3% and 4% respectively. What is the probability that a randomly selected item is defective?
假设一家工厂有三台机器 M₁、M₂ 和 M₃，它们分别生产总产量的 50%、30% 和 20%。次品率分别为 2%、3% 和 4%。随机抽取一件产品，它是次品的概率是多少？
Let D be the event that the item is defective. The machines form a natural partition since every item is produced by exactly one machine. Applying the law of total probability:
设 D 为产品是次品的事件。三台机器构成了自然的划分，因为每件产品恰好由一台机器生产。应用全概率公式：
P(D) = P(D|M₁)P(M₁) + P(D|M₂)P(M₂) + P(D|M₃)P(M₃)
P(D) = 0.02 × 0.50 + 0.03 × 0.30 + 0.04 × 0.20
P(D) = 0.010 + 0.009 + 0.008 = 0.027
So the probability that a randomly selected item is defective is 0.027, or 2.7%. Notice how each contribution is small, but together they give the overall probability.
因此，随机抽取一件产品为次品的概率是 0.027，即 2.7%。注意每一项的贡献都很小，但它们合在一起给出了总体概率。
6. Worked Example 2: Medical Testing | 例题二：医学检测
A diagnostic test for a certain disease has the following properties: if a person has the disease, the test is positive with probability 0.95; if a person does not have the disease, the test is positive with probability 0.03 (this is called a false positive). The prevalence of the disease in the population is 1%.
某种疾病的诊断检测具有以下性质：如果一个人患病，检测结果为阳性的概率是 0.95；如果一个人未患病，检测结果为阳性的概率是 0.03（这称为假阳性）。该疾病在人群中的患病率为 1%。
Let D represent the event of having the disease and T represent the event of a positive test result. Since D and D’ form a partition, we can find the overall probability of a positive test.
设 D 表示患病事件，T 表示检测结果为阳性的事件。由于 D 和 D’ 构成一个划分，我们可以求出检测结果为阳性的总体概率。
P(T) = P(T|D)P(D) + P(T|D’)P(D’)
P(T) = 0.95 × 0.01 + 0.03 × 0.99 = 0.0095 + 0.0297 = 0.0392
The probability of a positive test result is 0.0392, or 3.92%. Interestingly, most positive results actually come from people without the disease, since the false positive rate applies to the much larger healthy population.
检测结果为阳性的概率是 0.0392，即 3.92%。有趣的是，大多数阳性结果实际上来自未患病的人群，因为假阳性率作用于基数大得多的健康人群。
7. Relationship with Bayes’ Theorem | 与贝叶斯公式的关系
The law of total probability is closely linked to Bayes’ theorem. In fact, Bayes’ theorem can be derived by combining the definition of conditional probability with the law of total probability.
全概率公式与贝叶斯公式密切相关。事实上，贝叶斯公式可以通过将条件概率的定义与全概率公式结合起来推导。
Bayes’ theorem asks a reverse question: given that event A has occurred, what is the probability that it came from partition event Bᵢ? The denominator of Bayes’ theorem, P(A), is exactly the quantity computed by the law of total probability.
贝叶斯公式提出的是一个逆向问题：给定事件 A 已经发生，它来自划分事件 Bᵢ 的概率是多少？贝叶斯公式的分母 P(A) 正是通过全概率公式计算得出的量。
P(Bᵢ|A) = [P(A|Bᵢ)P(Bᵢ)] / [Σ P(A|Bⱼ)P(Bⱼ)]
In IB examinations, questions often ask you to first compute P(A) using the law of total probability, then use Bayes’ theorem in the next part. Mastering both is essential for full marks.
在IB考试中，题目常常要求你先用全概率公式计算 P(A)，然后在下一问中使用贝叶斯公式。熟练掌握这两者是获得满分的关键。
8. Using Tree Diagrams | 使用树状图
Tree diagrams are an excellent visual tool for solving problems involving the law of total probability. Each branch from the root represents a partition event Bᵢ, and each second-level branch represents the conditional event A given Bᵢ.
树状图是解决全概率公式问题的绝佳可视化工具。从根节点出发的每条分支代表一个划分事件 Bᵢ，每个第二层分支代表给定 Bᵢ 条件下的事件 A。
The probability of A is found by summing the products along all paths that lead to A. This is exactly the same operation as applying the law of total probability, but the visual format often helps avoid missing a branch or miscounting.
A 的概率通过沿所有通向 A 的路径将乘积求和得到。这与应用全概率公式的操作完全相同，但可视化格式往往有助于避免遗漏分支或算错。
| Path | Probability Calculation |
| B₁ then A | P(A|B₁)P(B₁) |
| B₂ then A | P(A|B₂)P(B₂) |
| Total | Σ P(A|Bᵢ)P(Bᵢ) |
9. Common Mistakes in IB Exams | IB考试中的常见错误
One common error students make is forgetting to check that the partition events are mutually exclusive and exhaustive. If the events overlap or do not cover the whole sample space, the formula produces an incorrect result.
学生常犯的一个错误是忘记检查划分事件是否互斥且完备。如果事件有重叠或没有覆盖整个样本空间，公式就会产生错误的结果。
Another frequent mistake is confusing P(A|B) with P(B|A). These two conditional probabilities are generally not equal, and switching them will lead to a completely different answer. Always read the question carefully to identify which conditional probability is given.
另一个常见错误是混淆 P(A|B) 与 P(B|A)。这两个条件概率通常不相等，互换它们会导致完全不同的答案。一定要仔细审题，确定题目给出的是哪个条件概率。
A third typical error is a calculation slip when multiplying decimals such as 0.03 × 0.99. Always double-check your arithmetic, especially in multi-part questions where an early numerical error propagates through the rest of the problem.
第三个典型错误是在计算小数乘法时出现失误，例如计算 0.03 × 0.99。务必复查你的算术，尤其是在多问答题中，早期的数值错误会传导到后续所有问题中。
10. Exam-Style Practice Question | 考试风格练习题
A bag contains marbles from two boxes. Box X contains 4 red and 6 blue marbles. Box Y contains 7 red and 3 blue marbles. A box is chosen at random with P(X) = 0.6 and P(Y) = 0.4, then a marble is drawn from the chosen box. Find the probability that a red marble is drawn.
一个袋子中的弹珠来自两个盒子。盒子 X 中有 4 个红色和 6 个蓝色弹珠。盒子 Y 中有 7 个红色和 3 个蓝色弹珠。随机选择一个盒子，P(X) = 0.6，P(Y) = 0.4，然后从选中的盒子中抽取一个弹珠。求抽到红色弹珠的概率。
Let R be the event of drawing a red marble. The boxes form a partition. Using the law of total probability:
设 R 为抽到红色弹珠的事件。两个盒子构成一个划分。应用全概率公式：
P(R) = P(R|X)P(X) + P(R|Y)P(Y) = (4/10)(0.6) + (7/10)(0.4)
P(R) = 0.24 + 0.28 = 0.52
The probability of drawing a red marble is 0.52, or 52%. Try working this out using a tree diagram on your own to confirm the same result.
抽到红色弹珠的概率是 0.52，即 52%。请尝试自己用树状图计算一遍，以验证相同的答案。
11. Strategic Advice for IB Exams | IB考试策略建议
When you see a probability question mentioning percentages of a population and a condition such as ‘randomly selected’, the law of total probability is almost certainly the intended method. Use a systematic approach: define events, list known probabilities, apply the formula, and check your answer for reasonableness.
当你看到一道概率题提到人群的百分数以及”随机选取”等条件时，全概率公式几乎可以肯定是所考察的方法。采用系统性的方法：定义事件、列出已知概率、套用公式、检查答案是否合理。
Always present your working clearly. In IB examinations, method marks are awarded even if your final numerical answer is slightly wrong. Writing down the formula in symbols first, then substituting numbers, is the safest way to secure those marks.
始终清晰地展示你的解题过程。在IB考试中，即使最终数值答案略有偏差，仍然会获得方法分。先用符号写出公式，然后代入数值，这是最稳妥的得分策略。
Finally, practise with a variety of contexts: factory production, medical testing, mixed populations, and two-stage experiments. Recognising the partition structure in different settings is a skill that improves with practice.
最后，在多种情境下进行练习：工厂生产、医学检测、混合人群和两阶段实验。在不同背景下识别划分结构是一项通过练习才能提高的技能。
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