Page 291 – Mastering Probability | 第291页 – 掌握概率

📚 Page 291 – Mastering Probability | 第291页 – 掌握概率

In your Cambridge KS3 Mathematics course, page 291 brings together key ideas about probability that you will need to master for Checkpoint and beyond. This article walks you through every concept step by step, from the probability scale to tree diagrams, making sure you are fully prepared for any question the exam might throw at you.

在剑桥 KS3 数学课程中，第 291 页汇集了从概率标度到树状图的一系列关键知识点，对 Checkpoint 考试和后续学习都至关重要。本文将带你逐步理解每一个概念，让你能够从容应对考试中可能出现的任何题型。

1. What Is Probability? | 什么是概率？

Probability is a measure of how likely an event is to happen. It can be written as a fraction, a decimal or a percentage, and it always lies between 0 and 1 inclusive. An event with probability 0 is impossible, while an event with probability 1 is certain. For example, when you toss a fair coin, the probability of landing on heads is 1/2, 0.5 or 50%.

概率是衡量事件发生可能性的指标。它可以写成分数、小数或百分数，并且总是在 0 到 1 之间（含 0 和 1）。概率为 0 的事件是不可能发生的，概率为 1 的事件是必然发生的。例如，抛一枚公平的硬币时，正面朝上的概率是 1/2、0.5 或 50%。

In any single trial, the sum of probabilities for all possible outcomes must equal 1. If the probability of rain tomorrow is 0.3, then the probability that it does not rain is 1 – 0.3 = 0.7. This is called the complement rule.

在任何一次试验中，所有可能结果的概率之和必须等于 1。如果明天下雨的概率是 0.3，那么不下雨的概率就是 1 – 0.3 = 0.7。这被称为互补法则。

2. The Probability Scale | 概率标度

The probability scale is a number line from 0 to 1 that helps you describe the likelihood of events using words such as impossible, unlikely, even chance, likely and certain. At KS3 you will be expected to place events accurately on this scale and to assign numerical values to everyday language.

概率标度是一条从 0 到 1 的数轴，可以帮助你用“不可能”“不太可能”“等可能性”“很可能”“必然”等词语来描述事件的可能性。在 KS3 阶段，你需要能够将事件准确地放置在这个标度上，并用数值来表达日常用语。

Word	Value	Example
Impossible	0	Rolling a 7 on a standard six-sided dice
Unlikely	between 0 and 1/2	Picking the only red sock from a drawer of 20 socks
Even chance	1/2	Getting heads when flipping a coin
Likely	between 1/2 and 1	Selecting a white bead from a bag containing 8 white and 2 black beads
Certain	1	The sun rising tomorrow

Being able to link numbers and words is something the Cambridge Checkpoint tests often look for. You will see statements such as ‘the probability it will snow is 0.9’ and need to choose the best description, which in this case is ‘very likely’.

剑桥 Checkpoint 考试经常考查数字与词语的对应关系。你会看到“下雪的概率是 0.9”这样的陈述，并需要选择最合适的描述，此处应为“非常可能”。

3. Systematic Listing of Outcomes | 系统列出结果

Before you can calculate probabilities, you need to know exactly what all the possible outcomes are. For simple experiments, such as flipping two coins, you can list outcomes systematically. The four outcomes are: HH, HT, TH, TT. Notice that HT and TH are different because the coins are distinct.

在计算概率之前，你需要确切地知道所有可能的结果有哪些。对于抛两枚硬币这样的简单试验，你可以系统地列出结果。四种结果是：HH、HT、TH、TT。请注意，HT 和 TH 是不同的，因为两枚硬币是独立的。

For three coins, there are 2³ = 8 outcomes. A sensible order helps you avoid missing any: HHH, HHT, HTH, HTT, THH, THT, TTH, TTT. When outcomes are equally likely, the probability of an event is the number of favourable outcomes divided by the total number of outcomes.

对于三枚硬币，则有 2³ = 8 种结果。有条理的顺序能帮助你避免遗漏：HHH、HHT、HTH、HTT、THH、THT、TTH、TTT。当所有结果等可能时，事件的概率等于有利结果的数量除以所有结果的总数。

If you are asked for the probability of getting exactly two heads from three coins, you count the outcomes with two Hs (HHT, HTH, THH) – that is 3 favourable outcomes out of 8, so the probability is 3/8.

如果要求抛三枚硬币恰好得到两个正面的概率，你需要数出有两个 H 的结果（HHT、HTH、THH）——即 8 种结果中有 3 种有利，因此概率是 3/8。

4. Sample Space Diagrams | 样本空间图

A sample space diagram is a table that shows all possible outcomes when two events happen together, such as rolling two dice. Draw a 6 by 6 grid, label rows 1–6 for the first dice and columns 1–6 for the second. Each cell represents one of the 36 equally likely outcomes.

样本空间图是一个显示两个事件同时发生时所有可能结果的表格，例如投掷两个骰子。画一个 6×6 的网格，用 1–6 标记第一枚骰子的行，用 1–6 标记第二枚骰子的列。每个格子代表 36 种等可能结果中的一种。

You can then count cells to find probabilities. For the sum of two dice being 7, the cells corresponding to (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) give 6 favourable outcomes. So P(sum = 7) = 6/36 = 1/6. This diagram is particularly helpful on page 291 style questions where a clear visual method is required.

然后你可以通过数格子来求概率。对于两个骰子点数之和为 7，对应 (1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1) 的格子共有 6 个有利结果，因此 P(和为 7) = 6/36 = 1/6。这种图表在第 291 页那种需要清晰视觉方法的题目中尤其有用。

Always write probabilities in their simplest form unless the question says otherwise. 6/36 simplifies to 1/6. This shows good mathematical practice.

除非题目另有要求，否则总是把概率化为最简形式。6/36 化简为 1/6，这体现了良好的数学规范。

5. Theoretical vs. Experimental Probability | 理论概率与实验概率

Theoretical probability is what we expect to happen based on equally likely outcomes – the kind you calculate using fractions. Experimental probability (or relative frequency) is what actually happens when you carry out an experiment multiple times. The formula is: experimental probability = number of times the event occurs ÷ total number of trials.

理论概率是我们基于等可能结果期望发生的情况——即用分数计算的那种概率。实验概率（或相对频率）是当你多次进行实验时实际发生的情况。公式为：实验概率 = 事件发生的次数 ÷ 试验总次数。

If you spin a spinner 50 times and it lands on red 12 times, the experimental probability of red is 12/50 = 0.24. The theoretical probability might be 0.2 if the spinner is fairly divided. The more trials you do, the closer experimental probability tends to get to the theoretical value. This is known as the law of large numbers.

如果你转动一个转盘 50 次，其中有 12 次停在红色上，那么红色的实验概率为 12/50 = 0.24。如果转盘是均匀分割的，理论概率可能是 0.2。你做的试验次数越多，实验概率就越接近理论值。这就是所谓的大数定律。

Page 291 often tests whether students can compare the two. For example, a question might give results of a dice-rolling experiment and ask you to explain why the experimental probability for getting a six is not exactly 1/6 after 30 rolls.

第 291 页常常考查学生是否能够比较这两种概率。例如，题目可能给出掷骰子实验的结果，并要求你解释为什么掷 30 次后得到 6 点的实验概率不正好是 1/6。

6. Introduction to Tree Diagrams | 树状图入门

A tree diagram is a branching diagram that shows all the possible outcomes of one or more events and their probabilities. Each branch represents a possible outcome, and you label it with the probability. At the end of each path, you can calculate the combined probability by multiplying along the branches.

树状图是一种显示一个或多个事件的所有可能结果及其概率的分支图。每条分支代表一个可能的结果，并标有概率。在每条路径的末端，你可以沿着分支将概率相乘，得到组合概率。

Tree diagrams are extremely useful for solving problems involving successive events, such as picking balls from a bag with or without replacement. They keep your work organised and help you avoid missing outcomes.

树状图对于解决涉及连续事件的问题非常有用，比如从袋子中有放回或无放回地取球。它们能让你的解答条理清晰，并帮助你避免遗漏结果。

When drawing a tree, always put the event that happens first on the left. For two events, you draw two sets of branches. Each set of branches must sum to 1 because one of those outcomes must happen.

在画树状图时，总是将先发生的事件放在左边。对于两个事件，你要画两组分支。每组分支的概率之和必须为 1，因为其中一个结果必定发生。

7. Tree Diagrams for Independent Events | 独立事件的树状图

Two events are independent if the outcome of the first does not affect the outcome of the second. An example is flipping a coin and rolling a dice. The probability of getting heads and then rolling a 4 is (1/2) × (1/6) = 1/12. For a tree diagram with independent events, the probabilities on the second set of branches stay the same regardless of the first outcome.

如果第一个事件的结果不影响第二个事件的结果，则这两个事件是独立的。例如抛硬币和掷骰子。得到正面并掷出 4 点的概率是 (1/2) × (1/6) = 1/12。对于独立事件的树状图，无论第一个结果如何，第二组分枝上的概率保持不变。

Suppose a bag contains 3 red and 5 blue counters. You pick one, replace it, then pick again. Because you replace the counter, the probabilities remain the same: P(red) = 3/8, P(blue) = 5/8 on both picks. The probability of picking two reds is 3/8 × 3/8 = 9/64.

假设一个袋子里有 3 个红色和 5 个蓝色筹码。你取出一个放回，然后再取一次。因为有放回，所以每次取的概率保持不变：P(红) = 3/8，P(蓝) = 5/8。取出两个红色筹码的概率为 3/8 × 3/8 = 9/64。

In a tree diagram, you would list all four final outcomes – RR, RB, BR, BB – and their probabilities. Always check that these add up to 1. In this case, 9/64 + 15/64 + 15/64 + 25/64 = 64/64 = 1.

在树状图中，你要列出所有四种最终结果——RR、RB、BR、BB——以及它们的概率。务必检查这些概率之和是否为 1。在此例中，9/64 + 15/64 + 15/64 + 25/64 = 64/64 = 1。

8. Tree Diagrams for Dependent Events | 相关事件的树状图

When events are dependent, the second outcome is affected by the first. This occurs when you do not replace the item after the first pick. The probabilities on the second branches must change to reflect the new situation.

当事件是相关的时，第二个结果会受第一个结果的影响。这种情况发生在第一次取出物品后不放回时。第二组分枝上的概率必须改变，以反映新的情形。

Using the same bag of 3 red and 5 blue counters without replacement: probability of red first is 3/8. If red is taken, there are now 2 red and 5 blue left, so P(red second) = 2/7. The probability of RR is (3/8) × (2/7) = 6/56 = 3/28.

用同一个装有 3 红 5 蓝筹码的袋子，这一次不放回：第一次取到红色的概率是 3/8。如果取走一个红色，剩下 2 红 5 蓝，所以 P(第二次是红) = 2/7。得到 RR 的概率是 (3/8) × (2/7) = 6/56 = 3/28。

For dependent events, the second set of branches depends on which path you are on. You must calculate the new total and the new number of favourable items after each first outcome. The tree helps you keep track of these changing probabilities, which is why it is so heavily featured on page 291 practice questions.

对于相关事件，第二组分枝取决于你当前所在的路径。你必须根据每种第一次结果计算新的总数和新的有利物品数量。树状图能帮助你跟踪这些变化的概率，这也是为什么第 291 页的练习题大量使用了树状图。

9. Calculating Probabilities from Tree Diagrams | 利用树状图计算概率

To find the probability of a combination of outcomes, you multiply along the branches. If the question asks for the probability of at least one of something, you often add the probabilities of the relevant final outcomes. For example, to find the probability of getting at least one red in two picks from the previous dependent tree, you could add the probabilities of RR, RB and BR.

要计算某一结果组合的概率，你需要沿着分支相乘。如果题目要求“至少有一次”的概率，你通常需要将相关最终结果的概率相加。例如，在前面相关事件的树中求两次取球中至少有一次是红色的概率，你需要将 RR、RB 和 BR 的概率相加。

Sometimes it is quicker to subtract from 1. P(at least one red) = 1 – P(no reds). No reds means BB. In the dependent bag, P(BB) = (5/8) × (4/7) = 20/56 = 5/14. Therefore P(at least one red) = 1 – 5/14 = 9/14.

有时用 1 减去反面概率来得更快。P(至少一个红) = 1 – P(没有红)。没有红就是 BB。在相关事件的袋子中，P(BB) = (5/8) × (4/7) = 20/56 = 5/14，因此 P(至少一个红) = 1 – 5/14 = 9/14。

The ‘subtract from 1’ method is a favourite in Cambridge assessments. Always check whether there are many paths to add or if one complementary path gives a simpler calculation.

“用 1 减”的方法是剑桥考试中的常用技巧。请始终检查是需要相加多条路径，还是用一条互补路径计算起来更简单。

10. Frequency Trees | 频率树

A frequency tree is similar to a probability tree but it shows the number of times each outcome happens, rather than probabilities. You start with a total number and break it down using the branches. No probabilities are written on the branches; instead you write frequencies.

频率树与概率树相似，但它显示的是每种结果发生的次数，而不是概率。你从一个总数出发，利用分支进行分解。分支上不写概率，而是写频数。

For example, 80 people were asked if they like tea or coffee. 50 said tea. Of those 50, 20 also like coffee. Of the 30 who did not choose tea, 5 like coffee. A frequency tree organises this information clearly and lets you answer questions such as ‘how many people like coffee?’ (20 + 5 = 25) or ‘what fraction of those who like coffee also like tea?’ (20/25 = 4/5).

例如，询问 80 人是否喜欢茶或咖啡。50 人表示喜欢茶。在这 50 人中，20 人也喜欢咖啡。在未选择茶的 30 人中，5 人喜欢咖啡。频率树能够清晰地组织这些信息，并让你回答诸如“有多少人喜欢咖啡？”（20 + 5 = 25）或“喜欢咖啡的人中，喜欢茶的人占多少比例？”（20/25 = 4/5）之类的问题。

Page 291 may include a frequency tree question to see if you can move between frequencies and probabilities. Remember, probability = frequency for that group / total in the relevant branch.

第 291 页可能会有一道频率树题目，用以考查你是否能够在频数和概率之间转换。记住，概率 = 该组的频数 / 相关分支的总数。

11. Common Mistakes on Page 291 | 第 291 页常见错误

One common mistake is not simplifying fractions. Writing 4/8 instead of 1/2 can lose marks in a ‘final answer’ box. Another is forgetting that probabilities on a tree branch must add up to 1 – always check that they do.

一个常见的错误是没有化简分数。在“最终答案”栏中写 4/8 而不是 1/2 会被扣分。另一个错误是忘记树状图每组分枝的概率相加必须等于 1——请务必检查这一点。

With dependent events, students often forget to change the denominator on the second pick. If the first sweet is eaten, the total goes down by one, and that must be reflected in the fraction. Drawing a clear tree with updated totals helps you avoid this error.

在处理相关事件时，学生常常忘记在第二次取物时改变分母。如果第一颗糖被吃掉，总数就减了一，这必须在分数中体现出来。画一个带有更新后总数的清晰树状图可以帮助你避免这一错误。

Another slip is confusing ‘at least one’ with ‘exactly one’. ‘At least one red’ includes two reds, whereas ‘exactly one red’ only looks at one red and one of the other colour. Read the question wording carefully.

另一个失误是把“至少一个”与“恰好一个”混淆。“至少一个红色”包含两个红色，而“恰好一个红色”只考虑一红一其他颜色。请仔细阅读题目的措辞。

12. Page 291 Practice Questions – Try It Yourself | 第 291 页练习题 – 亲自试一试

Let’s work through a typical page 291 question. A bag contains 4 red and 6 green buttons. Two buttons are picked at random without replacement. (a) Draw a tree diagram. (b) Find the probability that the two buttons are the same colour. (c) Find the probability that at least one is red.

我们来解决一道典型的第 291 页题目。一个袋子装有 4 个红色和 6 个绿色纽扣。随机抽取两个纽扣，不放回。(a) 画出树状图。(b) 求两个纽扣颜色相同的概率。(c) 求至少有一个是红色的概率。

For the tree, first pick: P(red) = 4/10, P(green) = 6/10. Second branches: after red, left 3 red, 6 green → total 9, so P(red) = 3/9, P(green) = 6/9. After green, left 4 red, 5 green → total 9, so P(red) = 4/9, P(green) = 5/9. Same colour means RR or GG. P(RR) = (4/10)×(3/9)=12/90=2/15. P(GG) = (6/10)×(5/9)=30/90=1/3. Total same colour = 2/15 + 1/3 = 2/15 + 5/15 = 7/15.

树状图：第一次取物，P(红) = 4/10，P(绿) = 6/10。第二组分枝：取出红色后，剩下 3 红 6 绿，总数 9，所以 P(红) = 3/9，P(绿) = 6/9。取出绿色后，剩下 4 红 5 绿，总数 9，所以 P(红) = 4/9，P(绿) = 5/9。相同颜色意味着 RR 或 GG。P(RR) = (4/10)×(3/9)=12/90=2/15。P(GG) = (6/10)×(5/9)=30/90=1/3。相同颜色的总概率 = 2/15 + 1/3 = 2/15 + 5/15 = 7/15。

For at least one red, use the complement: 1 – P(GG) = 1 – 1/3 = 2/3. Always write a concluding sentence in the context. This is the type of structured answer Cambridge examiners expect.

求至少一个红色，用互补法：1 – P(GG) = 1 – 1/3 = 2/3。永远要根据上下文写一句总结性的话。这正是剑桥考官期望的那种结构清晰的答案。

13. Probability and Real-Life Contexts | 概率与现实生活情境

Page 291 often connects maths to real life, such as weather forecasts, games, and risk assessments. When you see a percentage chance of rain, you are looking at a probability expressed as a percentage. A 40% chance means the relative frequency of similar weather conditions leading to rain is 0.4.

第 291 页经常将数学与现实生活联系起来，例如天气预报、游戏和风险评估。当你看到下雨的百分率时，你看到的是用百分数表示的概率。40% 的降雨几率意味着类似天气条件下导致下雨的相对频率是 0.4。

Lotteries and games of chance also use probability. If a spinner is split into 10 equal sections, the theoretical probability of landing on any section is 1/10. But a wonky spinner might not be fair – experimental probability would then differ. Recognising when games are fair or biased is a key Checkpoint skill.

彩票和概率游戏也用到概率。如果一个转盘被分成 10 个相等的扇形，那么转到任何一个扇形的理论概率都是 1/10。但一个不平衡的转盘可能是不公平的——那样的话实验概率就会不同。能够识别游戏是否公平或是偏向性的，是 Checkpoint 的一项重要技能。

Risk is another context. If a medical test is 99% accurate, the probability of an incorrect result is 1% or 0.01. Understanding this helps you make informed decisions.

风险是另一个应用场景。如果一项医学检测的准确率是 99%，那么结果出错的概率就是 1% 或 0.01。理解这一点有助于你作出明智的决定。

14. Key Takeaways for Page 291 Revision | 第 291 页复习要点

Probability is a measure of likelihood, from 0 to 1. List all outcomes systematically before calculating, and use sample space diagrams for two events. Tree diagrams are your friend for sequential events – multiply along branches and add the relevant outcomes. Always simplify fractions and check that your branch probabilities sum to 1.

概率是衡量可能性大小的指标，范围从 0 到 1。在计算之前，先系统地列出所有结果，对两个事件使用样本空间图。树状图是处理连续事件的好帮手——沿分支相乘，并将相关结果相加。务必化简分数，并检查分支概率之和是否为 1。

When faced with dependent events, update the probabilities after each pick. And don’t forget the ‘1 minus’ trick to save time on ‘at least one’ questions. These page 291 techniques will serve you well not just in KS3 but right through IGCSE and beyond.

面对相关事件时，要记得在每次选取后更新概率。不要忘记用“1 减”的技巧来节省“至少一次”类题目的时间。这些第 291 页的技巧不仅会在 KS3 阶段帮到你，在 IGCSE 乃至更高年级的学习中都大有裨益。

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Page 291 – Mastering Probability | 第291页 – 掌握概率