📚 IGCSE OCR Maths: Probability Key Points | IGCSE OCR 数学：概率 考点精讲
Probability is a measure of how likely an event is to occur. In the OCR IGCSE Mathematics specification, you are expected to calculate probabilities, construct and interpret tree diagrams, understand mutually exclusive and independent events, and apply conditional probability for Higher tier. This bilingual revision guide breaks down every key concept with clear examples, helping you master the topic for your exam.
概率是衡量某个事件发生可能性的量度。在 OCR IGCSE 数学考纲中，你需要计算概率，构建并解读树状图，理解互斥事件和独立事件，并在高级别要求中应用条件概率。这本双语复习指南将每个关键概念拆解成清晰的示例，助你掌握该主题，备考无忧。
1. The Probability Scale | 概率尺度
Probability is always expressed as a number between 0 and 1 inclusive. A probability of 0 means the event is impossible, while a probability of 1 means the event is certain. Values can be written as fractions, decimals or percentages.
概率始终表示为 0 到 1 之间的一个数（含 0 和 1）。概率为 0 表示事件不可能发生，概率为 1 表示事件必然发生。数值可以用分数、小数或百分数表示。
For example, on a fair six‑sided die, the probability of rolling a 7 is 0 (impossible) and the probability of rolling a number less than 7 is 1 (certain).
例如，掷一枚公平的六面骰子，掷出点数为 7 的概率是 0（不可能），掷出点数小于 7 的概率是 1（必然）。
Every probability can be placed on the scale; a probability of ½ indicates an event that is equally likely to happen or not happen.
每个概率都可以标在尺度上；概率为 ½ 表示某事件发生与不发生的可能性相同。
2. Basic Probability Formula | 基本概率公式
When all outcomes are equally likely, the probability of an event A is found using:
当所有结果等可能时，事件 A 的概率可用下式求得：
P(A) = number of favourable outcomes / total number of outcomes
sible. The probability of an event not happening (the complement) is given by P(not A) = 1 – P(A).
尽可能化简分数。事件不发生的概率（互补事件）由 P(非A) = 1 – P(A) 给出。
e probability of drawing an Ace is 4/52 = 1/13. The probability of not drawing an Ace is 1 – 1/13 = 12/13.
示例：一副 52 张扑克牌中，抽到一张 A 的概率是 4/52 = 1/13。抽不到 A 的概率是 1 – 1/13 = 12/13。
3. Sample Spaces and Diagrams | 样本空间与图示
A sample space is the set of all possible outcomes of an experiment. You can list outcomes systematically in a table, a list, or a diagram. For two coins flipped together, the sample space can be written as {HH, HT, TH, TT} with each outcome having probability ¼.
样本空间是一次试验所有可能结果的集合。你可以用表格、列表或图示系统地列出结果。两枚硬币同时抛掷，样本空间可写为 {HH, HT, TH, TT}，每个结果的概率为 ¼。
ually likely pairs, making it easy to find probabilities such as P(sum of 7) = 6/36 = 1/6.
结果网格对于掷两个骰子等组合事件特别有用。6×6 的表格显示了所有 36 个等可能的数对，从而可以轻松求出如 P(和为 7) = 6/36 = 1/6 的概率。
Venn diagrams can also represent a sample space, with the rectangle representing the whole set and circles representing specific events.
维恩图也可以表示样本空间，矩形代表全集，圆圈代表具体事件。
4. Mutually Exclusive Events | 互斥事件
Two events are mutually exclusive if they cannot happen at the same time. For mutually exclusive events A and B, the probability that either A or B occurs is the sum of their individual probabilities:
如果两个事件不能同时发生，则它们是互斥的。对于互斥事件 A 和 B，A 或 B 发生的概率是它们各自概率之和：
P(A or B) = P(A) + P(B)
because you cannot roll both numbers on a single throw. P(2 or 5) = 1/6 + 1/6 = 2/6 = 1/3.
这称为互斥事件的加法法则。例如，在一枚公平骰子上掷出 2 或 5 是互斥的，因为你不可能一次掷出两个数字。P(2 或 5) = 1/6 + 1/6 = 1/3。
5. Independent Events and the AND Rule | 独立事件与乘法法则
Two events are independent if the outcome of one does not affect the outcome of the other. For independent events A and B, the probability that both A and B happen is:
如果两个事件中一个的结果不影响另一个的结果，则它们是独立的。对于独立事件 A 和 B，A 与 B 都发生的概率为：
P(A and B) = P(A) × P(B)
coin and rolling a die). Flipping a head and rolling a 6: P(Head and 6) = ½ × 1/6 = 1/12.
只有在你确定事件独立时才使用此法则（例如抛硬币和掷骰子）。抛得正面且掷得 6：P(正面且6) = ½ × 1/6 = 1/12。
Do not assume independence for events like drawing counters without replacement, because the second draw’s probability depends on the first outcome.
对于不放回抽取筹码等事件，不要假设独立性，因为第二次抽取的概率取决于第一次的结果。
6. Tree Diagrams for Sequential Events | 序贯事件的树状图
Tree diagrams show all possible outcomes of two or more stages. Branches represent possible events, and the probability is written along each branch. To find the probability of a combined outcome, multiply the probabilities along the path. If more than one path leads to the same final outcome, add those path probabilities.
树状图展示两个或多个阶段的所有可能结果。分支代表可能的事件，概率写在每条分支上。要求出组合结果的概率，沿路径将概率相乘。如果有多条路径通向同一个最终结果，则将这些路径的概率相加。
nd 2 blue counters. Two counters are drawn without replacement. The tree diagram has first‑draw branches: P(R) = 3/5, P(B) = 2/5. The second‑ red and one blue (in any order) is (3/5 × 2/4) + (2/5 × 3/4) = 6/20 + 6/20 = 12/20 = 3/5.
示例：一个袋子装有 3 红 2 蓝筹码，无放回地抽取两次。树状图有第一次抽取分支：P(R) = 3/5, P(B) = 2/5。第二次抽取分支则取决于被移走的筹码。得一红一蓝（不计顺序）的概率是 (3/5 × 2/4) + (2/5 × 3/4) = 6/20 + 6/20 = 12/20 = 3/5。
Always check that the probabilities on branches from a single point add to 1, and the sum of all final outcome probabilities equals 1.
始终检查同一点出发的各分支概率之和为 1，并且所有最终结果的概率总和等于 1。
7. Conditional Probability (Higher Tier) | 条件概率（高级别）
Conditional probability is the probability of an event occurring given that another event has already occurred. It is written as P(A|B) and calculated using:
条件概率是指在已知另一事件已发生的条件下，某事件发生的概率。它写作 P(A|B)，并用下式计算：
P(A|B) = P(A and B) / P(B)
on the exam formula sheet. In the tree diagram for drawing without replacement, the second‑ ically conditional probabilities. For the bag with 3 red and 2 blue counters, P(second red | first blue) = 3/4 because after removing a blue, 3 red and 1 blue remain.
该公式在考试的公式表中提供。在不放回抽取的树状图中，第二层分支的概率自然就是条件概率。对于装有 3 红 2 蓝筹码的袋子，P(第二次红色 | 第一次蓝色) = 3/4，因为移除一个蓝色后，剩下 3 红 1 蓝。
Conditional probability questions often require you to identify the restricted sample space. Practice reading “given that” statements carefully.
条件概率题目常要求你识别受限的样本空间。仔细练习阅读“已知……”的语句。
8. General Addition Rule (OR Rule) | 一般加法法则（或法则）
When events are not mutually exclusive, simply adding probabilities would double‑count the intersection. The general addition rule applies to any two events A and B:
当事件不互斥时，简单相加概率会重复计算交集部分。一般加法法则适用于任意两个事件 A 和 B：
P(A or B) = P(A) + P(B) – P(A and B)
lusive, P(A and B) = 0, and the formula reduces to the simple sum. Example: In a class, 30% play football (F), 20% play tennis (T), and 10% play both. P(F or T) = 0.3 + 0.2 – 0.1 = 0.4.
如果 A 和 B 互斥，则 P(A 和 B) = 0，公式就退化为简单相加。示例：在一个班级中，30% 踢足球 (F)，20% 打网球 (T)，10% 两项都参加。P(F 或 T) =
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