v1.1 — 12 questions · 91 marks across IB Paper 1A / 1B / 2 / 3 HL. Bayes' Theorem questions marked HL only. Question-only practice set; companion Solutions file in Solutions/.12 道题 · 91 分，覆盖 IB Paper 1A / 1B / 2 / 3 HL。贝叶斯定理（Bayes' theorem）题目仅限 HL。仅含题目；配套解析文件位于 Solutions/ 目录。
PART I · PAPER 1 SECTION A第一部分 · 第一卷 A 节No calculator · short response · 22 marks不可使用计算器 · 简答题 · 22 分
Section A — Short ResponseA 节 —— 简答题
Show all working. Probabilities should be left as exact fractions where natural, or as decimals to 3 s.f. otherwise. No calculator permitted.写出完整解题过程。概率（probability）若自然可写为分数则保留精确分数，否则保留 3 位有效数字的小数。不可使用计算器。
A weather model gives $P(\text{rain on a given spring day}) = 0.4$. Different days are treated as independent.某天气模型给出 $P(\text{rain on a given spring day}) = 0.4$。不同的日子视为独立（independent）。
(a)Find the probability that a given day is not rainy.求某一天不下雨的概率（即余事件 complement）。[1]
(b)Find the expected number of rainy days in a $30$-day month.求 $30$ 天的一个月中下雨天数的期望（expected）。[2]
(c)Find the probability that there is at least one rainy day in a $3$-day weekend.求一个 $3$ 天的周末至少有一天下雨的概率。[2]
Q3MEDIUMPaper 1A4.6 Mutually Exclusive vs Independent[6 marks]
(d)State, with brief justification, whether "Pass" and "Male" are independent.简要说明并判断 "Pass" 与 "Male" 是否独立。[1]
PART II · PAPER 1 SECTION B第二部分 · 第一卷 B 节No calculator · extended response · 23 marks不可使用计算器 · 长答题 · 23 分
Section B — Extended ResponseB 节 —— 长答题
Set out clean diagrams (tree, Venn, or table) before computing. Marks are awarded for method, accuracy, and clear reasoning. No calculator permitted.动笔计算前先画清晰的图表（树状图 tree diagram、维恩图 Venn diagram 或表格）。得分项包括方法、准确性与清晰的推理。不可使用计算器。
Q5MEDIUMPaper 1B4.6 Tree Diagram[7 marks]
A bag contains $4$ red and $6$ blue balls. Two balls are drawn one after the other without replacement.一个袋中装有 $4$ 个红球与 $6$ 个蓝球。先后不放回（without replacement）地抽取两球。
(a)Draw and label a clear tree diagram for the two draws.为两次抽取画出清晰且带标签的树状图。[2]
In a survey of $100$ students about their IB subjects:在一项对 $100$ 名学生 IB 科目选择的调查中：
$65$ take Mathematics ($M$)$65$ 人选了数学（$M$）
$50$ take Physics ($P$)$50$ 人选了物理（$P$）
$25$ take both$25$ 人两门都选
(a)Sketch a Venn diagram showing the numbers in each region.画出维恩图，标出各区域内的人数。[2]
(b)Find the number of students who take neither Mathematics nor Physics.求既未选数学又未选物理的学生人数。[1]
(c)A student is chosen at random. Find the probability that the student takes exactly one of the two subjects.随机选一名学生，求该学生恰好选了两门科目中的一门的概率。[2]
(d)Find $P(P \mid M)$, the probability of taking Physics given that the student takes Mathematics.求 $P(P \mid M)$，即在学生选了数学的条件下，选物理的概率。[2]
(e)State, with a numerical comparison, whether $M$ and $P$ are independent.通过数值比较，判断 $M$ 与 $P$ 是否独立。[1]
Q7HARDPaper 1BAHL 4.10 Bayes[8 marks]HL
A factory uses two machines, $A$ and $B$. Machine $A$ produces $60\%$ of the daily output; Machine $B$ produces the remaining $40\%$. Of items produced by $A$, $2\%$ are defective. Of items produced by $B$, $5\%$ are defective.某工厂使用两台机器 $A$ 与 $B$。机器 $A$ 生产每日产量的 $60\%$，机器 $B$ 生产其余的 $40\%$。$A$ 生产的产品中 $2\%$ 不合格，$B$ 生产的产品中 $5\%$ 不合格。
(a)Draw a tree diagram showing source (A or B) and quality (defective $D$ or non-defective $D^{\,\prime}$).画出树状图，标明来源（$A$ 或 $B$）与质量（不合格 $D$ 或合格 $D^{\,\prime}$）。[2]
(b)Find the overall probability $P(D)$ that a randomly selected item is defective.求随机抽取一件产品为不合格的总概率 $P(D)$（全概率公式 law of total probability）。[2]
(c)A defective item is found in the warehouse. Apply Bayes' Theorem to find the probability that it came from Machine $A$. Show the formula and substitution.仓库中发现一件不合格品。利用贝叶斯定理（Bayes' theorem）求该产品来自机器 $A$ 的概率，写出公式与代入过程。[3]
(d)In one sentence, explain why $P(A \mid D) < P(A)$ in this scenario.用一句话说明：在本情境下为何 $P(A \mid D) < P(A)$。[1]
PART III · PAPER 2第三部分 · 第二卷Calculator (GDC) permitted · 18 marks允许使用图形计算器（GDC） · 18 分
Section C — Paper 2 (Calculator)C 节 —— 第二卷（允许计算器）
Use exact fractions where natural, decimals (3 s.f.) where called for. State key intermediate probabilities, not just GDC outputs.自然时使用精确分数，需要小数时保留 3 位有效数字。要写出关键的中间概率，不要只给计算器结果。
Q8EASYPaper 24.6 Two-Way Table[5 marks]
$200$ students from two year groups indicated their IB Math choice:两个年级共 $200$ 名学生填写了所选的 IB 数学课程：
(b)Find $P(\text{at least one is black})$.求 $P(\text{at least one is black})$（至少一次抽到黑球）。[2]
(c)Find $P(\text{second is white} \mid \text{first is black})$.求 $P(\text{second is white} \mid \text{first is black})$（在第一球是黑的条件下，第二球为白）。[2]
(d)A third ball is then drawn (still without replacement) given that the first two were both white. Find the probability this third ball is black.再抽取第三球（仍不放回），已知前两球都是白球。求第三球为黑球的概率。[1]
Q10HARDPaper 24.6 Tree + Conditional[6 marks]
A student takes the bus to school with probability $0.7$, and walks otherwise. If she takes the bus, the probability she arrives on time is $0.85$; if she walks, the probability she arrives on time is $0.6$.某学生以 $0.7$ 的概率乘公交去上学，否则步行。乘公交时按时到校的概率为 $0.85$；步行时按时到校的概率为 $0.6$。
(a)Find the probability that on a randomly chosen day she arrives on time.求随机一天她按时到校的概率。[3]
(b)Find $P(\text{took the bus} \cap \text{on time})$.求 $P(\text{took the bus} \cap \text{on time})$。[1]
(c)Given that she arrived on time, find the probability she took the bus.已知她按时到校，求她乘公交的概率。[2]
PART IV · PAPER 3 (HL ONLY)第四部分 · 第三卷（仅 HL）Calculator · extended modelling · 28 marks允许计算器 · 长题建模 · 28 分
Section D — Paper 3 (HL Extended Exploration)D 节 —— 第三卷（HL 长题探究）
Read the full scenario before starting. Each part builds on the previous. State Bayes' theorem in symbols once, then apply it consistently. AHL only.开始作答前先通读全部情境。各小问层层递进。先用符号写出一次贝叶斯定理（Bayes' theorem），随后保持一致地应用。仅限 AHL。
A diagnostic test for a rare disease is used in three regions whose disease prevalence and test characteristics differ:某罕见疾病的诊断检测在三个地区使用，各地区的疾病患病率（prevalence）与检测特性不同：
(a)In Region A, a randomly chosen person tests positive. Use Bayes' Theorem to find the probability that they actually have the disease. State $P(+)$ as an intermediate.在地区 A，随机抽取一人检测呈阳性。利用贝叶斯定理求其确实患病的概率，并将 $P(+)$ 作为中间量给出。[3]
(b)Repeat (a) for Region B and Region C, giving each posterior probability to 3 s.f.对地区 B 与地区 C 重复 (a) 的过程，给出每个后验（posterior）概率到 3 位有效数字。[3]
(c)Despite Region C having both the highest sensitivity and the highest specificity, its posterior $P(D \mid +)$ is the lowest of the three. Explain in one or two sentences why high test accuracy can still produce a low posterior probability.尽管地区 C 同时具有最高的灵敏度与最高的特异度，其后验 $P(D \mid +)$ 却是三地中最低的。用一两句话解释为何检测准确度高仍可能导致较低的后验概率。[2]
(d)A national programme samples patients from a population that is $40\%$ from Region A, $35\%$ from Region B, and $25\%$ from Region C. A randomly chosen patient tests positive. Find the probability that they came from Region B.某全国项目抽取的人群中 $40\%$ 来自地区 A、$35\%$ 来自地区 B、$25\%$ 来自地区 C。随机抽取一名患者检测呈阳性，求其来自地区 B 的概率。[4]
A screening test for a rare genetic marker has sensitivity $P(+ \mid D) = 0.99$ and specificity $P(- \mid D^{\,\prime}) = 0.95$. In the target population $P(D) = 0.01$. Successive tests on the same patient are conditionally independent given true status.一项针对罕见基因标志的筛查（screening test）的灵敏度为 $P(+ \mid D) = 0.99$，特异度为 $P(- \mid D^{\,\prime}) = 0.95$。目标人群中 $P(D) = 0.01$。对同一患者的多次检测在真实状态条件下相互独立（conditionally independent）。
(a)State Bayes' theorem in symbols for $P(D \mid +)$, and identify $P(D)$, $P(+\mid D)$, $P(+\mid D^{\,\prime})$ for this scenario.用符号写出关于 $P(D \mid +)$ 的贝叶斯定理，并在本情境中识别 $P(D)$、$P(+\mid D)$、$P(+\mid D^{\,\prime})$。[2]
(c)The same patient is retested and again tests positive. Treating the posterior from (b) as the new prior, find $P(D \mid +,+)$ after two consecutive positives, to 3 s.f.对同一患者再次检测仍为阳性。将 (b) 的后验作为新的先验（prior），求连续两次阳性后的 $P(D \mid +,+)$，保留 3 位有效数字。[3]
(d)Compare your answer in (c) to (b). In one sentence, explain in Bayesian terms why a second independent positive shifts the posterior so dramatically.将 (c) 与 (b) 比较。用一句话以贝叶斯语言说明：为何第二次独立的阳性结果会使后验概率大幅跃升。[2]
(e)Find $P(+,+ \mid D^{\,\prime})$ for a non-diseased patient, and explain in one sentence what this number means for false alarms in a large screening programme.对未患病者求 $P(+,+ \mid D^{\,\prime})$，并用一句话说明该数值在大规模筛查中对假阳性（false positive）警报的意义。[3]
(f)Suppose specificity drops to $0.90$ (false-positive rate $0.10$), sensitivity unchanged. Find the smallest number of consecutive independent positive tests $n$ such that the posterior $P(D \mid n \text{ positives}) > 0.95$. State $n$ and the resulting posterior, both from a GDC iteration.假设特异度降至 $0.90$（假阳性率为 $0.10$），灵敏度不变。求使后验 $P(D \mid n \text{ positives}) > 0.95$ 的最小独立连续阳性次数 $n$。用 GDC 迭代给出 $n$ 与对应的后验概率。[3]