📚 Pre-U OCR Statistics: Unit Test Mock Paper Analysis | Pre-U OCR 统计：单元测试模拟卷解析
This mock unit test covers the essential topics of Pre-U OCR Statistics: probability and conditional probability, discrete and continuous distributions, sampling theory and hypothesis testing. Every question is followed by a full step-by-step solution written in English and Chinese, so that you can check both your mathematics and your exam technique at the same time.
本模拟单元测试涵盖 Pre-U OCR 统计的必考专题：概率与条件概率、离散与连续分布、抽样理论以及假设检验。每道题后均配有完整的中英双语逐步解析，让你在核对数学过程的同时，一并检验答题技巧。
1. How the Mock Paper Is Structured | 模拟卷的结构与评分方式
The Cambridge Pre-U Statistics unit test mirrors the final examination in both style and demand: short procedural questions come first, then longer modelling and inference questions that carry the bulk of the marks. Marks are awarded for method, accuracy and, where a decision is required, a correctly worded conclusion.
剑桥 Pre-U 统计的单元测试在题型与难度上都与正式考试一致：先是短小的程序性题目，随后是占分较多的建模与推断题。评分分为方法分、精度分，以及在需要做出判断时的结论分（措辞必须准确）。
The paper below is worth 60 marks and should be completed in about 75 minutes. The distribution of marks is deliberately uneven, because inference questions almost always carry more credit than routine substitution.
下面这份试卷满分为 60 分，建议在约 75 分钟内完成。分值分布刻意保持不均，因为推断类题目的得分权重通常远高于套公式的常规计算。
| Question | Topic | Marks | Suggested time |
|---|---|---|---|
| 1 | Conditional probability | 8 | 10 min |
| 2 | Independence and tree diagrams | 8 | 10 min |
| 3 | Discrete random variables | 9 | 11 min |
| 4 | Binomial and Poisson models | 12 | 14 min |
| 5 | Normal distribution | 10 | 12 min |
| 6 | Sampling and hypothesis testing | 13 | 18 min |
Before you start, write down the distribution you are using for each question. Examiners repeatedly report that the most common cause of lost marks is an unstated or wrongly stated model.
开始作答前，请为每一题写下你所使用的分布。阅卷报告反复指出，失分最常见的原因就是没有写明或写错所用的概率模型。
2. Probability and Conditional Probability | 概率与条件概率
Question 1. A bag contains 5 red balls and 3 green balls. Two balls are drawn at random without replacement. (a) Find the probability that both balls are red. (b) Given that the second ball drawn is red, find the probability that the first ball drawn was also red.
第 1 题。一个袋子里有 5 个红球和 3 个绿球。从中不放回地随机抽取两个球。(a) 求两个都是红球的概率。(b) 已知第二个抽到的是红球，求第一个也是红球的概率。
Part (a) second is red. P(R₁) = 5/8. After one red is taken, 4 reds remain out of 7 balls, so P(R₂ | R₁) = 4/7. Hence P(R₁ ∩ R₂) = (5/8) × (4/7) = 20/56 = 5/14 ≈ 0.357.
(a) 部分 - 利用相依事件的乘法法则。设 R₁ 为第一次抽到红球，R₂ 为第二次抽到红球。P(R₁) = 5/8。在抽走一个红球后，剩下 7 个球中有 4 个红球，故 P(R₂ | R₁) = 4/7。因此 P(R₁ ∩ R₂) = (5/8) × (4/7) = 20/56 = 5/14 ≈ 0.357。
Notice how the denominator changes from 8 to 7. Writing "without replacement" in your working, or drawing the two branches of a tree diagram with 5/8 and 3/8 on the first level, earns the method mark even if the arithmetic slips.
注意分母从 8 变成 7。在解题过程中写明"不放回"，或在树状图第一层标出 5/8 与 3/8，即使后面算错，也能拿到方法分。
Part (b) - We need P(R₁ | R₂) = P(R₁ ∩ nd P(R₂), consider the two ways the second ball can be red: RR or GR. P(GR) = P(G₁) × P(R₂ | = (3/8) × (5/7) = 15/56. So P(R₂) = 5/14 + 15/56 = 20/56 + 15/56 = 35/56 = 5/8. Then P(R₁ | R₂) = (5/14) / (5/8) = (5/14) × (8/5) = 40/70 = 4/7 ≈ 0.571.
(b) 部分 - 需要求 P(R₁ | R₂) = P(R₁ ∩ R₂) / P(R₂)。分子已知。为求 P(R₂)，考虑第二次抽到红球的两种情形：先红后红 (RR) 或先绿后红 (GR)。P(GR) = P(G₁) × P(R₂ | G₁) = (3/8) × (5/7) = 15/56。所以 P(R₂) = 5/14 + 15/56 = 20/56 + 15/56 = 35/56 = 5/8。于是 P(R₁ | R₂) = (5/14) / (5/8) = (5/14) × (8/5) = 40/70 = 4/7 ≈ 0.571。
A useful check on part (b) is provided by symmetry: the probability that the first ball is red given the second is red must equal the probability that the second is red given the first is red, because the two draws are exchangeable. That common value is 4/7, which matches the conditional probability calculated in part (a).
(b) 部分有一个很好的检验方法：由于两次抽取具有可交换性，已知第二个是红球时第一个也是红球的概率，必须等于已知第一个是红球时第二个也是红球的概率，而后者正是 (a) 中的 4/7，与计算结果吻合。
= 1 - (3/8 × 2/7) = 1 - 6/56 = 50/56 = 25/28 ≈ 0.893. The long route, adding P(RG) + P(GR) + P(RR), gives the same answer and wastes two minutes.
把同一题拓展一步：求至少抽到一个红球的概率。最快的思路是用对立事件：P(至少一个红球) = 1 - P(GG) = 1 - (3/8 × 2/7) = 1 - 6/56 = 50/56 = 25/28 ≈ 0.893。若改用 P(RG) + P(GR) + P(RR) 逐项相加，答案相同但要多花两分钟。
3. Independence, Tree Diagrams and the Total Probability Rule | 独立性、树状图与全概率法则
Question 2. Events A and B satisfy P(A) = 0.6, P(B) = 0.5 and P(A ∪ B) = 0.8. (a) Show that A and B are independent. (b) Find P(A' ∩ B').
第 2 题。事件 A 与 B 满足 P(A) = 0.6、P(B) = 0.5、P(A ∪ B) = 0.8。(a) 证明 A 与 B 相互独立。(b) 求 P(A' ∩ B')。
Solution. From the addition rule, P(A ∪ B) = P(A) + P(B) - P(A ∩ B), so 0.8 = 0.6 + 0.5 - P(A ∩ B), giving P(A ∩ .3. Since P(A) × P(B) = 0.6 × 0.5 = 0.3, the two events are independent. For part (b), P(A' ∩ B') = 1 - P(A ∪ B) = 1 - 0.8 = 0.2.
解答。由加法法则 P(A ∪ B) = P(A) + P(B) - P(A ∩ B)，得 0.8 = 0.6 + 0.5 - P(A ∩ B)，故 P(A ∩ B) = 0.3。因为 P(A) × P(B) = 0.6 × 0.5 = 0.3，所以两事件相互独立。(b) 部分：P(A' ∩ B') = 1 - P(A ∪ B) = 1 - 0.8 = 0.2。
Two warnings. First, "independent" and "mutually exclusive" endent. Second, the independence test must always be performed explicitly by comparing P(A ∩ B) with P(A) × P(B); a numerical coincidence is not a proof.
两点提醒。第一，"独立"与"互斥"并不相同：概率均非零的互斥事件一定不独立。第二，独立性检验必须显式地把 P(A ∩ B) 与 P(A) × P(B) 进行比较；数值巧合不能算作证明。
em in Question 1. If B₁, B₂, ..., Bₙ partition the sample space, then P(A) = Σ P(Bᵢ) × P(A | does: multiply along the branches, then add the branches that produce the event you want.
全概率法则是第 1 题抽球问题的推广。若 B₁, B₂, ..., Bₙ 构成样本空间的一个划分，则 P(A) = Σ P(Bᵢ) × P(A | Bᵢ)。在两阶段试验中，这正是树状图所做的：沿树枝相乘，再把导致目标事件的所有树枝相加。
A typical examination question on this rule: machine X produces 60% of the output with a 2% defect rate, and machine Y produces the rest with a 5% defect rate. Then P(defective) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032.
该法则的典型考题：机器 X 生产 60% 的产品，次品率 2%；机器 Y 生产其余 40%，次品率 5%。则 P(次品) = 0.6 × 0.02 + 0.4 × 0.05 = 0.012 + 0.020 = 0.032。
4. Discrete Random Variables, Expectation and Variance | 离散随机变量、期望与方差
Question 3. , 2, 3, 4. (a) Find k. (b) Find E(X) and Var(X). (c) Hence find E(3X - 2) and Var(3X - 2).
第 3 题。离散随机变量 X 的概率分布为 P(X = x) = kx，其中 x = 1, 2, 3, 4。(a) 求 k。(b) 求 E(X) 与 Var(X)。(c) 进而求 E(3X - 2) 与 Var(3X - 2)。
Solution. and k = 0.1. This gives P(X = 1) = 0.1, P(X = 2) = 0.2, P(X = 3) = 0.3 and P(X = 4) = 0.4.
解答。概率之和必须为 1，故 k(1 + 2 + 3 + 4) = 10k = 1，得 k = 0.1。于是 P(X = 1) = 0.1、P(X = 2) = 0.2、P(X = 3) = 0.3、P(X = 4) = 0.4。
9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10.0, so Var(X) = E(X²) - [E(X)]² = 10 - 9 = 1.
于是 E(X) = Σ x P(X = x) = 1(0.1) + 2(0.2) + 3(0.3) + 4(0.4) = 0.1 + 0.4 + 0.9 + 1.6 = 3.0。又 E(X²) = 1(0.1) + 4(0.2) + 9(0.3) + 16(0.4) = 0.1 + 0.8 + 2.7 + 6.4 = 10.0，故 Var(X) = E(X²) - [E(X)]² = 10 - 9 = 1。
For part (c), use the linearity results E(aX + b
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