📚 IGCSE WJEC Maths: Probability Essentials | 概率考点精讲
Probability is a cornerstone of statistical reasoning in the IGCSE WJEC mathematics specification. This topic explores how to quantify uncertainty, model random events and make predictions based on logical rules. From basic ideas of equally likely outcomes to conditional probability and tree diagrams, mastering these concepts will build a strong foundation for data handling and later studies.
概率是IGCSE WJEC数学大纲中统计推理的基石。这一主题探讨如何量化不确定性、对随机事件建模，以及根据逻辑规则进行预测。从等可能结果的基本思想到条件概率和树形图，掌握这些概念将为数据处理和后续学习打下坚实基础。
1. Basic Probability Concepts | 基本概率概念
Probability measures the chance that a specific event will occur when an experiment is performed. The probability scale runs from 0 to 1 inclusive, where 0 represents an impossible event and 1 represents a certain event. An outcome is a single possible result, while an event is a set of one or more outcomes. The set of all possible outcomes is called the sample space, often denoted by S.
概率衡量在进行试验时某个特定事件发生的可能性。概率尺度从 0 到 1（含 0 和 1），其中 0 表示不可能事件，1 表示必然事件。结果是指一个单一的可能结果，而事件是由一个或多个结果组成的集合。所有可能结果的集合被称为样本空间，通常记为 S。
For example, when rolling a fair six-sided dice, the sample space is S = {1, 2, 3, 4, 5, 6}. If the event A is {rolling an even number}, then A = {2, 4, 6}. The probability that A occurs is written P(A).
例如，掷一枚公平的六面骰子时，样本空间为 S = {1, 2, 3, 4, 5, 6}。若事件 A 为「掷出偶数」，则 A = {2, 4, 6}。事件 A 发生的概率记作 P(A)。
2. Sample Spaces and Listing Outcomes | 样本空间与列出结果
To calculate probabilities correctly, you must be able to list all possible outcomes in a systematic way. Common methods include ordered lists, two-way tables and simple tree-structured lists. For two coins tossed simultaneously, the sample space can be written as {HH, HT, TH, TT}. For a dice thrown twice, a 6 × 6 table showing all ordered pairs (1,1) to (6,6) gives 36 equally likely outcomes.
为了正确计算概率，你必须能够系统地列出所有可能的结果。常用的方法包括有序列表、双向表和简单的树形图列表。对于同时抛掷两枚硬币，样本空间可以写为 {HH, HT, TH, TT}。对于一枚骰子掷两次，可以用一个 6 × 6 的表格显示从 (1,1) 到 (6,6) 的所有有序数对，共 36 个等可能结果。
When there are too many outcomes to list individually, you can use the counting principle: for a sequence of events, multiply the numbers of possibilities at each stage. The total number of outcomes for rolling a dice and flipping a coin is 6 × 2 = 12.
当结果太多而无法逐一列出时，可以使用计数的基本原理：对于一系列事件，将每一阶段的可能性数量相乘。掷一枚骰子并抛一枚硬币的总结果数为 6 × 2 = 12。
3. Theoretical Probability | 理论概率
Theoretical probability is based on the symmetry of an experiment and assumes that all outcomes in the sample space are equally likely. The basic formula is:
理论概率基于试验的对称性，并假设样本空间中的所有结果都是等可能的。基本公式为：
P(Event) = Number of favourable outcomes / Total number of outcomes
P(事件) = 有利结果的数量 / 可能结果的总数
For a fair dice, P(throwing a prime number) = P({2,3,5}) = 3/6 = 1/2. This formula works only when all outcomes are equally likely. If a spinner is biased, theoretical probability cannot be found by symmetry; you would rely on experimental data.
对于一枚公平的骰子，P(掷出质数) = P({2,3,5}) = 3/6 = ½。该公式仅在所有结果等可能时才有效。如果转盘是不均匀的，则无法通过对称性求出理论概率；你需要依靠实验数据。
4. Experimental Probability and Relative Frequency | 实验概率与相对频率
Experimental probability, also called relative frequency, is calculated from the results of an actual experiment or trial. If an event occurs r times in n trials, the relative frequency is r ÷ n. This value is an estimate of the true probability. The estimate tends to become more stable and approach the theoretical probability as the number of trials increases – this is known as the law of large numbers.
实验概率，也称为相对频率，是根据真实实验或试验的结果计算得出的。如果一个事件在 n 次试验中出现了 r 次，那么相对频率为 r ÷ n。这个值是对真实概率的估计。随着试验次数的增加，该估计值往往变得更加稳定并趋近于理论概率——这被称为大数定律。
For example, if a drawing pin lands point up 312 times in 500 drops, the experimental probability of ‘point up’ is 312/500 = 0.624. A greater number of drops should give a more reliable estimate.
例如，如果一枚图钉在 500 次掉落中有 312 次钉尖朝上，那么「钉尖朝上」的实验概率为 312/500 = 0.624。更多的掉落次数能得到更可靠的估计。
5. Mutually Exclusive Events | 互斥事件
Two events are mutually exclusive if they cannot happen at the same time. In set language, their intersection is empty: A ∩ B = ∅. For mutually exclusive events, the probability that either A or B occurs is simply the sum of their individual probabilities:
如果两个事件不可能同时发生，则它们互斥。用集合语言来说，它们的交集为空集：A ∩ B = ∅。对于互斥事件，A 或 B 发生的概率就是它们各自概率的和：
P(A ∪ B) = P(A) + P(B)
When rolling a dice, the events ‘rolling a 2’ and ‘rolling an odd number’ are mutually exclusive because a single roll cannot satisfy both. So P(2 or odd) = 1/6 + 3/6 = 4/6 = ⅔.
掷一枚骰子时，事件「掷出 2」和「掷出奇数」互斥，因为一次投掷不可能同时满足两者。因此 P(掷出2或奇数) = 1/6 + 3/6 = 4/6 = ⅔。
If events are mutually exclusive and exhaustive (they cover the whole sample space), their probabilities sum to 1. The complement of event A, written A′, has probability P(A′) = 1 − P(A).
如果事件互斥且穷举（它们涵盖了整个样本空间），它们的概率之和为 1。事件 A 的补事件记作 A′，其概率为 P(A′) = 1 − P(A)。
6. Independent Events | 独立事件
Two events are independent if the occurrence of one does not affect the probability of the other. For independent events, the probability that both A and B happen is the product of their probabilities:
如果两个事件的发生互不影响，则它们相互独立。对于独立事件，A 和 B 同时发生的概率是它们各自概率的乘积：
P(A ∩ B) = P(A) × P(B)
For instance, throwing a coin and rolling a dice are independent. P(head and 6) = ½ × 1/6 = 1/12. Independence is not always obvious; in exam questions you may need to check whether two events satisfy the multiplication rule before applying it. Be careful with scenarios involving selection without replacement – those events are generally dependent.
例如，抛一枚硬币和掷一枚骰子是独立的。P(正面且6) = ½ × 1/6 = 1/12。独立性并非总是显而易见的；在考题中，你可能需要先验证两个事件是否满足乘法法则，然后再加以应用。要注意涉及不放回抽样的情形——这类事件通常是相依的。
7. The Addition Rule for Probability | 概率的加法法则
When two events are not mutually exclusive, simply adding their probabilities would double-count the overlap. The general addition rule corrects this:
当两个事件不互斥时，直接把概率相加会重复计算重叠部分。一般的加法法则可以修正这一点：
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
This can be visualised using a Venn diagram with overlapping regions. Consider picking a card at random from a standard deck. Let A be ‘the card is a heart’ and B be ‘the card is a face card’. P(A) = 13/52, P(B) = 12/52, and P(A ∩ B) = 3/52 (three heart face cards). Using the rule: P(heart or face) = 13/52 + 12/52 − 3/52 = 22/52 = 11/26.
这可以用带重叠区域的韦恩图来直观展示。假设从一副标准扑克牌中随机抽取一张牌。设 A 为「这张牌是红桃」，B 为「这张牌是花牌」。P(A) = 13/52，P(B) = 12/52，P(A ∩ B) = 3/52（三张红桃花牌）。利用法则：P(红桃或花牌) = 13/52 + 12/52 − 3/52 = 22/52 = 11/26。
| For mutually exclusive events, P(A ∩ B) = 0, so the formula reduces to P(A) + P(B). | 对于互斥事件，P(A ∩ B) = 0，因此公式简化为 P(A) + P(B)。 |
| If you know P(A ∪ B) and the individual probabilities, you can rearrange to find P(A ∩ B). | 如果已知 P(A ∪ B) 和各自的概率，可以变形求出 P(A ∩ B)。 |
8. The Multiplication Rule and Combined Events | 乘法法则与复合事件
Combined events often involve the word ‘and’, meaning both events must happen. When events are independent, the multiplication rule P(A ∩ B) = P(A) × P(B) applies directly. However, when events are dependent, you need the general multiplication rule involving conditional probability. The product of the probability of one event and the conditional probability of the second event given the first gives the joint probability:
复合事件通常涉及「且」这个词，意味着两个事件必须同时发生。当事件独立时，可以直接使用乘法法则 P(A ∩ B) = P(A) × P(B)。然而，当事件相依时，你需要使用涉及条件概率的一般乘法法则。一个事件的概率乘以在第一个事件发生的条件下第二个事件发生的条件概率，即可得到联合概率：
P(A ∩ B) = P(A) × P(B|A)
Here P(B|A) is read as ‘the probability of B given A’. This general form is the foundation for solving complex probability problems that involve selections without replacement or sequences of dependent trials.
这里 P(B|A) 读作「在 A 发生的条件下 B 发生的概率」。这个一般形式是解决涉及不放回抽取或相依试验序列等复杂概率问题的基础。
9. Tree Diagrams | 树形图
Tree diagrams are powerful tools for illustrating multi-stage experiments. Each branch represents a possible outcome at a stage, and the probability of that outcome is written on the branch. Branch probabilities from the same node must sum to 1. To find the probability of a complete path, you multiply the probabilities along the branches. The final probability of an event is obtained by adding the probabilities of all relevant paths.
树形图是展示多阶段试验的强大工具。每条分支代表某一阶段的一个可能结果，该结果的概率就写在分支上。从同一个节点出发的各分支概率之和必须等于 1。要找到一条完整路径的概率，需将沿途各分支的概率相乘。某个事件的最终概率是通过将所有相关路径的概率相加得到的。
For example, a bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. On the first draw, P(red) = 5/8. If a red is taken, the bag has 4 red and 3 blue left, so P(red on second|red first) = 4/7. The path ‘red then red’ has probability 5/8 × 4/7 = 20/56 = 5/14. You can then work out ‘red then blue’, ‘blue then red’, ‘blue then blue’ and combine results to find P(second ball is red), which is 5/14 + 15/56 = 35/56 = 5/8. Notice that without replacement the events are dependent, and the total probability of a red on the second draw equals the original proportion, an interesting property.
例如，一个袋子里有 5 个红球和 3 个蓝球。不放回地抽取两个球。第一次抽取时，P(红) = 5/8。如果取出了红球，袋中剩余 4 红 3 蓝，因此 P(第二次出红|第一次出红) = 4/7。路径「红然后红」的概率为 5/8 × 4/7 = 20/56 = 5/14。然后你可以计算「红然后蓝」、「蓝然后红」、「蓝然后蓝」，并将结果组合起来求 P(第二个球是红球) = 5/14 + 15/56 = 35/56 = 5/8。注意，在不放回的情况下事件是相依的，而第二次抽到红球的总概率正好等于原来的比例，这是一个有趣的性质。
10. Conditional Probability | 条件概率
Conditional probability allows us to update the chance of an event given that another event has occurred. The defining formula is:
条件概率使我们能够在已知另一事件已经发生的情况下，更新某个事件发生的可能性。其定义公式为：
P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0
This formula can be rearranged to give the general multiplication rule P(A ∩ B) = P(B) × P(A|B). Conditional probability is essential when working with two-way tables and Venn diagrams, especially in higher-tier WJEC questions. For example, if 20% of students study both music and drama, and 50% study music, then the probability a music student also studies drama is P(drama|music) = 0.20 / 0.50 = 0.4.
这个公式可以变形为一般乘法法则 P(A ∩ B) = P(B) × P(A|B)。当使用双向表和韦恩图时，条件概率是必不可少的，尤其是在 WJEC 高级别考题中。举例来说，如果 20% 的学生同时学习音乐和戏剧，而 50% 的学生学习音乐，那么一个学习音乐的学生也学习戏剧的概率为 P(戏剧|音乐) = 0.20 / 0.50 = 0.4。
It is crucial to identify the ‘given’ event correctly and to realise that the sample space has been restricted to this event. Misinterpreting the conditioning event is a common error.
正确识别「给定」事件并意识到样本空间已缩小到该事件是至关重要的。误解条件事件是一个常见错误。
11. Expected Value and Expectation | 期望值与期望
In the context of repeated trials, the expected frequency of an event is found by multiplying the probability of the event by the number of trials:
在重复试验的语境中，一个事件的期望频数是通过将该事件的概率乘以试验次数来计算的：
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