📚 Probability of Combined Events | 复合事件的概率
When a Cambridge Checkpoint or KS3 question asks for the chance of two or more things happening together, you are working with combined events. This topic builds on basic probability by introducing organised lists, two-way tables and tree diagrams to find the probability of outcomes such as 'a head and a six'.
当剑桥 Checkpoint 或 KS3 的题目要求计算两个或多个事件一起发生的概率时，你正在处理复合事件。本主题在基础概率之上，通过引入有序列表、二维表和树形图来求解诸如"正面朝上且骰子为6"等结果的概率。
The ideas in this unit appear frequently in checkpoint assessments and in later IGCSE work. Being able to list outcomes systematically is just as important as getting the final fraction, because a correct method earns method marks even when an arithmetic slip changes the final answer.
本单元的知识在 Checkpoint 评估以及后续 IGCSE 学习中经常出现。能够系统地列出结果与得到最终分数同样重要，因为即使计算失误改变了最终答案，正确的方法依然能拿到方法分。
1. What is a combined event? | 什么是复合事件？
In probability, a combined event is an event made up of two or more separate events. For instance, if you roll a dice and flip a coin at the same time, the result 'a 6 on the dice and heads on the coin' is one combined outcome. The collection of all possible combined outcomes is called the sample space.
在概率中，复合事件是由两个或多个单一事件组成的事件。例如，如果同时掷一枚骰子并抛一枚硬币，结果"骰子为6且硬币正面朝上"就是一个复合结果。所有可能的复合结果的集合称为样本空间。
The key question to ask yourself is whether the two events happen together ('and') or whether either one of them is acceptable ('or'). Both are combined events, but they are calculated in completely different ways.
你需要问自己的关键问题是：这两个事件是同时发生（"and"，且），还是其中任意一个发生即可（"or"，或）。两者都是复合事件，但计算方法完全不同。
- 'And' combined events: both conditions must be satisfied, e.g. a 6 and a head. | 与型复合事件：两个条件都必须满足，例如骰子为6且硬币正面。
- 'Or' combined events: at least one of the conditions is satisfied, e.g. a 6 or a head. | 或型复合事件：至少满足其中一个条件，例如骰子为6或硬币正面。
- Independent events: the outcome of one event does not affect the other. | 独立事件：一个事件的结果不影响另一个事件。
- Dependent events: the outcome of one event changes the probability of the other, as happens when items are taken without replacement. | 相关事件：一个事件的结果会改变另一个事件的概率，例如不放回地取物时。
g that product down before you start is a quick way to check that you have listed everything.
一次掷骰子有 6 个结果，一次抛硬币有 2 个结果，因此复合样本空间包含 6 × 2 = 12 个等可能结果。在动笔之前先写出这个乘积，是检查自己是否列全了所有结果的快捷方法。
2. The sample space and systematic listing | 样本空间与系统列举
A systematic list or table of all possible outcomes is often called a sample space diagram. The word 'systematic' matters: examiners award marks for an organised approach, not for a random scatter of pairs that happens to be complete.
把所有可能结果系统列出的清单或表格通常称为样本空间图。"系统"二字很重要：考官给分看的是有条理的方法，而不是随意罗列、碰巧凑齐的成对结果。
For two coins, list the first coin's outcome and attach each outcome of the second coin to it. This gives HH, HT, TH, TT — four outcomes, not three.
对于两枚硬币，先列出第一枚硬币的结果，再把第二枚硬币的每个结果依次接上去，得到 HH、HT、TH、TT 共四个结果，而不是三个。
) = 1/4 P(exactly one head) = 2/4 = 1/2 P(no head) = 1/4
A very common error is to treat HT and TH as the same outcome. They are different, because the coins are separate objects and the order in which the results are recorded is fixed in advance. Only if the question asks about 'two identical objects' with no order at all should you collapse them together.
一个非常常见的错误是把 HT 与 TH 当成同一个结果。它们是不同的，因为两枚硬币是不同的物体，而且记录结果的顺序是事先固定的。只有当题目讨论"两个完全相同的物体"且完全不考虑顺序时，才可以把它们合并。
- Write the list in a fixed order (alphabetical, numerical, or first-event then second-event). | 按固定顺序书写清单（字母序、数字序，或先第一事件后第二事件）。
- Count the total number of outcomes before working out any probability. | 在计算任何概率之前，先数清结果的总数。
- Use the total as the denominator of every fraction. | 把这个总数作为每个分数的分母。
- Check your list against the multiplication rule, e.g. 6 × 2 = 12, 2 × 2 = 4. | 用乘法法则核对你的清单，例如 6 × 2 = 12、2 × 2 = 4。
3. Two-way tables | 二维表
A two-way table (also called a contingency table) sorts data by two categories at once. It is the fastest tool when a question gives you a group of people or objects and asks for the probability of a combination of two features.
二维表（也称列联表）同时按两个类别整理数据。当题目给出一组人或物，并要求求两个特征组合的概率时，它是最快的工具。
In the table below, 20 students are classified by gender and by whether they like mathematics. Notice that the row and column totals must agree with the grand total.
下表中，20 名学生按性别以及是否喜欢数学进行分类。注意各行各列的总计必须与总人数一致。
| Likes maths 喜欢数学 | Does not like 不喜欢 | Total 合计 | |
|---|---|---|---|
| Boys 男生 | 8 | 4 | 12 |
| Girls 女生 | 3 | 5 | 8 |
| Total 合计 | 11 | 9 | 20 |
nt is chosen at random, P(boy and likes maths) = 8/20 = 2/5, while P(likes maths) = 11/20.
读一个单元格得到联合概率，读一行或一列的总计得到边际概率。如果随机选一名学生，P(男生且喜欢数学) = 8/20 = 2/5，而 P(喜欢数学) = 11/20。
- Always fill in missing cells using the totals before answering. | 作答前先用总计把缺失的单元格补齐。
- Column headings and row headings must be clearly labelled. | 列标题和行标题必须清晰标注。
- The denominator is the grand total unless the question says 'given that...'. | 除非题目说"已知……"，否则分母取总计值。
- For 'or' questions, add the two totals and subtract the overlapping cell. | 遇到"或"的问题，把两个总计相加再减去重叠的单元格。
4. Tree diagrams | 树形图
A tree diagram shows the stages of a combined event one after another. Each branch carries the probability of that outcome, and the probabilities on the branches leaving any single point must add up to 1.
树形图按先后顺序展示复合事件的各个阶段。每条枝上标出该结果的概率，而从同一个分叉点出发的所有枝上概率之和必须等于 1。
To find the probability of a path, multiply along the branches. To find the probability of two or more different paths, add the results of those paths together.
要求某一条路径的概率，就沿着各枝相乘；要求两条或更多不同路径的合并概率，就把这些路径的结果相加。
P(path) = product of branch probabilities along that path
P(two or more paths) = sum of the path probabilities
pair of branches is each ½, and the second pair of branches is again each ½. The path for 'head then tail' has probability ½ × ½ = ¼, which matches the systematic list.
以抛两次硬币为例，第一层两条枝各为 ½，第二层两条枝同样各为 ½。"先正后反"这条路径的概率为 ½ × ½ = ¼，与系统列举的结果一致。
- Label every branch with a fraction, not just the final answers. | 每条枝都要标上分数，而不是只写最终答案。
- Write the outcome (e.g. HT) at the end of each path. | 在每条路径末端写出结果（如 HT）。
- Check that all final path probabilities add up to 1. This is a powerful self-check. | 检查所有末端路径概率之和是否为 1，这是非常有效的自检方法。
- Do not add along a single path, and do not multiply across different paths. | 不要沿单条路径相加，也不要跨不同路径相乘。
5. Independent events and the multiplication rule | 独立事件与乘法法则
Two events are independent when the outcome of the first has no effect on the probability of the second. Repeated rolls of a dice, repeated flips of a coin and spins of a spinner are all independent.
当第一个事件的结果不影响第二个事件的概率时，两个事件相互独立。重复掷骰子、重复抛硬币以及转动转盘都属于独立事件。
P(A and B) = P(A) × P(B) (A and B independent)
is agrees with the sample space of 12 equally likely outcomes, of which only one is (6, H).
因此 P(骰子为6且硬币正面) = 1/6 × 1/2 = 1/12。这与 12 个等可能结果的样本空间一致，其中只有 (6, H) 这一个结果符合要求。
B and C) = P(A) × P(B) × P(C). Rolling three sixes in a row gives 1/6 × 1/6 × 1/6 = 1/216.
对于三个独立事件，把法则推广即可：P(A 且 B 且 C) = P(A) × P(B) × P(C)。连续掷出三个 6 的概率为 1/6 × 1/6 × 1/6 = 1/216。
- Change percentages and decimals to fractions first if the question demands a fraction answer. | 如果题目要求分数答案，先把百分数和循环小数化成分数。
- Simplify the final fraction, e.g. 6/20 = 3/10. | 最后要化简分数，例如 6/20 = 3/10。
- Never add the probabilities for an 'and' question. | 遇到"且"的问题，绝不要把概率相加。
- Check that the answer is less than each individual probability. | 检查答案是否小于每个单独的概率。
6. Mutually exclusive events and the addition rule | 互斥事件与加法法则
Two events are mutually exclusive if they cannot both happen at the same time. On a single dice roll, 'getting a 2' and 'getting a 5' are mutually exclusive, because one roll cannot show both numbers.
如果两个事件不能同时发生，它们就是互斥的。在一次掷骰子中，"掷出 2"与"掷出 5"是互斥的，因为一次投掷不可能同时出现两个数字。
P(A or B) = P(A) + P(B) (A and B mutually exclusive)
he events can overlap, the general addition rule applies and you must subtract the overlap:
因此 P(掷出2或掷出5) = 1/6 + 1/6 = 2/6 = 1/3。如果两个事件可能重叠，则需要使用一般加法法则，并减去重叠部分：
P(A or B) = P(A) + P(B) − P(A and B)
For a fair dice, let A be 'an even number' and B be 'a number greater than 4' nly the 6). The answer is 3/6 + 2/6 − 1/6 = 4/6 = 2/3, which matches the set {2, 4, 5, 6}.
对于一枚公平骰子，设 A 为"偶数"，B 为"大于 4 的数"。则 P(A) = 3/6，P(B) = 2/6，P(A 且 B) = 1/6（只有 6）。答案为 3/6 + 2/6 − 1/6 = 4/6 = 2/3，与集合 {2, 4, 5, 6} 一致。
- Ask whether the two events could happen simultaneously before adding. | 相加之前先问：这两个事件可能同时发生吗？
- If they can overlap, always subtract the intersection. | 如果可以重叠，一定要减去交集。
- The answer must never exceed 1; if it does, you have forgotten to subtract. | 答案绝不能大于 1；若超过，说明你忘了减去重叠部分。
- 'Or' questions can also be solved by counting outcomes directly in the sample space. | "或"的问题也可以通过直接在样本空间中数结果来解决。
7. The complement rule and 'at least one' | 补集法则与"至少一次"
not A, written A′. Since the probabilities of all outcomes add up to 1, P(A′) = 1 − P(A).
事件 A 的补集是除 A 之外的所有情况，记作 A′。由于所有结果的概率之和为 1，所以 P(A′) = 1 − P(A)。
P(not A) = 1 − P(A)
The complement rule is the fastest way to answer 'at least one' questions. Instead of adding up all the cases with one, two or three successes, calculate the single case with no successes and subtract it from 1.
补集法则是回答"至少一次"问题的最快方法。与其把一次、两次、三次成功的所有情况相加，不如只计算完全没有成功的那一种情况，再用 1 减去它。
Example: a fair coin is flipped twice. Find the probability of at least one head.
P(no head) = ½ × ½ = ¼, so P(at least one head) = 1 − ¼ = ¾.
例：公平硬币抛两次，求至少出现一次正面的概率。
P(没有正面) = ½ × ½ = ¼，所以 P(至少一次正面) = 1 − ¼ = ¾。
- Words that signal the complement rule: 'at least one', 'not', 'none', 'neither'. | 提示使用补集法则的关键词：至少一个、不是、没有、两者都不。
- Be careful: 'at least one' means one or more, which is different from 'exactly one'. | 注意："至少一个"表示一个或更多，与"恰好一个"不同。
- 'Exactly one' must be counted path by path and then added. | "恰好一个"必须逐条路径计数后再相加。
- Write the complement calculation as a line of working to secure the method mark. | 把补集计算作为一行步骤写出，以拿到方法分。
8. Without replacement: dependent combined events | 不放回：相关复合事件
When objects are taken from a bag, box or deck and are not put back, the second probability changes. The denominator decreases by 1, and the numerator decreases if the same type of object was taken first.
当物体从袋子、盒子或一副牌中取出后不放回时，第二个概率会改变。分母减少 1；如果第一次取走的是同类物体，分子也随之减少。
Example: a bag contains 3 red and 2 blue balls. Two balls are taken without replacement. Find P(both red).
例：袋中有 3 个红球和 2 个蓝球，不放回地取出两个球，求 P(两个都是红球)。
P(both red) = 3/5 × 2/4 = 6/20 = 3/10
On a tree diagram, the second set of branches is different depending on which branch you are on: after a red ball there are 2 red and 2 blue left, while after a blue ball there are 3 red and 1 blue left. Both second layers must still sum to 1 along each branch point.
在树形图上，第二层枝的概率取决于你处在哪条枝上：取走红球后剩下 2 红 2 蓝，取走蓝球后剩下 3 红 1 蓝。两个第二层的每组枝在各自分叉点上的概率之和仍须为 1。
- Look for the phrase 'without replacement' or 'eats one' or 'does not put it back'. | 留意"不放回""吃掉一个""没有放回去"等表述。
- If the object is replaced, the events are independent and both fractions stay the same. | 如果放回，事件相互独立，两个分数保持不变。
- P(different colours) requires two paths: RB and BR, added together. | P(颜色不同) 需要两条路径：红蓝与蓝红，然后相加。
- Check totals: after drawing one red, there are 4 balls left, not 5. | 检查总数：取出一个红球后还剩 4 个球，而不是 5 个。
9. Common mistakes in Checkpoint questions | Checkpoint 题目中的常见错误
Most lost marks in this topic come from a small set of repeated errors. Reading through the list below before an assessment is worth several marks.
本主题中大部分失分都来自少数反复出现的错误。考前通读下面这份清单，往往能多拿好几分。
- Confusing 'and' with 'or'. 'And' usually means multiply; 'or' usually means add (with the overlap removed). | 混淆"且"与"或"。"且"通常用乘法；"或"通常用加法（并减去重叠）。
- Forgetting that HT and TH are different. Order matters unless the question says otherwise. | 忘记 HT 与 TH 是不同结果。除非题目另有说明，顺序是有意义的。
- Not changing the denominator without replacement. The second fraction must use the reduced total. | 不放回时没有改变分母。第二个分数必须使用减少后的总数。
- Adding branch probabilities along one path. Multiplication is needed along a path. | 沿单条路径把枝概率相加。沿路径应使用乘法。
- Forgetting to check that path probabilities total 1. This check catches most slips immediately. | 忘记检查路径概率之和是否为 1。这个检查能立刻发现大部分失误。
- Leaving the answer as an unsimplified fraction. Always simplify, e.g. 8/20 = 2/5. | 答案分数没有化简。一定要约分，例如 8/20 = 2/5。
- Writing a probability greater than 1. This signals an error, usually a forgotten subtraction. | 写出大于 1 的概率。这说明出错了，通常是忘记做减法。
10. Worked exam-style examples | 考试风格例题精讲
Work through the three examples below, covering a sample space, a tree diagram and a dependent event. Each one shows the working an examiner expects to see.
请完成下面三道例题，分别覆盖样本空间、树形图和相关事件。每题都展示了考官期望看到的步骤。
Example 1. A dice is rolled and a coin is flipped. Find P(a number greater than 4 and a tail).
n 4 are 5 and 6, so the probability is 2/6 = 1/3. P(tail) = 1/2. Then P = 1/3 × 1/2 = 1/6.
例 1. 掷一枚骰子并抛一枚硬币，求 P(数字大于 4 且硬币反面)。
大于 4 的结果是 5 和 6，概率为 2/6 = 1/3；P(反面) = 1/2。因此 P = 1/3 × 1/2 = 1/6。
Example 2. A fair coin is flipped three times. Find P(exactly two heads).
, so P = 3/8. Alternatively, add the three paths on a tree diagram: 1/8 + 1/8 + 1/8 = 3/8.
例 2. 公平硬币抛三次，求 P(恰好两次正面)。
共有 2 × 2 × 2 = 8 个等可能结果。符合条件的是 HHT、HTH、THH，所以 P = 3/8。也可以在树形图上把三条路径相加：1/8 + 1/8 + 1/8 = 3/8。
Example 3. A bag has 4 green and 6 yellow counters. Two are taken without replacement. Find P(both the same colour).
/9 = 12/90 = 2/15. P(YY) = 6/10 × 5/9 = 30/90 = 1/3. Add: 2/15 + 1/3 = 2/15 + 5/15 = 7/15.
例 3. 袋中有 4 个绿筹码和 6 个黄筹码，不放回地取出两个，求 P(两个颜色相同)。
P(绿绿) = 4/10 × 3/9 = 12/90 = 2/15；P(黄黄) = 6/10 × 5/9 = 30/90 = 1/3。相加：2/15 + 1/3 = 2/15 + 5/15 = 7/15。
11. Choosing the right method | 选择正确的方法
Deciding on a method before writing anything saves time and prevents muddled working. Use the table below as a rapid decision guide.
动笔之前先确定方法，可以节省时间并避免步骤混乱。下面的表格可作为快速决策指南。
| Question clue 题目线索 | Best method 最佳方法 |
|---|---|
| Two features, data given in a group 两个特征，给出整组数据 | Two-way table 二维表 |
| Two or three stages in a row 连续两三个阶段 | Tree diagram 树形图 |
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