📚 Probability Calculations: Basic Rules and Typical Examples | 概率计算:基本法则与典型例题
Probability is the mathematical language of uncertainty. In A-level mathematics, probability questions often appear in both pure statistics and mechanics contexts, and they require a clear understanding of sample spaces, events, and the rules for combining probabilities.
概率是描述不确定性的数学语言。在 A-level 数学中,概率问题常出现在统计与力学场景中,解答它们需要清晰理解样本空间、事件以及组合概率的基本法则。
1. Sample Space and Events | 样本空间与事件
The sample space, written as S, is the set of all possible outcomes of a probability experiment. Any subset of S is called an event.
样本空间用 S 表示,是某个随机试验所有可能结果的集合。S 的任意子集称为一个事件。
When every outcome in the sample space is equally likely, the probability of an event A is given by the ratio of favourable outcomes to the total number of outcomes.
当样本空间中每个结果等可能发生时,事件 A 的概率等于有利结果数与总结果数之比。
P(A) = n(A) / n(S)
Example: A fair six-sided die is rolled. The sample space is S = {1, 2, 3, 4, 5, 6}. Let A be the event that the score is even. Then A = {2, 4, 6}, so P(A) = 3/6 = 1/2.
例如:掷一枚均匀的六面骰子,样本空间为 S = {1, 2, 3, 4, 5, 6}。设 A 表示“掷出偶数”,则 A = {2, 4, 6},因此 P(A) = 3/6 = 1/2。
2. Complement Rule | 补事件法则
The complement of an event A, written A′, contains all outcomes that are not in A. Since A and A′ together make up the whole sample space, the sum of their probabilities is 1.
事件 A 的补事件记作 A′,包含所有不在 A 中的结果。由于 A 和 A′ 合起来构成整个样本空间,所以它们的概率之和为 1。
P(A′) = 1 − P(A)
Example: The probability of rolling a 6 on a fair die is 1/6. The probability of not rolling a 6 is therefore 1 − 1/6 = 5/6.
例如:掷均匀骰子得到 6 的概率为 1/6,因此没有得到 6 的概率为 1 − 1/6 = 5/6。
The complement rule is especially useful for questions involving the phrase “at least one”.
补事件法则在遇到“至少一个”这类表述时尤其有用。
3. Addition Rule for Mutually Exclusive Events | 互斥事件的加法法则
Two events are mutually exclusive if they cannot occur at the same time. For mutually exclusive events A and B, the probability that A or B occurs is the sum of their individual probabilities.
如果两个事件不可能同时发生,则称它们为互斥事件。对于互斥事件 A 和 B,A 或 B 发生的概率等于它们各自概率之和。
P(A ∪ B) = P(A) + P(B)
Example: A single card is drawn from a standard deck. Let A be “the card is a King” and B be “the card is a Queen”. These events are mutually exclusive, so P(A ∪ B) = 4/52 + 4/52 = 8/52 = 2/13.
例如:从一副标准扑克牌中抽取一张。设 A 表示“抽到 K”,B 表示“抽到 Q”。这两个事件互斥,所以 P(A ∪ B) = 4/52 + 4/52 = 8/52 = 2/13。
4. General Addition Rule | 一般加法法则
When two events can happen at the same time, they have an overlap A ∩ B. The simple addition rule must be corrected by subtracting the overlapping probability.
当两个事件可能同时发生时,它们存在交集 A ∩ B。简单加法法则必须减去重叠部分的概率来加以修正。
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Example: In a class of 30 students, 18 study Mathematics, 14 study Physics, and 8 study both. Let M be “studies Mathematics” and P be “studies Physics”. Then
例如:某班有 30 名学生,其中 18 人学习数学,14 人学习物理,8 人两科都学。设 M 表示“学习数学”,P 表示“学习物理”。则
P(M ∪ P) = 18/30 + 14/30 − 8/30 = 24/30 = 4/5
This means that a randomly chosen student studies at least one of the two subjects with probability 4/5.
这表示随机选一名学生,他至少学习其中一门科目的概率为 4/5。
5. Conditional Probability | 条件概率
Conditional probability measures the probability of one event occurring given that another event has already occurred. It is written P(A | B), read as “the probability of A given B”.
条件概率度量的是在某个事件已经发生的条件下,另一个事件发生的概率,记作 P(A | B),读作“在 B 发生的条件下 A 的概率”。
P(A | B) = P(A ∩ B) / P(B), provided P(B) > 0
Example: In a group of 20 students, 12 are girls, 5 are left-handed, and 3 students are left-handed girls. If a student is chosen at random and is known to be a girl, the probability that she is left-handed is
例如:某组有 20 名学生,其中 12 名是女生,5 名是左撇子,3 名是左撇子女生。如果随机选一名学生且已知她是女生,则她为左撇子的概率为
P(L | G) = P(L ∩ G) / P(G) = (3/20) / (12/20) = 3/12 = 1/4
6. Multiplication Rule and Independent Events | 乘法法则与独立事件
If two events A and B are independent, the occurrence of one does not affect the probability of the other. For independent events, the probability that both occur is the product of their probabilities.
如果两个事件 A 和 B 独立,那么一个事件的发生不会影响另一个事件发生的概率。对于独立事件,二者同时发生的概率等于它们各自概率的乘积。
P(A ∩ B) = P(A) × P(B)
In general, without independence, we use the multiplication rule with conditional probability.
在不满足独立性的一般情形下,我们使用包含条件概率的乘法法则。
P(A ∩ B) = P(A) × P(B | A)
Example: Two fair coins are tossed. The outcomes on different coins are independent. The probability of getting two heads is 1/2 × 1/2 = 1/4.
例如:抛掷两枚均匀硬币,不同硬币的结果相互独立。得到两个正面的概率为 1/2 × 1/2 = 1/4。
7. Independence and Mutual Exclusivity | 独立与互斥的区别
Students often confuse independent events with mutually exclusive events, but they are very different ideas.
学生常常把独立事件与互斥事件混淆,但二者是完全不同的概念。
| Property | Mutually Exclusive | Independent |
| Meaning | Cannot happen together | One result does not affect the other |
| Key condition | P(A ∩ B) = 0 | P(A ∩ B) = P(A) × P(B) |
If A and B are mutually exclusive and both have positive probability, then they cannot be independent, because P(A ∩ B) = 0 while P(A) × P(B) > 0.
如果 A 和 B 互斥,且两者概率均为正,那么它们不可能独立,因为此时 P(A ∩ B) = 0,但 P(A) × P(B) > 0。
8. Law of Total Probability | 全概率法则
Suppose events B₁, B₂, …, Bₙ form a partition of the sample space. Then the unconditional probability of any event A can be found by weighting the conditional probabilities by the probabilities of the partition events.
设事件 B₁, B₂, …, Bₙ 构成样本空间的一个分割。那么任意事件 A 的无条件概率可以通过对各个分割事件的条件概率以其概率为权重求和得到。
P(A) = Σ P(Bᵢ) × P(A | Bᵢ)
Example: Factory X produces 60% of the items and has a defect rate of 2%. Factory Y produces 40% of the items and has a defect rate of 5%. If an item is chosen at random, the probability that it is defective is
例如:工厂 X 生产 60% 的产品,次品率为 2%;工厂 Y 生产 40% 的产品,次品率为 5%。随机选择一件产品,它为次品的概率为
P(D) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.02 = 0.032
9. Bayes’ Theorem | 贝叶斯定理
Bayes’ Theorem enables us to reverse a conditional probability: it tells us the probability of Bᵢ given that A has occurred.
贝叶斯定理使我们能够反转条件概率:它给出在 A 已发生的条件下 Bᵢ 发生的概率。
P(Bᵢ | A) = P(Bᵢ) × P(A | Bᵢ) / Σ P(Bⱼ) × P(A | Bⱼ)
Using the previous factory example, suppose a randomly chosen item is found to be defective. The probability that it came from Factory X is
使用上面工厂的例子:假设随机抽取的一件产品被发现是次品。它来自工厂 X 的概率为
P(X | D) = (0.6 × 0.02) / 0.032 = 0.012 / 0.032 = 0.375
So despite Factory X producing 60% of the items, only 37.5% of all defective items come from Factory X because Factory Y has a higher defect rate.
由此可见,尽管工厂 X 生产了 60% 的产品,但由于工厂 Y 的次品率更高,所以所有次品中只有 37.5% 来自工厂 X。
10. Tree Diagrams | 树形图
Tree diagrams are a powerful way to organise multi-stage probability problems. Each branch represents an outcome, and probabilities along branches are multiplied.
树形图是整理多阶段概率问题的有力工具。每条分支代表一个结果,沿分支的概率相乘。
Example: A bag contains 4 red and 6 blue balls. Two balls are drawn at random without replacement. The probability that the first ball is red is 4/10. Given that the first ball is red, the probability that the second ball is also red is 3/9. Hence
例如:袋子中有 4 个红球和 6 个蓝球。不放回地随机抽取两个球。第一个球为红球的概率为 4/10。在第一个球为红球的条件下,第二个球也为红球的概率为 3/9。因此
P(RR) = 4/10 × 3/9 = 12/90 = 2/15
Tree diagrams are particularly useful when a problem has two or three stages and the probabilities change after each stage.
当问题包含两到三个阶段,并且每个阶段之后概率发生变化时,树形图尤其有用。
11. Typical Exam-Style Example | 典型综合例题
A bag contains 3 red and 5 green counters. Two counters are drawn at random without replacement. Find the probability that:
袋子中有 3 个红色计数器与 5 个绿色计数器。不放回地随机抽取两个,求下列事件的概率:
(a) both counters are red.
(a) 两个都是红色。
P(RR) = 3/8 × 2/7 = 6/56 = 3/28
(b) at least one counter is red.
(b) 至少有一个是红色。
P(at least one red) = 1 − P(GG) = 1 − (5/8 × 4/7) = 1 − 20/56 = 36/56 = 9/14
(c) the two counters are of different colours.
(c) 两个计数器颜色不同。
P(RG) + P(GR) = (3/8 × 5/7) + (5/8 × 3/7) = 15/56 + 15/56 = 30/56 = 15/28
12. Common Mistakes and Tips | 常见错误与建议
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For “at least one” questions, use the complement rule: P(at least one) = 1 − P(none).
遇到“至少一个”时,使用补事件法则:P(至少一个) = 1 − P(一个都没有)。
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Never add probabilities of events that are not mutually exclusive.
不要将不互斥的事件的概率直接相加。
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Check whether events are independent before multiplying probabilities.
相乘概率之前,先检查事件是否独立。
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Pay close attention to “with replacement” versus “without replacement”.
认真区分“有放回”和“无放回”。
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Always define the events clearly before starting a calculation.
开始计算前,先清楚定义所涉及的事件。
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In conditional probability, be careful about which event is the condition; P(A | B) and P(B | A) are generally not equal.
在条件概率中,注意哪个事件是条件;P(A | B) 与 P(B | A) 通常不相等。
Mastering these core rules, and practising with structured examples, will give you confidence in solving probability problems in your examinations.
掌握这些核心法则,并通过结构化的例题进行练习,将帮助你在考试中自信地解决概率问题。
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