Probability概率
Sub-topics 4.5, 4.6 (SL core) and 4.10 (AHL — HL only) of IB AA HL Topic 4. Sample spaces, combined events, conditional probability, independence, Bayes' theorem.
IB AA HL Topic 4 的子主题 4.5、4.6（SL 核心内容）以及 4.10（AHL，仅 HL）。覆盖样本空间（sample space）、复合事件（combined events）、条件概率（conditional probability）、独立性（independent events）以及贝叶斯定理（Bayes' theorem）。
How to use this guide本指南使用说明
Read only the Unit Cram Cheat-Sheet at the top and the gold-dashed Cram-Mode Cheat at the head of each section. Memorize four formulas: $P(A) = n(A)/n(U)$, $P(A\cup B) = P(A)+P(B)-P(A\cap B)$, $P(A\mid B) = P(A\cap B)/P(B)$, and (HL) Bayes. Skim one worked example per section. Take the practice quiz.
只看顶部的 Unit Cram Cheat-Sheet（单元速记小抄）和每节开头金色虚线框的 Cram-Mode Cheat。背熟四条公式：$P(A) = n(A)/n(U)$、$P(A\cup B) = P(A)+P(B)-P(A\cap B)$、$P(A\mid B) = P(A\cap B)/P(B)$，以及（HL）贝叶斯定理（Bayes' theorem）。每节扫一道例题，做练习测验。
Open every ▸ Going deeper. Probability is the topic where Paper 2 sets a Venn / tree / table scenario and then layers conditional and Bayes parts on top — the marks come from fluently translating between the diagram and the formula. Practice the Bayes partition setup until it's a reflex.
把每个 ▸ Going deeper（深入）都展开。概率是 Paper 2 最爱出复合题的主题——先设一个韦恩图（Venn diagram）/ 树状图（tree diagram）/ 表格情境，再叠加条件概率与贝叶斯。得分点在于在图与公式之间流畅切换。把贝叶斯划分（partition）的套路练到下意识。
Unit Cram Cheat-Sheet单元速记小抄
Core (SL 4.5, 4.6 — always-on):核心内容（SL 4.5、4.6——必考）：
$$ P(A) = \frac{n(A)}{n(U)} \qquad 0 \le P(A) \le 1 \qquad P(A') = 1 - P(A) $$ $$ P(A \cup B) = P(A) + P(B) - P(A \cap B) $$ $$ \text{Mutually exclusive: } P(A \cap B) = 0,\;\; P(A \cup B) = P(A) + P(B) $$ $$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} \quad (P(B) > 0) $$ $$ \text{Independent: } P(A \cap B) = P(A)\,P(B) \;\Longleftrightarrow\; P(A \mid B) = P(A) $$ $$ \text{Expected number of occurrences of } A \text{ in } n \text{ trials} = n\,P(A) $$HL only (AHL 4.10 — Bayes, up to 3 events):仅 HL（AHL 4.10——贝叶斯，最多 3 个事件）：
$$ P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{\displaystyle\sum_{j} P(B \mid A_j)\,P(A_j)} \quad \text{for partition } \{A_1, A_2, A_3\} $$Decision tree: "and" between events → intersection (multiply if independent or use conditional); "or" → union (subtract the overlap unless mutually exclusive); "given" → conditional; "reverse the conditional" → Bayes.
关键词决策树：事件之间出现"and"（且）→ 交集（intersection）（独立就相乘，否则用条件概率）；"or"（或）→ 并集（union）（除非互斥，否则要减去重叠）；"given"（在……条件下）→ 条件概率；要"反转条件"→ 贝叶斯。
Sample Spaces and Events样本空间与事件 SL 4.5
trial）指随机实验的一次执行；可能的单个结果称为结果（outcome）；所有结果的集合就是样本空间（sample space）$U$；事件（event）$A$ 是 $U$ 的一个子集。当所有结果是等可能的（equally likely outcomes）时，
$$ P(A) = \frac{n(A)}{n(U)}, $$
其中 $n(\cdot)$ 表示元素个数。概率（probability）取值在 $[0, 1]$ 之间。必然事件概率为 $1$，不可能事件概率为 $0$。
Valid whenever all outcomes in $U$ are equally likely. If they aren't, fall back to relative frequency: $P(A) \approx \dfrac{\text{number of times } A \text{ occurs}}{\text{number of trials}}$ as the number of trials grows large.
前提是 $U$ 中所有结果等可能。若不等可能，则退而求其次用相对频率（relative frequency）估计：当试验次数足够大时，$P(A) \approx \dfrac{\text{number of times } A \text{ occurs}}{\text{number of trials}}$。
Outcome. A single element of $U$ (HEADS, the integer $4$, the queen of spades).
Sample space $U$. The set of all outcomes. For one die: $U = \{1, 2, 3, 4, 5, 6\}$, $n(U) = 6$.
Event. A subset of $U$. "Even score on a die" is $A = \{2, 4, 6\}$, $n(A) = 3$.
Equally likely. Each outcome has the same probability $1/n(U)$. State this assumption explicitly when you use $P(A) = n(A)/n(U)$.
Relative frequency. Empirical estimate from observed data; the IB uses it to motivate probability but expects exact answers wherever possible. 试验（
trial）。 一次执行（抛一次硬币、掷一次骰子、抽一张牌）。结果（
outcome）。 $U$ 中的单个元素（HEADS、整数 $4$、黑桃皇后）。样本空间 $U$（
sample space）。 所有结果组成的集合。掷一次骰子：$U = \{1, 2, 3, 4, 5, 6\}$，$n(U) = 6$。事件（
event）。 $U$ 的子集。"骰子点数为偶数"对应 $A = \{2, 4, 6\}$，$n(A) = 3$。等可能（
equally likely）。 每个结果概率都等于 $1/n(U)$。每次用 $P(A) = n(A)/n(U)$ 都要明示这条假设。相对频率（
relative frequency）。 来自观测数据的经验估计；IB 用它来引出概率概念，但凡能给出精确答案就要给精确答案。
Worked Example — Two dice, sum equals 7例题——两骰之和为 7
Problem: Two fair six-sided dice are rolled. Find the probability the sum is $7$.
Sample space. Ordered pairs $(d_1, d_2)$ with each $d_i \in \{1, \ldots, 6\}$, so $n(U) = 36$. Each pair is equally likely.
Event. Sum $= 7$ gives $A = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$, $n(A) = 6$.
$$ P(A) = \frac{6}{36} = \frac{1}{6}. $$题目：掷两枚均匀的六面骰子。求两数之和为 $7$ 的概率。
样本空间。有序对 $(d_1, d_2)$，其中每个 $d_i \in \{1, \ldots, 6\}$，所以 $n(U) = 36$，每一对等可能。
事件。"和为 $7$"对应 $A = \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$，$n(A) = 6$。
$$ P(A) = \frac{6}{36} = \frac{1}{6}. $$Worked Example — Drawing a card例题——抽一张牌
Problem: One card is drawn from a standard $52$-card deck. What is the probability the card is a face card (J, Q, K)?
$n(U) = 52$. Face cards: $3$ ranks $\times$ $4$ suits $= 12$. So $n(A) = 12$ and
$$ P(A) = \frac{12}{52} = \frac{3}{13}. $$Notice we keep the answer as an exact fraction — Paper 1 is non-calculator, and decimals lose marks when an exact form was available.
题目：从一副标准 $52$ 张牌中抽一张。求抽到人头牌（J、Q、K）的概率。
$n(U) = 52$。人头牌：$3$ 种点数 $\times$ $4$ 种花色 $= 12$ 张。所以 $n(A) = 12$，
$$ P(A) = \frac{12}{52} = \frac{3}{13}. $$注意答案保留为精确分数——Paper 1 不可使用计算器，若可写成精确分数却用小数，是会扣分的。
▸ Going deeper — The three Kolmogorov axioms深入——柯尔莫哥洛夫三公理
Every probability statement in IB AA HL ultimately rests on three axioms:
Axiom 1 (non-negativity). $P(A) \ge 0$ for every event $A$.
Axiom 2 (unit measure). $P(U) = 1$.
Axiom 3 (additivity for disjoint events). If $A \cap B = \varnothing$, then $P(A \cup B) = P(A) + P(B)$.
The complement rule, the addition rule for overlapping events, and the bounds $0 \le P(A) \le 1$ are all derived from these three. For instance, $P(A) + P(A') = P(U) = 1$ follows from Axiom 3 (since $A$ and $A'$ are disjoint and their union is $U$) and Axiom 2.
You won't be asked to recite the axioms, but Paper 2 "show that" questions sometimes ask you to justify a step — quoting "by the addition rule for mutually exclusive events" is exactly Axiom 3 in disguise.
IB AA HL 中所有概率陈述最终都建立在三条公理之上：
公理 1（非负性）。对任意事件 $A$，$P(A) \ge 0$。
公理 2（规范性）。$P(U) = 1$。
公理 3（互斥事件可加性）。若 $A \cap B = \varnothing$，则 $P(A \cup B) = P(A) + P(B)$。
补集（complement）法则、可重叠事件的加法定律（addition rule），以及 $0 \le P(A) \le 1$ 都是从这三条推导而来。例如 $P(A) + P(A') = P(U) = 1$，就是先用公理 3（$A$ 与 $A'$ 互斥且并起来等于 $U$），再用公理 2 得到的。
考试不会要求你背公理，但 Paper 2 的 "show that" 题有时要你说明某一步的依据——写 "by the addition rule for mutually exclusive events" 实际上就是公理 3 的另一种表达。
Complementary Events and Expected Frequency互补事件与期望频数 SL 4.5
complement）$A'$ 包含所有不属于 $A$ 的结果：
$$ P(A') = 1 - P(A). $$
当题目问"$P(\text{at least one})$"（至少一个）或"$P(\text{not } X)$"（不发生 $X$）时，从补集入手通常比直接枚举更快。对 $n$ 次独立试验（independent events），$A$ 发生次数的期望（expected number of occurrences）为 $n\,P(A)$。
Worked Example — At least one of something例题——至少出现一次
Problem: A bag contains $5$ red and $3$ blue marbles. Three marbles are drawn without replacement. Find $P(\text{at least one blue})$.
Use the complement. "At least one blue" $=$ NOT "all red":
$$ P(\text{all red}) = \frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} = \frac{60}{336} = \frac{5}{28}. $$ $$ P(\text{at least one blue}) = 1 - \frac{5}{28} = \frac{23}{28}. $$题目：袋中有 $5$ 个红球与 $3$ 个蓝球。不放回地抽 $3$ 个球。求 $P(\text{at least one blue})$（至少一个蓝球）。
用补集。"至少一个蓝球" $=$ "不全是红球"的补：
$$ P(\text{all red}) = \frac{5}{8} \cdot \frac{4}{7} \cdot \frac{3}{6} = \frac{60}{336} = \frac{5}{28}. $$ $$ P(\text{at least one blue}) = 1 - \frac{5}{28} = \frac{23}{28}. $$Worked Example — Expected frequency例题——期望频数
Problem: A spinner shows red with probability $0.3$. It is spun $200$ times. How many times is it expected to land on red?
The expected number of occurrences is $n\,P(A)$:
$$ E = 200 \times 0.3 = 60. $$This is the theoretical expectation, not a prediction of the exact count. In a real $200$-spin trial the count will fluctuate around $60$.
题目：一个旋转盘指向红色的概率为 $0.3$。共旋转 $200$ 次。期望落在红色上的次数是多少？
期望发生次数（expected number of occurrences）等于 $n\,P(A)$：
这是理论期望，不是对实际次数的预测。在真实的 $200$ 次试验中，实际计数会在 $60$ 附近波动。
▸ Going deeper — Why $E = nP(A)$深入——为什么 $E = nP(A)$
Let $X_i$ be the indicator of $A$ on trial $i$ ($X_i = 1$ if $A$ occurs, $0$ otherwise). Then $E[X_i] = 1 \cdot P(A) + 0 \cdot P(A') = P(A)$. The total count $X = X_1 + X_2 + \cdots + X_n$ has expectation
$$ E[X] = E[X_1] + E[X_2] + \cdots + E[X_n] = n\,P(A), $$using linearity of expectation. Linearity holds even when the $X_i$ are not independent, which is why $E[X] = nP(A)$ also works for drawing without replacement (where each trial has the same marginal probability $P(A)$ even though the trials are dependent). This is the symptom-free way to read IB problems that say "in the long run."
设 $X_i$ 为第 $i$ 次试验中 $A$ 的指示函数（$A$ 发生则 $X_i = 1$，否则为 $0$）。则 $E[X_i] = 1 \cdot P(A) + 0 \cdot P(A') = P(A)$。总计数 $X = X_1 + X_2 + \cdots + X_n$ 的期望为
$$ E[X] = E[X_1] + E[X_2] + \cdots + E[X_n] = n\,P(A), $$这用到了期望的线性性。即便 $X_i$ 不独立，线性性仍然成立。这就是为什么 $E[X] = nP(A)$ 对"不放回抽样"同样适用：虽然各次试验相互依赖，但每次的边际概率仍为 $P(A)$。IB 题目里出现 "in the long run"（长期来看）的提法时，正是这个意思。
Combined Events and the Addition Rule复合事件与加法定律 SL 4.6
mutually exclusive events，不能同时发生）时，$P(A \cap B) = 0$，公式简化为 $P(A) + P(B)$。读题时盯紧关键词："and"（交，intersection）、"or"（并，union）。
Mutually exclusive. $A$ and $B$ cannot both occur. $P(A \cap B) = 0$.
Independent. Knowing one does not change the probability of the other. $P(A \cap B) = P(A)\,P(B)$.
Two non-trivial events ($P(A), P(B) > 0$) cannot be both mutually exclusive and independent — if they were exclusive then $P(A\cap B) = 0 \ne P(A)P(B)$. 这两个词在英文里听起来很像，但在概率里含义完全相反：
互斥（
mutually exclusive）。$A$ 与 $B$ 不能同时发生。$P(A \cap B) = 0$。独立（
independent）。知道其中一个发生与否，不会改变另一个的概率。$P(A \cap B) = P(A)\,P(B)$。若两个事件概率都不为零（$P(A), P(B) > 0$），它们不可能既互斥又独立——若互斥则 $P(A\cap B) = 0 \ne P(A)P(B)$。
Worked Example — Card from a deck例题——从一副牌中抽一张
Problem: One card is drawn from a standard deck. Let $A = \{\text{card is a heart}\}$, $B = \{\text{card is a face card}\}$. Find $P(A \cup B)$.
$P(A) = 13/52$, $P(B) = 12/52$. Overlap (hearts that are face cards): J$\heartsuit$, Q$\heartsuit$, K$\heartsuit$, so $P(A \cap B) = 3/52$.
$$ P(A \cup B) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}. $$题目：从标准 52 张牌中抽一张。设 $A = \{\text{card is a heart}\}$（红心），$B = \{\text{card is a face card}\}$（人头牌）。求 $P(A \cup B)$。
$P(A) = 13/52$，$P(B) = 12/52$。重叠部分（既是红心又是人头牌）：J$\heartsuit$、Q$\heartsuit$、K$\heartsuit$，因此 $P(A \cap B) = 3/52$。
$$ P(A \cup B) = \frac{13}{52} + \frac{12}{52} - \frac{3}{52} = \frac{22}{52} = \frac{11}{26}. $$Worked Example — Disjoint cases例题——互斥情形
Problem: A single ball is drawn from an urn containing $4$ red, $6$ green and $5$ blue balls. Find $P(\text{red or green})$.
Red and green are mutually exclusive (the ball has one colour), so:
$$ P(R \cup G) = P(R) + P(G) = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. $$Equivalently, complement of blue: $1 - 5/15 = 2/3$.
题目：从装有 $4$ 红、$6$ 绿、$5$ 蓝共 $15$ 个球的瓮中抽一个球。求 $P(\text{red or green})$。
红与绿互斥（一个球只有一种颜色），所以：
$$ P(R \cup G) = P(R) + P(G) = \frac{4}{15} + \frac{6}{15} = \frac{10}{15} = \frac{2}{3}. $$等价做法：取蓝色的补集，$1 - 5/15 = 2/3$。
▸ Going deeper — Inclusion-exclusion for three events深入——三事件的容斥原理
The two-event addition rule generalizes. For three events,
$$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C). $$Reading. Add the singletons, subtract every pairwise overlap (each was added twice in the singletons but should appear once), then add back the triple overlap (it was added three times and subtracted three times, so we need it once more). This pattern is called inclusion-exclusion; the IB usually presents it as a Venn-diagram problem.
Example. Of $100$ students, $50$ study French ($F$), $40$ study Spanish ($S$), $30$ study German ($G$). $20$ study French and Spanish, $15$ French and German, $10$ Spanish and German; $5$ study all three. How many study at least one language?
$$ |F \cup S \cup G| = 50 + 40 + 30 - 20 - 15 - 10 + 5 = 80. $$So $80$ study at least one and $20$ study none.
两事件加法定律可以推广。对于三事件，
$$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C). $$解读：先加单个事件，再减去每一对的两两重叠（每对在单个加和里被算了两次，应保留一次），最后加回三者的共同部分（在前两步里被加三次、减三次，需要再加一次）。这就是容斥原理（inclusion-exclusion）。IB 常以韦恩图问题考查这条公式。
示例：$100$ 名学生中，$50$ 学法语 ($F$)，$40$ 学西语 ($S$)，$30$ 学德语 ($G$)。$20$ 同时学法、西，$15$ 同时学法、德，$10$ 同时学西、德，$5$ 三种都学。问至少学一门的有多少人？
$$ |F \cup S \cup G| = 50 + 40 + 30 - 20 - 15 - 10 + 5 = 80. $$所以有 $80$ 人至少学一门，$20$ 人一门都没学。
Venn, Tree, and Sample-Space Diagrams韦恩图、树状图与样本空间表 SL 4.6
Tree diagram — sequential events; multiply along branches, add over leaves.
Sample-space table — two independent trials with small outcome sets (dice, coins).
The IB markscheme awards setup marks for the correct diagram; draw it before computing. 韦恩图（
Venn diagram）——事件之间存在重叠，按区域读人数/计数。树状图（
tree diagram）——顺序发生的事件；沿分支相乘，按叶子求和。样本空间表（
sample-space table）——两次独立试验且结果集合较小（骰子、硬币）。IB 评分方案会给"正确画出图"这一步独立的设置分；务必先画图再计算。
Venn diagram — two events韦恩图——两个事件
Sample-space table — two dice样本空间表——两枚骰子
| 1 | 2 | 3 | 4 | 5 | 6 | |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Each of the $36$ cells is equally likely. Read off probabilities by counting cells: $P(\text{sum} = 7) = 6/36 = 1/6$; $P(\text{sum} \ge 10) = 6/36 = 1/6$; $P(\text{sum is prime}) = 15/36 = 5/12$.
$36$ 个格子每一个都等可能。数格子就能直接读概率：$P(\text{sum} = 7) = 6/36 = 1/6$；$P(\text{sum} \ge 10) = 6/36 = 1/6$；$P(\text{sum is prime}) = 15/36 = 5/12$。
Tree diagram — two-stage experiment树状图——两阶段试验
Bag: $3$ red, $2$ blue. Draw two without replacement.
袋中：$3$ 红、$2$ 蓝。不放回地抽两个。
-
Draw 1第 1 次: R ($3/5$)
- Draw 2第 2 次: R ($2/4$) → RR : $3/5 \cdot 2/4 = 6/20$
- Draw 2第 2 次: B ($2/4$) → RB : $3/5 \cdot 2/4 = 6/20$
-
Draw 1第 1 次: B ($2/5$)
- Draw 2第 2 次: R ($3/4$) → BR : $2/5 \cdot 3/4 = 6/20$
- Draw 2第 2 次: B ($1/4$) → BB : $2/5 \cdot 1/4 = 2/20$
Leaves sum to $6/20 + 6/20 + 6/20 + 2/20 = 20/20 = 1$.
叶子概率之和 $6/20 + 6/20 + 6/20 + 2/20 = 20/20 = 1$。
Multiply along a path. Each leaf probability is the product of the branches that reach it.
Add over leaves. To find $P(\text{event})$, sum the leaf probabilities for every path that satisfies the event.
For with replacement, every branch repeats the same probabilities. For without replacement, the second-stage probabilities depend on the first-stage outcome (as above). 同一节点的分支之和为 $1$。动手计算前先核对——一个分支错，叶子全错。
沿路径相乘。每个叶子的概率等于到达它所经过的所有分支概率的乘积。
按叶子求和。要算 $P(\text{event})$，把满足该事件的所有路径的叶子概率相加。
有放回（with replacement）时，各分支概率重复出现；不放回（without replacement）时，第二阶段的概率取决于第一阶段的结果（如上图所示）。
Worked Example — Tree, two-colour draw例题——树状图，两色抽取
Problem: Using the bag $3$ R, $2$ B above, find $P(\text{exactly one red})$.
"Exactly one red" $=$ RB or BR — two leaves:
$$ P(\text{exactly one R}) = \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}. $$题目：沿用上面的袋子（$3$ R、$2$ B），求 $P(\text{exactly one red})$（恰好一个红球）。
"恰好一红" $=$ RB 或 BR——两片叶子：
$$ P(\text{exactly one R}) = \frac{6}{20} + \frac{6}{20} = \frac{12}{20} = \frac{3}{5}. $$▸ Going deeper — Why tree diagrams encode the multiplication rule深入——树状图为何对应乘法定律
A path from root to leaf is a sequence of outcomes. The probability that all of those outcomes happen is, by the multiplication rule for conditional probabilities (chain rule),
$$ P(E_1 \cap E_2 \cap \cdots \cap E_k) = P(E_1)\,P(E_2 \mid E_1)\,P(E_3 \mid E_1 \cap E_2)\cdots $$Each branch probability is exactly the conditional one needed for the next step. Multiplying along the path therefore gives the joint probability of that whole path. Summing over disjoint paths gives the probability of any union of paths — which is what events on tree diagrams always are.
从根到叶的一条路径就是一串结果。这些结果全部发生的概率，由条件概率的乘法定律（链式法则，chain rule）给出：
每条分支的概率正好是下一步所需的条件概率。因此沿路径相乘就得到这条路径的联合概率。把若干条互斥路径的概率相加，就得到任意路径并的概率——而树状图上的事件本质上正是路径之并。
Conditional Probability条件概率 SL 4.6
Worked Example — Two-way table例题——双向表
Problem: A school surveys $200$ students about sports preferences:
| Soccer (S) | Not soccer ($S'$) | Total | |
|---|---|---|---|
| Boys (B) | 60 | 40 | 100 |
| Girls (B') | 30 | 70 | 100 |
| Total | 90 | 110 | 200 |
A student is chosen at random. Find (a) $P(S \mid B)$ and (b) $P(B \mid S)$.
(a) Restrict to boys ($n = 100$); $60$ of them prefer soccer:
$$ P(S \mid B) = \frac{60}{100} = \frac{3}{5}. $$(b) Restrict to soccer ($n = 90$); $60$ of them are boys:
$$ P(B \mid S) = \frac{60}{90} = \frac{2}{3}. $$Note that $P(S \mid B) \ne P(B \mid S)$ — the order of the conditional matters. This is the seed of Bayes' theorem.
题目：某校对 $200$ 名学生做体育偏好调查：
| Soccer (S) | Not soccer ($S'$) | Total | |
|---|---|---|---|
| Boys (B) | 60 | 40 | 100 |
| Girls (B') | 30 | 70 | 100 |
| Total | 90 | 110 | 200 |
随机选一名学生。求 (a) $P(S \mid B)$ 与 (b) $P(B \mid S)$。
(a) 限制到男生（$n = 100$），其中 $60$ 人喜欢足球：
$$ P(S \mid B) = \frac{60}{100} = \frac{3}{5}. $$(b) 限制到喜欢足球的学生（$n = 90$），其中 $60$ 人是男生：
$$ P(B \mid S) = \frac{60}{90} = \frac{2}{3}. $$注意 $P(S \mid B) \ne P(B \mid S)$——条件的方向至关重要。这正是贝叶斯定理（Bayes' theorem）的起点。
Worked Example — From the formula例题——直接套公式
Problem: $P(A) = 0.6$, $P(B) = 0.4$, $P(A \cap B) = 0.24$. Find $P(A \mid B)$ and $P(B \mid A)$.
$$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.4} = 0.6, $$ $$ P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.24}{0.6} = 0.4. $$Notice $P(A \mid B) = P(A)$ and $P(B \mid A) = P(B)$ — the events are independent. (Equivalent check: $P(A)P(B) = 0.24 = P(A \cap B)$.)
题目：$P(A) = 0.6$，$P(B) = 0.4$，$P(A \cap B) = 0.24$。求 $P(A \mid B)$ 与 $P(B \mid A)$。
$$ P(A \mid B) = \frac{P(A \cap B)}{P(B)} = \frac{0.24}{0.4} = 0.6, $$ $$ P(B \mid A) = \frac{P(A \cap B)}{P(A)} = \frac{0.24}{0.6} = 0.4. $$注意到 $P(A \mid B) = P(A)$、$P(B \mid A) = P(B)$——两个事件相互独立。（等价验证：$P(A)P(B) = 0.24 = P(A \cap B)$。）
▸ Going deeper — The multiplication rule and the chain rule深入——乘法定律与链式法则
Rearranging the conditional formula gives the multiplication rule:
$$ P(A \cap B) = P(B)\,P(A \mid B). $$By symmetry, the same intersection equals $P(A)\,P(B \mid A)$. Equating gives Bayes' theorem in its smallest form (see Section 2.7).
Extending to three events:
$$ P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B). $$This is the chain rule and it underwrites every tree diagram: each branch probability is a conditional, and a leaf is a product of conditionals along its path.
把条件概率公式整理一下就是乘法定律：
$$ P(A \cap B) = P(B)\,P(A \mid B). $$由对称性，该交集也等于 $P(A)\,P(B \mid A)$。两者相等就给出贝叶斯定理（Bayes' theorem）的最小形式（见 2.7 节）。
推广到三个事件：
$$ P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B). $$这就是链式法则（chain rule），也是每张树状图背后的原理：每条分支概率都是一个条件概率，每片叶子就是这条路径上所有条件概率的乘积。
Independent Events独立事件 SL 4.6
independent events）是一种数值意义上的性质——必须把等式算出来验证，单凭"看起来没关系"是不能下结论的。
Independence is not "disjoint." Two disjoint non-trivial events are dependent, not independent (knowing one happened tells you the other did not).
With replacement ⇒ independent across trials. Without replacement ⇒ dependent.
"Show that $A$ and $B$ are independent" means compute both sides of $P(A)P(B) = P(A \cap B)$ and confirm equality. 独立不等于"不同事件"。从一张牌中抽到 K 与抽到红心是独立事件（$P(K \cap H) = 1/52 = (4/52)(13/52)$ ✓），尽管它们看似"不同"。
独立不等于互斥。两个互不相交且概率都不为零的事件其实是相依的，不是独立的（知道一个发生，就立刻知道另一个一定没发生）。
有放回 ⇒ 各次试验相互独立；不放回 ⇒ 相依。
"Show that $A$ and $B$ are independent"（证明 $A$ 与 $B$ 独立）就是要把 $P(A)P(B)$ 与 $P(A \cap B)$ 都算出来，确认两边相等。
Worked Example — Verifying independence例题——验证独立性
Problem: A single card is drawn from a $52$-card deck. Let $A = \{\text{ace}\}$, $B = \{\text{spade}\}$. Are $A$ and $B$ independent?
$P(A) = 4/52 = 1/13$; $P(B) = 13/52 = 1/4$; $P(A \cap B) = P(\text{ace of spades}) = 1/52$. Check:
$$ P(A)\,P(B) = \frac{1}{13} \cdot \frac{1}{4} = \frac{1}{52} = P(A \cap B). $$So $A$ and $B$ are independent.
题目：从 $52$ 张牌中抽一张。设 $A = \{\text{ace}\}$（A 牌），$B = \{\text{spade}\}$（黑桃）。问 $A$、$B$ 是否独立？
$P(A) = 4/52 = 1/13$；$P(B) = 13/52 = 1/4$；$P(A \cap B) = P(\text{ace of spades}) = 1/52$。验证：
$$ P(A)\,P(B) = \frac{1}{13} \cdot \frac{1}{4} = \frac{1}{52} = P(A \cap B). $$因此 $A$、$B$ 独立。
Worked Example — Sequential independent trials例题——序贯独立试验
Problem: A biased coin shows heads with probability $0.6$. It is tossed three times. Find $P(\text{exactly two heads})$.
Each toss is independent. Three orderings of two heads: HHT, HTH, THH. Each has probability $0.6 \cdot 0.6 \cdot 0.4 = 0.144$.
$$ P(\text{exactly 2 H}) = 3 \cdot 0.144 = 0.432. $$This is the binomial probability $\binom{3}{2}\,p^2\,(1-p)^{1} = 3 \cdot 0.36 \cdot 0.4 = 0.432$ — the bridge to Unit D3.
题目：一枚有偏硬币正面朝上概率为 $0.6$。抛 3 次。求 $P(\text{exactly two heads})$（恰好两次正面）。
各次抛掷相互独立。"两正一反"有三种排列：HHT、HTH、THH，每种概率都是 $0.6 \cdot 0.6 \cdot 0.4 = 0.144$。
$$ P(\text{exactly 2 H}) = 3 \cdot 0.144 = 0.432. $$这其实就是二项概率 $\binom{3}{2}\,p^2\,(1-p)^{1} = 3 \cdot 0.36 \cdot 0.4 = 0.432$——通向 D3 单元的桥梁。
▸ Going deeper — Pairwise vs. mutual independence深入——两两独立与相互独立
For three or more events, pairwise independence ($P(A_i \cap A_j) = P(A_i)P(A_j)$ for every pair) is not the same as mutual independence (every subset factors). The IB AA HL syllabus only requires the two-event criterion, so this is mostly trivia — but it's a classic counterexample worth knowing.
Counterexample. Toss two fair coins. Let $A = \{\text{first is H}\}$, $B = \{\text{second is H}\}$, $C = \{\text{exactly one H}\}$. Then $P(A) = P(B) = P(C) = 1/2$ and every pairwise intersection equals $1/4 = (1/2)(1/2)$. But $A \cap B \cap C = \varnothing$ (if both are H, you don't have exactly one), so $P(A \cap B \cap C) = 0 \ne 1/8$. Pairwise yes, mutually no.
当涉及三个或更多事件时，两两独立（任意一对都满足 $P(A_i \cap A_j) = P(A_i)P(A_j)$）不等于相互独立（任意子集的概率都等于各事件概率之积）。IB AA HL 大纲只考查两事件判定，所以这只是补充知识，但反例值得知道。
反例：抛两枚均匀硬币。设 $A = \{\text{first is H}\}$、$B = \{\text{second is H}\}$、$C = \{\text{exactly one H}\}$。则 $P(A) = P(B) = P(C) = 1/2$，且任何两两交集都等于 $1/4 = (1/2)(1/2)$。但 $A \cap B \cap C = \varnothing$（若两次都正面，就不可能恰好一次正面），所以 $P(A \cap B \cap C) = 0 \ne 1/8$。两两独立成立，相互独立不成立。
Bayes' Theorem贝叶斯定理 AHL 4.10 HL
partition）$\{A_1, A_2, A_3\}$ 与任意满足 $P(B) > 0$ 的事件 $B$，
$$ P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{\displaystyle\sum_{j} P(B \mid A_j)\,P(A_j)}. $$
IB AA HL 大纲规定，贝叶斯定理（Bayes' theorem）中划分最多含三个事件。分子是联合概率 $P(A_i \cap B) = P(B\mid A_i)P(A_i)$；分母是 $B$ 的全概率 $P(B)$，在所有 $A_j$ 上求和。
For two events ($\{A, A'\}$):
对两个事件的划分（$\{A, A'\}$）：
$$ P(A \mid B) = \frac{P(B \mid A)\,P(A)}{P(B \mid A)\,P(A) + P(B \mid A')\,P(A')} $$Setup ritual. Identify the partition $\{A_1, \ldots, A_k\}$ (the causes). Write the prior $P(A_i)$ for each. Write the likelihood $P(B \mid A_i)$ for each. Plug into the formula. 题目给的是正向条件概率——$P(\text{positive test} \mid \text{disease})$、$P(\text{defect} \mid \text{factory})$、$P(\text{spam word} \mid \text{spam})$——而问的是反向：$P(\text{disease} \mid \text{positive test})$、$P(\text{factory} \mid \text{defect})$、$P(\text{spam} \mid \text{contains word})$。只要题目问"已知一个[下游观察]，求 $P([\text{上游原因}])$"，答案就是贝叶斯。
套路化设定：先确定划分 $\{A_1, \ldots, A_k\}$（即所有可能"原因"）。为每个原因写出先验（
prior）$P(A_i)$。再写出每个的似然（likelihood）$P(B \mid A_i)$。最后代入公式。
Worked Example — Medical test (two events)例题——医学检测（两个事件）
Problem: A disease affects $1\%$ of a population. A test has $95\%$ sensitivity ($P(+ \mid D) = 0.95$) and $90\%$ specificity ($P(- \mid D') = 0.90$, so $P(+ \mid D') = 0.10$). A randomly chosen person tests positive. Find $P(D \mid +)$.
Partition: $\{D, D'\}$ with priors $P(D) = 0.01$, $P(D') = 0.99$.
Likelihoods: $P(+ \mid D) = 0.95$, $P(+ \mid D') = 0.10$.
Total probability of testing positive:
$$ P(+) = P(+ \mid D)P(D) + P(+ \mid D')P(D') = (0.95)(0.01) + (0.10)(0.99) = 0.0095 + 0.099 = 0.1085. $$Bayes:
$$ P(D \mid +) = \frac{(0.95)(0.01)}{0.1085} = \frac{0.0095}{0.1085} = \frac{19}{217} \approx 0.0876. $$Even a $95\%$-sensitive test on a rare disease gives only about $8.8\%$ posterior — the base rate dominates. This is the classic "base-rate" lesson Bayes teaches.
题目：某疾病在人群中患病率为 $1\%$。检测的灵敏度为 $95\%$（$P(+ \mid D) = 0.95$），特异度为 $90\%$（$P(- \mid D') = 0.90$，故 $P(+ \mid D') = 0.10$）。随机选一人检测呈阳性，求 $P(D \mid +)$。
划分：$\{D, D'\}$，先验 $P(D) = 0.01$，$P(D') = 0.99$。
似然：$P(+ \mid D) = 0.95$，$P(+ \mid D') = 0.10$。
检测呈阳性的全概率（law of total probability）：
贝叶斯：
$$ P(D \mid +) = \frac{(0.95)(0.01)}{0.1085} = \frac{0.0095}{0.1085} = \frac{19}{217} \approx 0.0876. $$对一种罕见病，即便检测灵敏度高达 $95\%$，后验（posterior）也只有约 $8.8\%$——基线患病率才是主导因素。这正是贝叶斯告诉我们的"基率"经典教训。
Worked Example — Three factories例题——三家工厂
Problem: A company sources widgets from three factories: $A_1$ supplies $50\%$, $A_2$ supplies $30\%$, $A_3$ supplies $20\%$. The defect rates are $1\%$, $2\%$, $5\%$ respectively. A widget is selected at random and found to be defective ($B$). Find $P(A_2 \mid B)$.
Partition: $\{A_1, A_2, A_3\}$ with $P(A_1) = 0.5$, $P(A_2) = 0.3$, $P(A_3) = 0.2$.
Likelihoods: $P(B \mid A_1) = 0.01$, $P(B \mid A_2) = 0.02$, $P(B \mid A_3) = 0.05$.
Total probability of a defective widget:
$$ P(B) = (0.01)(0.5) + (0.02)(0.3) + (0.05)(0.2) = 0.005 + 0.006 + 0.010 = 0.021. $$Posterior for factory 2:
$$ P(A_2 \mid B) = \frac{(0.02)(0.3)}{0.021} = \frac{0.006}{0.021} = \frac{6}{21} = \frac{2}{7}. $$So about $28.6\%$ of defective widgets come from factory $2$. Sanity check: $P(A_1 \mid B) + P(A_2 \mid B) + P(A_3 \mid B) = 5/21 + 6/21 + 10/21 = 21/21 = 1$. ✓
题目：某公司从三家工厂进货：$A_1$ 占 $50\%$，$A_2$ 占 $30\%$，$A_3$ 占 $20\%$。三家次品率分别为 $1\%$、$2\%$、$5\%$。随机抽一件产品发现是次品（$B$），求 $P(A_2 \mid B)$。
划分：$\{A_1, A_2, A_3\}$，$P(A_1) = 0.5$，$P(A_2) = 0.3$，$P(A_3) = 0.2$。
似然：$P(B \mid A_1) = 0.01$，$P(B \mid A_2) = 0.02$，$P(B \mid A_3) = 0.05$。
抽到次品的全概率：
$$ P(B) = (0.01)(0.5) + (0.02)(0.3) + (0.05)(0.2) = 0.005 + 0.006 + 0.010 = 0.021. $$来自工厂 2 的后验：
$$ P(A_2 \mid B) = \frac{(0.02)(0.3)}{0.021} = \frac{0.006}{0.021} = \frac{6}{21} = \frac{2}{7}. $$因此次品中约有 $28.6\%$ 来自工厂 2。验算：$P(A_1 \mid B) + P(A_2 \mid B) + P(A_3 \mid B) = 5/21 + 6/21 + 10/21 = 21/21 = 1$。 ✓
▸ Going deeper — Deriving Bayes from the chain rule深入——从链式法则推导贝叶斯
From the multiplication rule, the joint probability of $A_i$ and $B$ can be written two ways:
$$ P(A_i \cap B) = P(A_i)\,P(B \mid A_i) = P(B)\,P(A_i \mid B). $$Solve for $P(A_i \mid B)$:
$$ P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{P(B)}. $$The denominator $P(B)$ comes from the law of total probability: if $\{A_1, A_2, A_3\}$ partitions $U$ then
$$ P(B) = P(B \cap A_1) + P(B \cap A_2) + P(B \cap A_3) = \sum_j P(B \mid A_j)\,P(A_j). $$Substituting gives the full Bayes formula. The IB syllabus stops at three partition elements, but the same derivation generalizes to any number.
由乘法定律，$A_i$ 与 $B$ 的联合概率可以用两种方式写出：
$$ P(A_i \cap B) = P(A_i)\,P(B \mid A_i) = P(B)\,P(A_i \mid B). $$解出 $P(A_i \mid B)$：
$$ P(A_i \mid B) = \frac{P(B \mid A_i)\,P(A_i)}{P(B)}. $$分母 $P(B)$ 来自全概率公式（law of total probability）：若 $\{A_1, A_2, A_3\}$ 是 $U$ 的划分，则
代回去就得到完整的贝叶斯公式。IB 大纲只考到三个划分元素，但同样的推导对任意数量都成立。
▸ Going deeper — "Likelihood times prior" reading深入——把贝叶斯读成"似然 × 先验"
Three pieces of vocabulary worth knowing:
- Prior $P(A_i)$ — what you believed about cause $A_i$ before seeing $B$.
- Likelihood $P(B \mid A_i)$ — how plausible the observation $B$ is under cause $A_i$.
- Posterior $P(A_i \mid B)$ — updated belief about $A_i$ after observing $B$.
The formula reads posterior $\propto$ likelihood $\times$ prior, normalized by the total probability of the evidence. This reading is what makes Bayes intuitive: rare causes (small prior) can still dominate if their likelihood for the evidence is much higher than competitors' — and vice versa, common causes win even with mediocre likelihoods.
有三个词值得记住：
- 先验（
prior）$P(A_i)$——在观察到 $B$ 之前，你对原因 $A_i$ 的信念。 - 似然（
likelihood）$P(B \mid A_i)$——在 $A_i$ 为真的前提下，观察到 $B$ 的合理程度。 - 后验（
posterior）$P(A_i \mid B)$——观察到 $B$ 之后对 $A_i$ 更新过的信念。
公式可以读作 后验 $\propto$ 似然 $\times$ 先验，再用证据的全概率归一化。这个读法让贝叶斯变得直观：罕见原因（先验小）只要其似然远高于其它竞争原因，仍可能在后验中占主导；反过来，常见原因即便似然平平也能胜出。
Exam Strategy & Common Pitfalls考试策略与常见陷阱
- $P(A) = n(A)/n(U)$ and $P(A') = 1 - P(A)$
- $P(A) = n(A)/n(U)$ 与 $P(A') = 1 - P(A)$
- $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
- $P(A \cup B) = P(A) + P(B) - P(A \cap B)$
- $P(A \mid B) = P(A \cap B)/P(B)$
- $P(A \mid B) = P(A \cap B)/P(B)$
- Independent $\Leftrightarrow P(A \cap B) = P(A)P(B)$
- 独立 $\Leftrightarrow P(A \cap B) = P(A)P(B)$
- $E[\text{occurrences}] = n\,P(A)$
- $E[\text{occurrences}] = n\,P(A)$（期望发生次数）
- Bayes' theorem (up to 3 events) HL
- 贝叶斯定理（最多 3 个事件）HL
- "and" $\to$ intersection; "or" $\to$ union; "given" $\to$ conditional
- "and" $\to$ 交集；"or" $\to$ 并集；"given" $\to$ 条件概率
- Mutually exclusive $\ne$ independent — opposite ideas
- 互斥 $\ne$ 独立——两者含义截然相反
- Tree diagrams encode the chain rule (multiply along, add across)
- 树状图本质上就是链式法则：沿路径相乘，按叶子相加
- When complement is faster than direct enumeration
- 何时用补集比直接枚举更快
- Why $P(A \mid B) \ne P(B \mid A)$ in general
- 为什么一般 $P(A \mid B) \ne P(B \mid A)$
- Posterior $\propto$ likelihood $\times$ prior (Bayes intuition) HL
- 后验 $\propto$ 似然 $\times$ 先验（贝叶斯直觉）HL
Common Pitfalls常见陷阱
2. Forgetting to subtract the overlap in $P(A \cup B)$ when the events are not exclusive.
3. Inverting the conditional: writing $P(A \mid B)$ when the question asks for $P(B \mid A)$.
4. Treating "without replacement" trials as independent — the second-stage probabilities change.
5. Not labelling tree branches, then summing the wrong leaves.
6. Filling in a Venn diagram from outside in. Always start with the central intersection.
7. For Bayes (HL): forgetting the denominator is the total probability of the evidence, summed over the whole partition — not just one branch.
8. Giving decimal answers on Paper 1 when an exact fraction was reachable. 1. 把"互斥（
mutually exclusive）"与"独立（independent）"混为一谈——两者几乎从不共存。2. 当 $A$、$B$ 并非互斥时，$P(A \cup B)$ 忘记减去重叠部分。
3. 把条件方向写反：题目要 $P(B \mid A)$ 却写成 $P(A \mid B)$。
4. 把"不放回"的多次试验当成独立试验——后一阶段的概率会变化。
5. 树状图分支没标好，最后把不该相加的叶子加进去。
6. 韦恩图从外往里填。一律先填中心的交集。
7. 贝叶斯（HL）：忘记分母是证据在整个划分上的全概率，而不是单一分支。
8. Paper 1 给小数答案，但题目允许保留精确分数——这会扣分。
Paper 2 (calc): Longer multi-part scenarios that combine several sub-topics — e.g. set up a Venn, then ask for a conditional, then ask for "given $B$, find $P(A)$" (Bayes for HL). Decimal answers acceptable but write the exact form first if available. Paper 1（不可使用计算器）：大部分概率题出在这里。答案要写精确分数。树状图、韦恩图、双向表为主，HL 学生还会有短小的贝叶斯部分，题目数字会刻意设计成相消干净的样子。
Paper 2（可使用计算器）：较长的多问情境，会同时考好几个子主题——例如先建立韦恩图，再算条件概率，最后问"已知 $B$ 求 $P(A)$"（HL 即贝叶斯）。允许小数答案，但能写出精确形式时务必先写。
Flashcards闪卡
Multiply along a path
Add across leaves求和为 $1$
沿路径相乘
按叶子相加
then $A$-only, $B$-only
then "neither"先填 $A\cap B$
再填只属于 $A$、只属于 $B$
最后填"都不属于"
Unit D2 — Practice QuizD2 单元——练习测验
Ten mixed-difficulty items in IB Paper-style flavour. Your score updates in real time at the top of the page. Aim for 8/10 before exam day. Items tagged HL are AHL 4.10 (Bayes).
10 道难度混合的题目，仿 IB Paper 风格。得分会在页面顶部实时更新。考前目标至少 8/10。带 HL 标记的是 AHL 4.10（贝叶斯）题。
Readiness Checklist备考清单
Click each item you've mastered. Aim for 100% before exam day. Items marked HL are AHL 4.10 (Bayes) and apply only to HL students.
每掌握一条就点一下。考试前争取 100%。带 HL 标记的是 AHL 4.10（贝叶斯），仅 HL 学生需要。
- Define trial, outcome, sample space $U$, and event; compute $P(A) = n(A)/n(U)$能定义试验、结果、样本空间 $U$ 与事件，并计算 $P(A) = n(A)/n(U)$ SL 4.5
- Apply the complement rule $P(A') = 1 - P(A)$ and recognize "at least one" cues能用补集法则 $P(A') = 1 - P(A)$，并识别"at least one"的信号 SL 4.5
- Compute the expected number of occurrences in $n$ trials using $n\,P(A)$能用 $n\,P(A)$ 计算 $n$ 次试验中 $A$ 的期望发生次数 SL 4.5
- Apply the addition rule $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ for any two events能对任意两个事件套用加法定律 $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ SL 4.6
- Identify mutually exclusive events and use the collapsed addition rule能识别互斥事件并使用简化版加法定律 SL 4.6
- Build and interpret Venn diagrams; fill in the intersection first能构建并解读韦恩图；一律先填交集 SL 4.6
- Build and interpret tree diagrams; multiply along paths and add over leaves能构建并解读树状图；沿路径相乘、按叶子相加 SL 4.6
- Use sample-space tables for two-trial experiments (dice, coins)能用样本空间表处理两阶段试验（骰子、硬币） SL 4.6
- Apply the conditional probability formula $P(A \mid B) = P(A\cap B)/P(B)$能套用条件概率公式 $P(A \mid B) = P(A\cap B)/P(B)$ SL 4.6
- Read two-way tables and extract conditional probabilities by restricting the sample space能读双向表，并通过限制样本空间得出条件概率 SL 4.6
- Test independence using $P(A \cap B) = P(A)P(B)$ and distinguish it from "mutually exclusive"能用 $P(A \cap B) = P(A)P(B)$ 检验独立，并把它与"互斥"区分开 SL 4.6
- State Bayes' theorem for a partition of up to three events and identify priors / likelihoods / posteriors能写出最多三事件划分下的贝叶斯定理，并指认先验 / 似然 / 后验 HL AHL 4.10
- Apply Bayes to two-event problems (e.g. medical tests) and derive the posterior given an observation能把贝叶斯用于两事件问题（如医学检测），在给定观察下求后验 HL AHL 4.10
- Apply Bayes to three-event partition problems (e.g. multiple factories / sources) and sanity-check that posteriors sum to $1$能把贝叶斯用于三事件划分问题（如多家工厂 / 来源），并验证后验之和为 $1$ HL AHL 4.10
IB Paper-Style PracticeIB 试卷风格练习
IB exam-style questions across Paper 1A (short response, no calc), Paper 1B (extended response, no calc), Paper 2 (calculator), and Paper 3 HL (Bayes extended exploration). EMH difficulty mix. Mark-by-mark solutions live in the separate solutions file. Use this after the in-page quiz and flashcards.
IB 考试风格题，涵盖 Paper 1A（短答，无计算器）、Paper 1B（长答，无计算器）、Paper 2（可用计算器），以及 Paper 3 HL（贝叶斯长题探究）。难度按 EMH 分级。逐分解答见独立的解答文档。建议在做完本页测验与闪卡后再来。