📚 Probability – Sample Space Diagrams and Tree Diagrams | 概率 – 样本空间图与树状图
Page 193 of the Cambridge KS3 Mathematics series introduces a fundamental skill: working with sample space diagrams and tree diagrams to calculate probabilities. Understanding how to list outcomes systematically and how to represent combined events visually is essential for solving probability problems accurately. This article breaks down each sub‑topic with clear English and Chinese explanations, examples that mirror the style of the textbook, and practical tips to avoid common mistakes.
剑桥 KS3 数学系列第 193 页介绍了一项基本技能：利用样本空间图和树状图计算概率。理解如何系统列出结果，以及如何用图表直观表示复合事件，是准确解决概率问题的关键。本文逐一分解各个子主题，配有与教材风格一致的英文和中文解释、例题，以及避免常见错误的实用提示。
1. What is a Sample Space? | 什么是样本空间？
A sample space is the complete list of all possible outcomes of a probability experiment. For a single flip of a fair coin, the sample space is {Heads, Tails}. For rolling an ordinary six‑sided die once, the sample space can be written as {1, 2, 3, 4, 5, 6}. Being able to identify the sample space is the first step towards finding exact probabilities, because probability is calculated as the number of favourable outcomes divided by the total number of outcomes in the sample space.
样本空间是一个概率实验中所有可能结果的完整列表。抛一枚均匀硬币时，样本空间为 {正面, 反面}。掷一个普通六面骰子一次，样本空间可写为 {1, 2, 3, 4, 5, 6}。能够确定样本空间是求出精确概率的第一步，因为概率等于有利结果的数量除以样本空间中结果的总数。
2. Listing Outcomes Systematically | 系统列出结果
When an experiment involves two or more events, it is crucial to list all possible outcomes without missing any. A systematic approach helps. For instance, when flipping a coin and rolling a die together, we can pair each side of the coin with each number on the die: H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6. This systematic listing confirms there are 12 equally likely outcomes in total.
当实验涉及两个或多个事件时，必须无遗漏地列出所有可能结果。采用系统方法十分必要。例如，同时抛一枚硬币和掷一个骰子时，我们可以将硬币的每一面与骰子的每个点数配对：H1、H2、H3、H4、H5、H6、T1、T2、T3、T4、T5、T6。这种系统列出方式可确认总共存在 12 个等可能的结果。
3. Sample Space Diagrams for Two Events | 两个事件的样本空间图
A sample space diagram is a visual way to show all outcomes for two events. The diagram is usually drawn as a table, with the outcomes of the first event in the left‑hand column and the outcomes of the second event across the top row. Each cell inside the table represents one combined outcome. This format is extremely useful when calculating probabilities for ‘and’ combined events, such as “rolling a 5 and getting tails”.
样本空间图是展示两个事件所有结果的可视化工具。该图通常以表格形式绘制，左侧一列写出第一个事件的结果，顶部一行列出第二个事件的结果。表中的每个单元格代表一个组合结果。在计算“并且”复合事件的概率时（例如“掷出 5 且得到反面”），这种格式非常有用。
4. Using Tables to Represent Sample Spaces | 使用表格表示样本空间
Consider rolling two fair six‑sided dice. The sample space can be displayed in a 6 × 6 table. The top row shows the numbers on die 1, and the leftmost column shows the numbers on die 2. Each inner cell contains the ordered pair (die 1, die 2).
考虑掷两个均匀的六面骰子。样本空间可显示在一个 6 × 6 的表格中。顶行表示第一个骰子的点数，最左列表示第二个骰子的点数。每个内部单元格包含有序数对（骰子1, 骰子2）。
| + | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| 2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| 3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| 4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| 5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| 6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
This table makes it clear that there are 36 equally likely joint outcomes. To find the probability that the sum of the two dice is 7, for example, you simply count the number of cells where the sum equals 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) → six outcomes. Therefore P(sum = 7) = 6/36 = 1/6.
该表清楚地表明共有 36 个等可能的联合结果。例如，要计算两个骰子点数之和为 7 的概率，只需数出和为 7 的单元格数量：(1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1) → 六种结果。因此 P(和为7) = 6/36 = 1/6。
5. Calculating Probabilities from a Sample Space | 从样本空间计算概率
Once the sample space is constructed, the probability of any event A is given by: P(A) = (number of outcomes in A) ÷ (total number of outcomes in the sample space). Always check that outcomes are equally likely. If the events are equally likely, this ratio gives the exact theoretical probability. For the coin and die example, P(Tails and even number) = 3/12 = 1/4 because the successful outcomes are T2, T4, T6.
一旦构建了样本空间，任何事件 A 的概率可通过以下公式给出：P(A) =（事件 A 中的结果数）÷（样本空间中的结果总数）。务必检查各结果是否等可能。如果事件等可能，该比值就能给出精确的理论概率。以硬币和骰子为例，P(反面且偶数) = 3/12 = 1/4，因为成功的结果为 T2、T4、T6。
P(Event) = Number of favourable outcomes / Total number of outcomes
P(事件) = 有利结果数 / 总结果数
6. Mutually Exclusive Events | 互斥事件
Two events are mutually exclusive if they cannot happen at the same time. For a single roll of a die, the events “rolling a 3” and “rolling an even number” are mutually exclusive because 3 is not even. When events are mutually exclusive, the probability of either event occurring is the sum of their individual probabilities. This is written as P(A or B) = P(A) + P(B). The table makes it easy to see that no outcome belongs to both events simultaneously.
如果两个事件不可能同时发生，则称它们为互斥事件。对于掷一次骰子，“掷出 3”和“掷出偶数”这两个事件是互斥的，因为 3 不是偶数。当事件互斥时，任一事件发生的概率等于它们各自概率之和，即 P(A 或 B) = P(A) + P(B)。从表格可以清楚地看到，没有任何结果同时属于两个事件。
7. The Sum of Probabilities | 概率之和
A basic rule in probability is that the sum of the probabilities of all mutually exclusive and exhaustive outcomes in a sample space equals 1. For the roll of a die: P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1. This fact can be used to find missing probabilities. If P(rain tomorrow) = 0.3, then P(no rain tomorrow) = 1 – 0.3 = 0.7.
概率的一个基本规则是：样本空间中所有互斥且完备的结果的概率之和等于 1。以掷骰子为例：P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 1。利用这一事实可以求出缺失的概率。如果 P(明天下雨) = 0.3，那么 P(明天不下雨) = 1 – 0.3 = 0.7。
8. Tree Diagrams – Independent Events | 树状图 – 独立事件
A tree diagram shows all possible outcomes of a sequence of events, branching out from a single starting point. For independent events (where one event does not affect the probability of the other), the probabilities along each branch are multiplied. Consider flipping a fair coin twice. The first set of branches gives P(H) = 1/2 and P(T) = 1/2. From each of these, further branches represent the second flip, again with P(H) = 1/2 and P(T) = 1/2. The probability of obtaining two heads is P(H,H) = 1/2 × 1/2 = 1/4.
树状图能展示一系列事件的所有可能结果，从单个起点开始分支。对于独立事件（即一个事件不影响另一个事件的概率），沿各分支的概率应相乘。考虑抛一枚均匀硬币两次。第一组分叉给出 P(H) = 1/2 和 P(T) = 1/2。从每一个分支再分出代表第二次抛掷的分叉，同样 P(H) = 1/2 和 P(T) = 1/2。得到两个正面的概率为 P(H,H) = 1/2 × 1/2 = 1/4。
9. Tree Diagrams – Dependent Events (Without Replacement) | 树状图 – 相关事件（无放回）
When events are dependent, the probability on the second branch changes based on the outcome of the first event. A classic example is drawing two counters from a bag without replacement. Suppose a bag contains 3 red and 2 blue counters. The probability of drawing a red first is 3/5. If a red is taken and not replaced, the bag then contains 2 red and 2 blue, so P(red on second draw) changes to 2/4 = 1/2. The probability of red then red is 3/5 × 2/4 = 6/20 = 3/10. Tree diagrams help visualise these changing probabilities on each branch.
当事件相关时，第二个分支的概率会根据第一个事件的结果而变化。一个经典例子是从袋中无放回地抽取两个筹码。假设袋中有 3 个红色和 2 个蓝色筹码。第一次抽到红色的概率为 3/5。若抽出一个红色且不放回，则袋中剩余 2 红 2 蓝，因此 P(第二次抽到红色) 变为 2/4 = 1/2。连续两次抽到红色的概率为 3/5 × 2/4 = 6/20 = 3/10。树状图有助于直观呈现每个分支上不断变化的概率。
10. Fraction, Decimal and Percentage Probabilities | 分数、小数和百分比表示概率
In KS3, probabilities are expressed as fractions, decimals or percentages. A probability of 1/4 can be written as 0.25 or 25%. All three forms are acceptable, but you must be able to convert between them fluently. When solving problems, leaving the answer as a simplified fraction is often preferred because it shows exactness. For example, P(sum of two dice = 8) = 5/36, which is about 0.1389, but writing 5/36 is more precise and easier to interpret in a tree diagram.
在 KS3 阶段，概率可以用分数、小数或百分比表示。概率 1/4 可写成 0.25 或 25%。这三种形式均可接受，但你必须能够熟练地在它们之间相互转换。解题时，将答案保留为最简分数通常更受欢迎，因为它体现了精确性。例如，P(两个骰子之和为 8) = 5/36，约为 0.1389，但写成 5/36 更为精确，且在树状图中更容易解释。
11. Common Misconceptions | 常见误解
One frequent mistake is adding probabilities when events are not mutually exclusive, resulting in double‑counting. Another is thinking that past outcomes affect future ones in independent trials – for example, believing that after five tails a coin is ‘due’ to land on heads. This is the gambler’s fallacy. In a fair coin toss, P(Heads) remains 1/2 every time. Additionally, always check that your total probability for all possible outcomes adds up to 1; if it does not, a mistake has been made.
一个常见错误是在事件不互斥时将概率相加，导致重复计算。另一个错误是认为在独立试验中过去的结果会影响未来的结果——例如，以为抛硬币连续出现五次反面后，下一次“理应”是正面。这就是赌徒谬误。抛一枚均匀硬币时，每次 P(正面) 始终是 1/2。此外，一定要检查所有可能结果的概率总和是否为 1；如果不是，就说明哪里出错了。
12. Exam Tips and Summary | 考试技巧与小结
When faced with a probability question involving two events, first decide whether a sample space table or a tree diagram is more appropriate. Use tables for two separate items like two dice or a coin and a spinner. Use tree diagrams for sequences of events, especially when ‘without replacement’ is involved. Always label branches clearly with probabilities, and for tree diagrams remember the AND rule (multiply along branches) and the OR rule (add probabilities of different branches). Writing out the sample space fully – even for a small number of outcomes – can prevent careless errors. Finally, simplify fractions and double‑check that your probabilities make sense in the context of the problem.
遇到涉及两个事件的概率问题时，首先要决定样本空间表格还是树状图更合适。对于两个独立的工具，如两个骰子或一枚硬币与一个转盘，适合使用表格；对于事件的先后顺序，尤其是涉及“无放回”时，适合使用树状图。始终在分支上清晰地标注概率，并且对于树状图，记住“且”法则（沿分支相乘）和“或”法则（不同分支的概率相加）。完整写出样本空间——即使结果数量很少——可以避免因粗心导致的错误。最后，化简分数，并检验所得概率在问题情境中是否合理。
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