AS Mathematics Unit 5 (9709/52) June 2019 Common Mistakes Summary | AS 数学 9709/52 2019年6月真题易错点总结

📚 AS Mathematics Unit 5 (9709/52) June 2019 Common Mistakes Summary | AS 数学 9709/52 2019年6月真题易错点总结

The CAIE AS Mathematics Paper 5 (Probability & Statistics 1) from June 2019 covers a wide range of topics, including permutations and combinations, normal distribution, coded data, conditional probability, and discrete random variables. Many students lose marks not because they lack knowledge, but because of recurring small mistakes in interpretation, notation, and calculation. This article summarises the most common pitfalls seen in that examination session, with clear explanations to help you avoid them in the future.

2019年6月CAIE AS数学试卷5（概率与统计1）涵盖了排列组合、正态分布、编码数据、条件概率和离散随机变量等众多主题。很多学生失分并非因为知识欠缺，而是反复出现理解、符号和计算上的小错误。本文总结了该次考试中最常见的易错点，并附上清晰的解释，帮助你在未来避开这些陷阱。

1. Misreading ‘not all together’ in permutations | 排列问题中“不全在一起”的误读

A classic question asked for the number of arrangements where three particular letters are not all next to each other. Many learners wrongly calculated the number of ways where no two of these letters are adjacent, which is a stricter condition. The correct approach is to subtract the arrangements where the three letters are all together (treating them as a single block) from the total unrestricted arrangements. Then adjust for internal arrangements within the block.

一道经典题是求三个特定字母不全相邻的排列数。很多学习者错误地计算了这三个字母中任意两个都不相邻的情况数，这是个更严格的条件。正确的做法是用无限制条件的总排列数减去三个字母全部相邻的排列数（将它们当作一个整体块处理），再乘以块内字母的排列。

English: Total ways without restriction: n! ; ways with the 3 letters as one block: (n-2)! × 3! ; answer = n! − (n-2)! × 3!
中文：无限制的总排列数：n! ；三个字母作为一个块的方法数：(n-2)! × 3! ；答案 = n! − (n-2)! × 3!

2. Indirect probability using the complement | 利用互补事件求概率的间接方法

When finding the probability that at least one event occurs, many students dive into summing the probabilities of ‘one occurs’, ‘two occur’, etc. This quickly becomes messy and error‑prone. The smarter route is to compute the probability of the complement – that none of the events happen – and subtract from 1. In the June 2019 exam, a question involving multiple independent selections caught out candidates who tried direct addition.

在求至少一个事件发生的概率时，许多学生一头扎进“一个发生”、“两个发生”等概率的求和里，很快就变得混乱且易错。更聪明的路径是计算互补事件的概率——即所有事件都不发生——然后用 1 减去该值。在2019年6月的考试中，一道涉及多次独立选择的题目让尝试直接相加的考生翻了车。

P(at least one) = 1 − P(none)

P(至少有一个) = 1 − P(一个都没有)

3. Normal distribution – continuity correction confusion | 正态分布——连续性校正混淆

The exam required using a normal approximation to a binomial. A major stumbling block was applying the continuity correction incorrectly. Remember: when you approximate a discrete binomial with a continuous normal, you adjust by 0.5. For example, P(X ≥ 10) becomes P(X > 9.5). Many candidates either omitted the correction entirely or added 0.5 instead of subtracting, especially with strict versus non‑strict inequalities.

考试要求用正态分布近似二项分布。一个主要的绊脚石是连续性校正应用错误。记住：用连续正态分布近似离散的二项分布时，要调整 0.5。例如，P(X ≥ 10) 变成 P(X > 9.5)。很多考生要么完全遗漏校正，要么错误地加了 0.5，特别是在面对带等号和不带等号的不等式时。

English: Binomial P(X ≤ a) → Normal P(X < a+0.5); P(X ≥ a) → P(X > a−0.5)
中文：二项分布 P(X ≤ a) → 正态 P(X < a+0.5)；P(X ≥ a) → P(X > a−0.5)

4. Coded data – misapplying mean and standard deviation | 编码数据——均值和标准差的错误应用

A question gave a summary of coded values y = (x − a)/b, and students had to find the mean and standard deviation of the original x. Several errors appeared: dividing the coded mean by b instead of multiplying, forgetting to add a after scaling, and treating the coded standard deviation as the variance of x. Remember the rules: if y = (x − a)/b, then x = a + b y, so mean(x) = a + b×mean(y), and std dev(x) = b × std dev(y). The variance scales by b².

一道题给出了编码数据 y = (x − a)/b 的汇总值，考生需要求出原始 x 的均值和标准差。出现了几种错误：用编码均值除以 b 而不是乘以 b，调整尺度后忘记加 a，以及把编码的标准差直接当作 x 的方差。记住规则：若 y = (x − a)/b，则 x = a + b y，因此 mean(x) = a + b×mean(y)，标准差 std dev(x) = b × std dev(y)。方差则缩放 b² 倍。

S.D.(x) = b × S.D.(y), Var(x) = b² × Var(y)

标准差(x) = b × 标准差(y)，方差(x) = b² × 方差(y)

5. Discrete random variable – forgetting to sum probabilities to 1 | 离散随机变量——忘记概率和为1

In probability distribution tables, an unknown constant k often appears. To find k, you must set the sum of all probabilities equal to 1. In the rush of the exam, students sometimes set up the equation correctly but then make simple algebraic slips, or they forget to check that all the given expressions indeed represent probabilities. Always verify that each probability derived from k lies between 0 and 1 once k is found; negative or greater‑than‑one probabilities indicate an error.

在概率分布表中经常出现未知常数 k。要求解 k，必须使所有概率之和等于1。考试匆忙中，学生有时正确列出了方程，却犯下简单的代数错误，或者忘记检查所有给出的表达式是否确实代表概率。求出 k 后，务必验证每个根据 k 算出的概率值都在 0 到 1 之间；出现负值或大于 1 的概率就说明有错。

6. Conditional probability – reversing the condition | 条件概率——条件颠倒

One question gave P(A|B) and asked for P(A ∩ B) or P(B|A). A very typical mistake is to treat P(A|B) as P(B|A). These are generally not equal. Use the definition: P(A|B) = P(A ∩ B) / P(B). In the 2019 paper, students who confused the given direction of conditioning lost marks even when they could perform the correct multiplication.

P(A ∩ B) = P(A|B) × P(B)

7. Combinations vs permutations in selection | 选择问题中的组合与排列混淆

When selecting a committee or picking items without replacement where order does not matter, you must use combinations (nCr). A number of candidates used permutations (nPr) and thus inflated their answer. The exam question had a selection from groups – e.g., choosing men and women – and the intended method was multiplication of combinations. Permutations are only needed when the arrangement of the chosen individuals matters.

当组建委员会或不放回地选取物品且次序无关时，必须使用组合 (nCr)。不少考生使用了排列 (nPr)，从而多算了结果。考试题目涉及从组中选取——例如选择男女——预期的方法是组合相乘。只有当选出个体的排列顺序重要时才需要使用排列。

English: Combinations for unordered selection; Permutations for ordered arrangements.
中文：无序选取用组合；有序排列用排列。

8. Finding the mean given a probability (inverse normal) | 给定概率求均值（逆正态）的常见错误

A normal distribution with an unknown mean μ and known standard deviation was given, along with a probability statement like P(X > k) = 0.8. The challenge is to find μ. Many students mis‑signed the z‑value or set up the standardising equation incorrectly. The key steps: standardise to z = (k − μ)/σ ; from P(X > k) = 0.8, note that P(X < k) = 0.2, so the corresponding z is negative (since 0.2 < 0.5). Using a positive z‑value instead leads to a wrong μ.

题目给出一个均值 μ 未知但标准差已知的正态分布，以及一个概率描述如 P(X > k) = 0.8，要求解出 μ。许多学生弄错了 z 值的正负号，或者将标准化方程列错。关键步骤：标准化为 z = (k − μ)/σ ；由 P(X > k) = 0.8，可知 P(X < k) = 0.2，因此对应的 z 为负（因为 0.2 < 0.5）。若误用为正 z 值就会得到错误的 μ。

z = (k − μ) / σ , with z negative for left‑tail probability < 0.5

z = (k − μ) / σ ，当左侧概率 < 0.5 时 z 取负值

9. Variance of aX+b – sign errors | aX+b 的方差——符号错误

Questions on the expectation and variance of a linear function often appear, e.g., if Y = 3 − 2X, find Var(Y). Many students mistakenly use Var(Y) = 3 − 2² Var(X) or forget that adding a constant b has no effect on variance. The correct rule is Var(aX + b) = a² Var(X). The constant shift disappears, and the coefficient a is squared regardless of its sign. In 2019, some candidates subtracted the variance term from the constant, losing easy marks.

关于线性函数的期望与方差的题目经常出现，例如若 Y = 3 − 2X，求 Var(Y)。许多学生错误地用 Var(Y) = 3 − 2² Var(X) 或忘记加常数 b 对方差没有影响。正确的规则是 Var(aX + b) = a² Var(X)。常数偏移消失，系数 a 不论正负号均被平方。2019年的考试中，有些考生从常数项中减去方差项，白白丢掉了容易的分数。

Var(aX + b) = a² Var(X)

10. Tree diagrams – missing branches or wrong multiplication | 树状图——遗漏分支或错误乘法

A conditional probability tree diagram was required to solve a multi‑stage probability problem. Errors that repeatedly surfaced included: forgetting that the second‑stage probabilities must sum to 1 along each branch set; multiplying probabilities that were for mutually exclusive outcomes instead of along branches; and adding probabilities across different paths when they should be multiplied. In the Jun 2019 paper, a subtle tree for ‘with replacement’ vs ‘without replacement’ confused many.

有一道条件概率题需要用树状图来求解多阶段概率问题。反复出现的错误包括：忘记第二阶段各分支的概率之和必须为1；将本应沿着分支相乘的概率误作为互斥结果的概率来相加；以及在应相乘时却跨路径相加。在2019年6月试卷中，关于“有放回”与“无放回”的微妙树状图让许多人困惑。

English: Multiply along branches; add across different paths for the same final outcome.
中文：沿着分支相乘；对相同最终结果的不同路径则相加。

11. Binomial distribution – using tables incorrectly | 二项分布——错误使用表格

The exam provided cumulative binomial tables for some parts. A common blunder was reading P(X ≤ r) from the table when the question asked for P(X < r). Remember: for a discrete binomial, P(X < r) = P(X ≤ r − 1). Another error was overlooking that the probability in the table was for ‘less than or equal to’, and directly using the tabled value as P(X = r). Always convert the required event into cumulative form before looking up tables.

考试为某些部分提供了二项分布累积概率表。一个常见的疏忽是题目要求 P(X < r) 时却从表中读取了 P(X ≤ r)。记住：对于离散的二项分布，P(X < r) = P(X ≤ r − 1)。另一个错误是忽略了表中给出的是“小于等于”的概率，而直接将表值用作 P(X = r)。在查表前务必将所求事件转化为累积形式。

P(X < r) = P(X ≤ r − 1); P(X = r) = P(X ≤ r) − P(X ≤ r − 1)

P(X < r) = P(X ≤ r − 1)；P(X = r) = P(X ≤ r) − P(X ≤ r − 1)

12. Interpretation of ‘exactly’, ‘at least’, ‘between’ | “恰好”、“至少”、“之间”的解读错误

Ambiguities in wording like ‘more than 4’, ‘no more than 3’, ‘between 2 and 5 inclusive’ caused many candidates to mis‑set their probability expressions. For instance, ‘between 2 and 5 inclusive’ means 2, 3, 4, and 5. ‘Exactly two’ is just P(X = 2). The June 2019 paper had a part requiring P(2 < X < 6) which, for a discrete variable, is P(X = 3, 4, 5). Students who mistakenly included 2 or 6 lost the marks.

诸如“多于4个”、“不超过3个”、“在2到5之间（含）”之类措辞的歧义导致许多考生写错了概率表达式。例如，“在2到5之间（含）”意味着 2, 3, 4, 5。“恰好两个”仅仅是 P(X = 2)。2019年6月的试卷有一个部分要求计算 P(2 < X < 6)，对于离散变量，这意味着 P(X = 3, 4, 5)。错误地包含了 2 或 6 的考生丢了分。

English: Inclusive uses ≤ or ≥; strict inequalities require adjustment for discrete data.
中文：包含端点使用 ≤ 或 ≥；严格不等式对于离散数据需要调整。

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AS Mathematics Unit 5 (9709/52) June 2019 Common Mistakes Summary | AS 数学 9709/52 2019年6月真题易错点总结