📚 Review set 22A – NON-CALCULATOR | 复习题集22A – 非计算器部分
Review set 22A in many IB Mathematics textbooks is designed as a non-calculator assessment of probability concepts, counting principles and discrete distributions. This set usually includes a mix of straightforward probability calculations, tree diagrams, permutations, combinations, Venn diagrams and simple expectation problems. The absence of a calculator tests your ability to handle fractions, factorials and arithmetic fluently by hand, while maintaining a deep conceptual understanding. By working through this article, you will revisit the core ideas underpinning such a review set and learn to approach each problem type with confidence.
许多 IB 数学教材中的复习题集 22A 旨在以非计算器的方式检测概率概念、计数原理和离散分布。该题集通常包含基础概率计算、树状图、排列、组合、韦恩图以及简单的期望值问题。不允许使用计算器，正是为了检验你手工处理分数、阶乘和四则运算的熟练程度，同时也要保持深刻的概念理解。通过阅读本文，你将重温此类复习集所依赖的核心思想，并学会自信地应对各类题型。
1. Overview of Review Set 22A | 复习题集22A概述
A typical non-calculator Review set 22A targets the probability chapter (often Chapter 22) and requires you to solve problems without any electronic aid. The questions are crafted to assess your fluency with foundational probability rules, combinatorial reasoning, and the ability to interpret events using diagrams. You can expect to see items on sample spaces, mutually exclusive events, independent events, conditional probability, Bayes’ theorem in simple contexts, and discrete random variables such as uniform or binomial-like setups with small numbers.
典型的非计算器复习题集 22A 针对概率章节（常见第22章），要求你在不使用任何电子设备的情况下解题。这些题目旨在检测你对基础概率法则、组合推理以及运用图形解释事件的熟练度。可预期的题型包括样本空间、互斥事件、独立事件、条件概率、简单情境下的贝叶斯定理，以及小数字情形的离散随机变量（如均匀分布或类似二项分布的结构）。
2. Fundamental Counting Principles | 基本计数原理
The multiplication principle states that if one task can be performed in m ways and a second task in n ways, then the total number of ways to perform both tasks in sequence is m × n. This extends naturally to multiple stages. The addition principle applies when tasks are alternatives: if Task A can be done in m ways and Task B in n ways, and they cannot occur together, the total number of ways to choose one of them is m + n. A large share of non-calculator probability items begins with identifying which counting principle to apply.
乘法原理指出，若一项任务有 m 种完成方式，另一项任务有 n 种方式，则依次完成这两项任务的方式数为 m × n。该原理自然可推广到多个阶段。加法原理适用于互斥选择的情形：若任务 A 有 m 种方法，任务 B 有 n 种方法，且两者不能同时发生，则选择其中一种任务的总方法数为 m + n。大量非计算器概率题的第一步就是确定该运用哪种计数原理。
You will often need to combine these principles. For example, a student selects one main course and one dessert from a menu, but some combinations are unavailable. Multiply first, then subtract the invalid cases, or split the situation into disjoint scenarios and add their counts. In a non-calculator setting, keeping numbers small and using systematic listing can prevent errors.
你经常需要结合使用这两个原理。例如，某学生从菜单中选一份主菜和一份甜点，但某些组合不可用。可先乘法计算总数，再减去无效组合，或者将情形拆分成互斥的几类再分别相加。在非计算器环境中，控制数量规模并使用系统列举可以有效避免错误。
3. Permutations and Combinations De-mystified | 排列组合不再神秘
Permutations count arrangements where order matters. The number of permutations of n distinct objects taken r at a time is denoted P(n, r) or nPr and is given by P(n, r) = n! / (n – r)!. Combinations count selections where order does not matter. The number of combinations is C(n, r) = n! / [r! (n – r)!]. In non-calculator problems, factorials are usually small (up to around 7! = 5040) so that simplification by cancellation is manageable.
排列计算的是顺序有要求的安排方式。从 n 个不同物体中取 r 个的排列数记作 P(n, r) 或 nPr，公式为 P(n, r) = n! / (n – r)!。组合计算的是顺序无关的选取方式数，公式为 C(n, r) = n! / [r! (n – r)!]。在非计算器题目中，阶乘数值通常较小（如最多算到 7! = 5040），因此可以通过约分轻松处理。
A common task is to identify whether the scenario requires permutations or combinations. If selecting a committee of 3 from 10 people, use combinations: C(10,3) = 120. If arranging 3 of them in a president, vice-president and secretary hierarchy, use permutations: P(10,3) = 720. When the same items are reused (e.g., digits with repetition), the situation calls for a product of counts rather than factorial formulas. Keep an eye on key words like ‘arrange’, ‘select’, ‘order matters’ and ‘repetition allowed’.
常见任务是判断情境应使用排列还是组合。若从10人中选出3人组成委员会，用组合：C(10,3) = 120。若安排他们担任主席、副主席和秘书这3个有顺序的职位，则用排列：P(10,3) = 720。当同一元素可重复使用时（如数字可重复），应当使用计数的乘积而不是阶乘公式。请留意‘安排’、‘选出’、‘顺序重要’和‘允许重复’等关键词。
4. Basic Probability: Sample Space and Events | 基本概率：样本空间和事件
Probability is defined over a sample space S, the set of all possible outcomes. An event A is a subset of S, and its probability is P(A) = |A| / |S|, provided all outcomes are equally likely. This equally likely assumption holds in many Review set 22A problems, such as rolling a fair die or drawing cards from a well-shuffled deck. In a non-calculator context, you are expected to compute these fractions exactly and simplify them.
概率定义在样本空间 S 上，S 是所有可能结果的集合。事件 A 是 S 的子集，其概率为 P(A) = |A| / |S|，前提是所有结果等可能。这一等可能假设在许多复习题集22A的问题中都成立，例如掷公平骰子或从洗匀的扑克牌中抽牌。在非计算器中，你需要精确计算这些分数并进行约简。
The complement rule is indispensable: P(A’) = 1 – P(A). When direct counting of a desired event is cumbersome, counting the complement often simplifies the arithmetic. For instance, the probability of getting at least one head in three coin tosses is 1 – P(all tails) = 1 – 1/8 = 7/8. Questions also frequently ask for probabilities of unions and intersections that can be found with Venn diagrams or the inclusion–exclusion formula.
补集法则不可或缺：P(A’) = 1 – P(A)。当直接点数所需事件比较繁琐时，计数补集往往能简化运算。例如，掷三枚硬币至少出现一次正面的概率为 1 – P(全是反面) = 1 – 1/8 = 7/8。题目也常要求计算并集和交集的概率，这些可借助韦恩图或容斥公式求出。
P(A ∪ B) = P(A) + P(B) – P(A ∩ B)
掌握这个公式后，很多涉及‘或’的概率题都可以手算完成。
5. Mutually Exclusive vs Independent | 互斥与独立的区别
Two events are mutually exclusive if they cannot occur simultaneously, meaning P(A ∩ B) = 0. They are independent if the occurrence of one does not affect the probability of the other, which is expressed as P(A ∩ B) = P(A)·P(B). These concepts are easily confused. A classic non-calculator question gives a scenario and asks you to determine whether events are mutually exclusive, independent, both, or neither. Always check the intersection probability carefully.
两个事件若不能同时发生，则互斥，即 P(A ∩ B) = 0。若一个事件的发生不影响另一个事件发生的概率，则它们独立，表示为 P(A ∩ B) = P(A)·P(B)。这两个概念容易被混淆。经典的非计算器题目会给出情境，要求判断事件是互斥、独立、两者皆是，还是两者皆非。务必仔细检查交集概率。
Consider rolling a fair six-sided die: let A be ‘rolling an even number’ and B be ‘rolling a 3’. Here P(A ∩ B)=0, so they are mutually exclusive. But P(A)=1/2, P(B)=1/6, and 1/2 × 1/6 ≠ 0, so they are not independent. If instead B is ‘rolling a number greater than 3’, then A and B are not mutually exclusive (4 and 6 are in the intersection) but checking independence requires computing P(A ∩ B) = P({4,6}) = 2/6 = 1/3, while P(A)P(B) = (1/2)×(1/2)=1/4, so they are dependent.
考虑掷一颗公平的六面骰子：设 A 为‘掷出偶数’，B 为‘掷出3点’。这时 P(A ∩ B)=0，因此二者互斥。但 P(A)=1/2，P(B)=1/6，而 1/2 × 1/6 ≠ 0，故不独立。若将 B 改为‘掷出大于3的点数’，则 A 与 B 不互斥（4和6在交集中），但检验独立性需要计算 P(A ∩ B) = P({4,6}) = 2/6 = 1/3，而 P(A)P(B) = (1/2)×(1/2)=1/4，因此它们是相关事件。
6. Conditional Probability Formula | 条件概率公式
The probability of event A given that event B has occurred is P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0. This formula is the cornerstone of many non-calculator problems. You may be given a two-way table, a Venn diagram with counts, or a simple text description and asked to compute conditional probabilities without a calculator. All you need is to identify the restricted sample space (B) and count the favourable outcomes within it.
在事件 B 已发生的条件下，事件 A 的概率为 P(A|B) = P(A ∩ B) / P(B)，只要 P(B) > 0。该公式是许多非计算器题目的基石。题目可能给出一个双向表格、附带频数的韦恩图或简单的文字描述，要求你在没有计算器的情况下计算条件概率。你只需确定缩减后的样本空间 (B)，然后从中数出有利结果即可。
For example, from a standard deck of 52 playing cards, the probability that a card is a heart given that it is a red card is P(heart | red) = (13/52) / (26/52) = 13/26 = 1/2. In problems without replacement, conditional probabilities change with each selection. A tree diagram can systematically track these changing probabilities, and multiplying along the branches gives the probability of each intersection.
例如，从一副标准的52张扑克牌中，已知抽到的是一张红色牌，问它是红心的概率：P(红心 | 红色) = (13/52) / (26/52) = 13/26 = 1/2。在不放回抽样问题中，条件概率会随每次抽取而变化。此时树状图可以系统地跟踪这些变动的概率，沿着树枝相乘即得到每个交集的概率。
7. Tree Diagrams and Systematic Listing | 树状图和系统列举
A probability tree diagram is particularly helpful for multi-stage experiments, especially when the events are conditional (e.g., drawing without replacement). Each branch is labelled with a probability; the outcomes at the tips represent intersections, and their probabilities are the products along the paths. When several paths lead to the same event, you sum their probabilities. In a non-calculator setting, keep all probabilities as unsimplified fractions to make addition easy, then simplify at the end.
概率树状图特别适用于多阶段试验，尤其是事件为条件相关的情形（如不放回抽取）。每条枝干标有概率；末梢的结果代表交集，其概率为路径上概率的乘积。当多条路径指向同一事件时，需将这些概率相加。在非计算器环境中，可将所有概率保留为未约分的分数以方便相加，最后再进行约简。
Systematic listing (or sample space enumeration) is another powerful non-calculator tool. For experiments with small total outcomes, writing down the complete sample space eliminates guesswork. For instance, listing all 36 outcomes of rolling two dice allows you to compute any probability directly. In combinatorially more complex situations, a partially ordered listing according to a pattern (e.g., arranging letters) ensures you count all possibilities without double-counting.
系统列举（或称样本空间枚举）是另一个强大的非计算器工具。对于结果总数较小的试验，写出完整的样本空间可以避免猜测。例如，列出掷两颗骰子的全部36种结果，就可以直接计算任何概率。在组合较为复杂的情况中，按某种模式（如排列字母）进行部分有序的列举，可确保数出所有可能性且不重复计算。
8. Bayes’ Theorem
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