📚 Probability – Tree Diagrams for Combined Events | 复合事件的概率树状图
Imagine you flip a coin and roll a dice at the same time. How many different results can you get, and what is the chance of getting a head and a six? Tree diagrams are powerful tools that help us organise and calculate probabilities for such combined events. In this article, you will learn how to draw tree diagrams, use the AND and OR rules, and solve problems involving independent events, all with plenty of examples that match the Cambridge KS3 Mathematics syllabus.
想象一下，你同时抛一枚硬币并掷一个骰子。你可能得到多少种不同的结果？得到正面朝上和数字六的概率又是多少？树状图是一种强大的工具，它能够帮助我们理清并计算这类复合事件的概率。在这篇文章中，你将学会如何绘制树状图，使用“与”规则和“或”规则，并解决涉及独立事件的问题，所有例子都与剑桥 KS3 数学大纲紧密贴合。
1. What Are Combined Events? | 什么是复合事件？
A combined event happens when we perform two or more single events together, such as rolling two dice, flipping two coins, or picking coloured balls from a bag twice. The outcome of the whole experiment is made up of the outcomes of each individual event. In KS3 probability, we often need to list all possible outcomes or calculate the chance of a particular combination occurring.
复合事件是指我们将两个或多个单一事件联合起来进行的实验，例如同时掷两颗骰子、抛两枚硬币，或者从一个袋子里取两次彩色球。整个实验的结果由每个单独事件的结果组合而成。在 KS3 的概率学习中，我们经常需要列出所有可能的结果，或者计算某一特定组合发生的概率。
2. Representing Outcomes with Sample Spaces | 用样本空间表示结果
Before drawing tree diagrams, it is useful to understand sample spaces. A sample space is the set of all possible outcomes. For a single coin flip, the sample space is {Head, Tail}. For rolling a fair six-sided dice, it is {1, 2, 3, 4, 5, 6}. When we combine two such events, we can list outcomes in a two-way table or a grid. However, tree diagrams often give a clearer picture, especially when events happen one after another.
在绘制树状图之前，了解样本空间是很有用的。样本空间是所有可能结果的集合。对于单次抛硬币，样本空间是{正面，反面}。对于掷一个均匀的六面骰子，样本空间是{1, 2, 3, 4, 5, 6}。当我们组合两个这样的事件时，可以用双向表格或网格来列出结果。然而，树状图通常能给出更清晰的示意，特别是当事件连续发生时。
3. Introducing Tree Diagrams | 树状图简介
A tree diagram uses branches to show the possible outcomes of each stage in a sequence of events. Each branch represents a single outcome from that stage, and we label it with its probability. The diagram grows from left to right, with new branches for the next event. The final outcomes are shown at the tips of the last set of branches. Tree diagrams make it very easy to apply the multiplication and addition rules of probability.
树状图使用分支来表示一连串事件中每一步可能的结果。每个分支代表该阶段的一个单一结果，并在其上标注它的概率。图表从左到右展开，为下一事件增添新的分支。最终的结果显示在最后一组分支的末端。树状图使得概率的乘法和加法规则应用起来非常容易。
4. How to Draw a Tree Diagram | 如何绘制树状图
To draw a tree diagram for two events: first, identify the outcomes of the first event and draw a branch for each outcome, writing the probability along the branch. From the end of each first-stage branch, draw branches for all the outcomes of the second event, again with their probabilities. For example, when flipping a coin twice, the first flip has two branches (Head with probability ½, Tail with probability ½). From each of these, two branches show the second flip (again Head or Tail, each ½). The diagram ends with four final outcomes: HH, HT, TH, TT.
绘制两个事件的树状图：首先，确认第一个事件的所有结果，为每个结果画一条分支，沿着分支写出概率。然后，从每个第一阶段分支的末端，为第二个事件的所有可能结果画出分支，同样标注概率。例如，在抛两次硬币的实验中，第一次抛有两条分支（正面概率½，反面概率½）。从这两条分支各自再延伸出两条分支代表第二次抛（还是正面或反面，各½）。图表最终显示四种结果：HH，HT，TH，TT。
5. Calculating Probabilities from a Tree Diagram | 从树状图计算概率
To find the probability of a specific final outcome, multiply the probabilities along the branches that lead to that outcome. For the coin example, the probability of getting two Heads (HH) is ½ × ½ = ¼. This works because the two coin flips are independent events – one does not affect the other. When several paths give the same type of outcome (e.g., exactly one Head), you add the probabilities of those paths together.
要计算某个特定最终结果的概率，就将通往该结果的分支上所标注的概率乘起来。在抛硬币的例子里，得到两个正面的概率是½ × ½ = ¼。这样计算是可行的，因为两次抛硬币是独立事件——一次的结果不影响另一次。当有几条路径产生同类型的结果时（比如恰好出现一次正面），你需要把这些路径的概率相加。
6. The AND Rule (Multiplication) | “与”规则（乘法）
When we want both event A and event B to happen in a sequence, we multiply the probabilities along the relevant branches. This is called the AND rule or multiplication rule. It is used for independent events (where the outcome of A does not change the probability of B). The general form is: P(A and B) = P(A) × P(B). On a tree diagram, you simply read along one complete path from start to finish.
当我们希望在一连串事件中事件A与事件B同时发生时，我们将相关分支上的概率相乘。这被称为“与”规则或乘法规则。它用于独立事件（即A的结果不改变B的概率的情形）。通用公式为：P(A与B) = P(A) × P(B)。在树状图上，你只需要沿着一条从起点到终点的完整路径读出并相乘即可。
7. The OR Rule (Addition) | “或”规则（加法）
Sometimes we want to know the probability of outcome X or outcome Y occurring, where X and Y are different final outcomes that cannot happen at the same time (mutually exclusive). For example, in two coin flips, what is the chance of getting exactly one Tail? This could happen via HT (Head then Tail) or TH (Tail then Head). These paths are separate, so we add their probabilities: ¼ + ¼ = ½. This is the OR rule: P(X or Y) = P(X) + P(Y), provided X and Y are mutually exclusive.
有时我们想知道结果X或结果Y发生的概率，而X和Y是不同的最终结果且不能同时发生（互斥）。例如在两次抛硬币中，刚好得到一次反面的概率是多少？这可以通过 HT（先正后反）或 TH（先反后正）实现。这些路径是分开的，因此我们把它们的概率加起来：¼ + ¼ = ½。这就是“或”规则：P(X或Y) = P(X) + P(Y)，前提是X与Y为互斥事件。
8. Independent Events and Replacement | 独立事件与放回问题
When dealing with items like coloured counters in a bag, we must check whether we replace the counter after the first pick. If we replace it (put it back), the probabilities for the second pick remain exactly the same as the first – the events are independent. A tree diagram for “with replacement” will show identical probabilities on the second-stage branches for each first-stage outcome. For instance, picking a red counter from a bag with 3 red and 2 blue is 3/5. If we replace it, the second pick is still 3/5 for red, regardless of the first result.
在处理诸如袋中彩色筹码之类的问题时，我们必须检查第一次取后是否放回。如果我们放回去，第二次取时的概率与第一次完全一样——事件是独立的。一个“放回”的树状图会在每个第一阶段结果之后的第二阶段分支上显示相同的概率。例如，从一个装有3红2蓝的袋子里取出一枚红色筹码的概率是3/5。如果放回，无论第一次结果如何，第二次取到红色的概率依然是3/5。
9. Conditional Probability (Without Replacement) | 条件概率（不放回）
If we do not replace the first item, the probabilities for the second event change depending on what was taken out. This is called conditional probability – the chance of event B given that event A has happened. The tree diagram still works, but you must update the fractions on the second-stage branches. In the bag of 3 red and 2 blue, if you take one red and keep it out, only 2 red and 2 blue remain. So the probability of red on the second pick becomes 2/4 = ½, whereas if a blue was taken first, the second pick probabilities are 3/4 red and 1/4 blue. Understanding these changes is crucial for accurate calculations.
如果我们不放回第一个取出的物品，第二次事件的概率会根据第一次取出的结果而变化。这被称为条件概率——在事件A已经发生的情况下事件B的概率。树状图依然有效，但你必须更新第二阶段分支上的分数。在装有3红2蓝的袋子里，如果你取出一个红色并放在一边，就只剩下2红2蓝。所以第二次取到红色的概率变为2/4 = ½，而如果第一次取出的是蓝色，第二次的概率就是3/4红色和1/4蓝色。理解这些变化对于准确计算至关重要。
10. Real-Life Examples | 实际生活中的例子
Tree diagrams appear in many everyday situations. A restaurant menu might offer a choice of starter and main course; the probability that a customer who chooses at random picks soup and then steak can be found using a tree diagram. In sports, the chance of a football team winning a match and then winning a penalty shootout can be modelled. Even weather forecasting uses tree-like structures: the probability of rain today and rain tomorrow is a simple combined event.
树状图出现在许多日常情境中。一份餐厅菜单可能提供前菜和主菜的选择；一位随机选择的顾客同时点了汤和牛排的概率就可以用树状图来求。在体育中，一支足球队赢得比赛又赢下点球大战的机会可以用树状图建模。甚至天气预报也用到类似树状的结构：今天下雨并且明天也下雨的概率就是一个简单的复合事件。
11. Common Mistakes to Avoid | 常见错误要避免
Many students forget to update probabilities when there is no replacement, leading to wrong answers. Others add all branch probabilities at the final tips instead of multiplying along a path. Make sure you only add probabilities of different paths when you are looking for an “OR” scenario. Also, always check that the probabilities on branches coming from a single point add up to 1. If they do not, there is an error. Finally, be careful not to double count outcomes when using the addition rule – the outcomes must be mutually exclusive.
很多学生忘记在不放回的情况下更新概率，从而导致答案错误。还有一些学生把末端的所有分支概率直接相加，而不是沿着一条路径相乘。务必确保只在寻找“或”的情境时才将不同路径的概率相加。另外，始终检查从一个节点发出的所有分支的概率之和是否为1。如果不是，说明有错误。最后，小心在使用加法规则时不要重复计算——结果必须是互斥的。
12. Practice Question and Solution | 练习题与解答
Question: A bag contains 4 white and 6 black balls. Two balls are drawn one after the other without replacement. Draw a tree diagram and find the probability that both balls are white, and the probability that exactly one is black.
Solution: First draw: White = 4/10 = 2/5, Black = 6/10 = 3/5. If first is White, remaining: 3 white, 6 black, so P(White second) = 3/9 = 1/3, P(Black second) = 6/9 = 2/3. If first is Black, remaining: 4 white, 5 black, so P(White second) = 4/9, P(Black second) = 5/9. P(both white) = (2/5) × (1/3) = 2/15. Exactly one black can happen via (White then Black) or (Black then White): (2/5 × 2/3) = 4/15 and (3/5 × 4/9) = 12/45 = 4/15. Total = 4/15 + 4/15 = 8/15.
题目：一个袋子里有4个白球和6个黑球。连续抽出两个球，不放回。画出树状图，并求出两个都是白球的概率，以及恰好有一个是黑球的概率。
解答：第一次抽：白球 = 4/10 = 2/5，黑球 = 6/10 = 3/5。若第一次是白球，剩余：3白6黑，所以P(第二次白)=3/9=1/3，P(第二次黑)=6/9=2/3。若第一次是黑球，剩余：4白5黑，所以P(第二次白)=4/9，P(第二次黑)=5/9。P(两个都是白球) = (2/5) × (1/3) = 2/15。恰好有一个黑球可以通过（先白后黑）或（先黑后白）实现：(2/5 × 2/3) = 4/15，以及(3/5 × 4/9) = 12/45 = 4/15。总和 = 4/15 + 4/15 = 8/15。
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