📚 A-Level Maths Conditional Probability: Common Mistakes Worksheet | A-Level 数学条件概率易错点总结
Conditional probability is a cornerstone of A-Level Mathematics, yet it consistently trips up even well-prepared students. The subtlety of ‘given that’ changes everything: the sample space shrinks, dependencies matter, and formulas must be applied with absolute precision. This worksheet-style revision article identifies the most frequent errors made in conditional probability and explains how to avoid them, using typical exam–style scenarios. Work through each section to diagnose your own blind spots.
条件概率是A-Level数学的基石，却总能让准备充分的学生马失前蹄。’在已知……的条件下’的微妙之处改变了一切：样本空间缩小了，依赖关系至关重要，公式必须绝对精准地使用。这篇类似练习题的复习文章，指出了条件概率中最常见的错误，并通过典型的考试题型，解释如何避免这些错误。逐一学习每个部分，以诊断你自己的知识盲点。
1. Reversing the Condition: P(A|B) vs P(B|A) | 混淆条件方向：P(A|B) 与 P(B|A)
The most classic trap is swapping the event and the condition. Students often read ‘probability that a person has a disease given they tested positive’ and accidentally calculate ‘probability of testing positive given they have the disease’. These are not equal unless the two unconditional probabilities happen to be the same, which is rarely true. In symbols, P(A|B) = P(B|A) only if P(A) = P(B) and the events are mutually inclusive in a very specific way. Always underline the wording: the event after ‘given that’ is the condition, it goes in the denominator.
最经典的陷阱就是把事件和条件互换。学生常把’已知检测为阳性，该人患病的概率’误算成’已知患病，检测为阳性的概率’。除非两个无条件概率恰好相等，且事件满足特定的包容关系，否则这两者并不相等，而这种情况极少发生。用符号表示，P(A|B) = P(B|A) 仅当 P(A) = P(B) 且事件处于极特殊的关系时才成立。务必在题目措辞下划线：’已知……’后面的事件是条件，要放在分母位置。
For example, if 1% of the population has disease D, and a test is 95% accurate (both sensitivity and specificity), then P(D|positive) is about 16%, whereas P(positive|D) is 95%. Many students will mistakenly quote 95% as the answer. Always use the definition:
例如，如果总人口中有1%患有疾病D，检测准确率为95%（灵敏度和特异度均为95%），那么 P(D|阳性) 约为 16%，而 P(阳性|D) 为 95%。很多学生会错误地引用 95% 作为答案。永远使用定义式：
P(A | B) = P(A ∩ B) / P(B)
Before jumping in, list both the event you need and the condition. Highlighter on the word ‘given’ helps.
在计算前，列出你需要的事件和你已知的条件。用荧光笔标出’已知’一词会很有帮助。
2. Misapplying the Multiplication Rule | 乘法法则的误用
The general multiplication rule is P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B). A common error is to write P(A ∩ B) = P(A) × P(B) when events are not independent. This completely ignores the conditional relationship. Students sometimes assume independence because the problem doesn’t state dependence, or they find it easier. Only use the product P(A)P(B) if independence is explicitly given or proven. When in doubt, use the chain rule with the appropriate conditional probability.
一般乘法法则是 P(A ∩ B) = P(A) × P(B|A) = P(B) × P(A|B)。一个常见错误是，在事件不独立时写成 P(A ∩ B) = P(A) × P(B)，这完全忽略了条件关系。学生有时会因为题目没有明确说明相依性，或者因计算简单而错误地假定独立。只有在明确给出或已证独立性的情况下，才能使用 P(A)P(B) 的乘积。拿不准时，始终使用包含恰当条件概率的链式法则。
For instance, in a bag with 5 red and 3 blue marbles, drawing two without replacement: the probability both are red is P(R₁ ∩ R₂) = (5/8) × (4/7), not (5/8)². The condition changes the second probability.
例如，一个袋子中有5个红球和3个蓝球，不放回地抽取两次：两球都是红色的概率是 P(R₁ ∩ R₂) = (5/8) × (4/7)，而不是 (5/8)²。条件改变了第二个概率。
3. Independence: False Assumptions & Misinterpreting Tests | 独立性：错误假设与检验的误读
Two events are independent if P(A|B) = P(A), or equivalently P(A ∩ B) = P(A)P(B). Many mistakes arise from assuming independence because events ‘feel’ unrelated, or from failing to test for it. In exam questions, you often need to verify independence by comparing P(A) with P(A|B). A subtle error is to assume that mutually exclusive events are independent – they are not. If A and B are mutually exclusive, P(A ∩ B) = 0, so P(A|B) = 0, which differs from P(A) unless P(A) = 0.
若 P(A|B) = P(A) 或等价地 P(A ∩ B) = P(A)P(B)，则两事件独立。许多错误源于凭’感觉’认为事件不相关而假定独立，或者疏于检验。考试题常要求通过比较 P(A) 与 P(A|B) 来验证独立性。一个微妙的错误是认为互斥事件是独立的——实际上它们不独立。若 A 与 B 互斥，P(A ∩ B) = 0，因此 P(A|B) = 0，除非 P(A) = 0，否则与 P(A) 不同。
In tree diagrams, the second set of branch probabilities is always conditional on the first event. Do not confuse these conditional branches with unconditional probabilities when testing independence.
在树形图中，第二层分支的概率总是以第一层事件为条件的。在检验独立性时，不要把这些条件分支概率与无条件概率混淆。
4. Tree Diagram Branch Probability Errors | 树形图分支概率错误
When drawing a tree for conditional probability problems, the probabilities on the first set of branches are unconditional: P(A) and P(A’). The probabilities on the second set of branches must be conditional: P(B|A), P(B’|A), P(B|A’) and P(B’|A’). A frequent mistake is to write P(B) on the second branches. This is wrong because the overall probability of B depends on which first branch occurred. Always label the second branches with the ‘given’ notation explicitly. Also, probabilities along any single path multiply to give an intersection probability; the sum of probabilities on all complete paths equals 1.
为条件概率问题绘制树形图时，第一层分支上的概率是无条件的：P(A) 和 P(A’)。第二层分支上的概率必须是有条件的：P(B|A), P(B’|A), P(B|A’) 和 P(B’|A’)。一个常见错误是在第二层分支上写 P(B)。这是错的，因为 B 的总体概率取决于发生了哪个第一层事件。务必明确用’条件’符号标记第二层分支。此外，任何一条路径上的概率相乘得到的是一个交事件的概率；所有完整路径的概率之和等于 1。
Another pitfall is omitting the complement branches: if you show P(B|A) = 0.3, you must also include P(B’|A) = 0.7 on the same fork. Incomplete branches cause flawed calculations when finding total probabilities.
另一个陷阱是遗漏补集分支：如果你标明 P(B|A) = 0.3，你必须在同一分叉处也标明 P(B’|A) = 0.7。分支不完整会导致在计算总概率时出错。
5. Confusing P(B) with P(B|A) in Bayes–type Questions | 在贝叶斯类问题中混淆 P(B) 与 P(B|A)
Bayes’ theorem is simply a rearrangement of the conditional probability definition, but its misuse is rampant. The theorem states:
贝叶斯定理不过是条件概率定义的重新整理，但滥用现象十分普遍。定理表述为：
P(A | B) = [P(B | A) × P(A)] / P(B)
where P(B) = P(B|A)P(A) + P(B|A’)P(A’). The key error is plugging P(B|A) in the denominator instead of the total probability P(B). Some students try to use P(B) = 0.5 by default or misinterpret the partition. Always compute the denominator using the Law of Total Probability with all mutually exclusive ways B can occur.
其中 P(B) = P(B|A)P(A) + P(B|A’)P(A’)。关键错误是将 P(B|A) 填入分母而不是总概率 P(B)。有些学生默认用 P(B) = 0.5，或者误解了分割。务必利用全概率公式，用 B 发生的所有互斥途径来计算分母。
A typical exam blunder: a factory has machines A and B producing items. 30% from A with 2% defective, 70% from B with 5% defective. Find probability an item was from A given it is defective. The incorrect approach: 0.02 / (0.02+0.05) ≈ 0.286; the correct denominator is 0.3×0.02 + 0.7×0.05 = 0.006 + 0.035 = 0.041, giving P(A|defective) = 0.006/0.041 ≈ 0.146. Weighting matters.
一个典型的考试错误：某工厂有机器 A 和 B 生产零件。A 生产 30% 的零件，次品率 2%；B 生产 70%，次品率 5%。求在已知零件为次品的条件下，它来自 A 的概率。错误做法：0.02/(0.02+0.05) ≈ 0.286；正确的分母是 0.3×0.02 + 0.7×0.05 = 0.006 + 0.035 = 0.041，得到 P(A|次品) = 0.006/0.041 ≈ 0.146。权重很重要。
6. Ignoring the Reduced Sample Space | 忽略缩减的样本空间
Conditional probability can be thought of as zooming in on the part of the Venn diagram where the condition holds. The new sample space is B, not the whole S. An intuitive mistake is to use P(A ∩ B) as the final answer for P(A|B) without dividing by P(B). Remember: P(A|B) is the proportion of A within B. In a contingency table, P(A|B) is the cell count for A∩B divided by the total of column B (or row B), not the grand total. Practise rewriting the table with the condition’s row/column as the new total of 1.
条件概率可以理解为将视角缩小到凡恩图中条件成立的那部分。新的样本空间是 B，而不是整个 S。一个直觉性错误是把 P(A ∩ B) 当作 P(A|B) 的最终答案，而没有除以 P(B)。记住：P(A|B) 是 A 在 B 内部的比例。在列联表中，P(A|B) 是 A∩B 的单元格计数除以 B 列（或 B 行）的总和，而不是总和。练习将表格中作为条件的那行或列改写为新的总和 1。
Example: a two–way table showing gender and subject choice. If asked ‘Given a student studies Maths, probability they are male’, you take the male–maths cell divided by the total maths students. If you mistakenly divide by total students, the answer is wrong and often far too small.
举例：一个显示性别与选课的二维表。如果问’已知某学生学数学，该生为男性的概率’，你应取男性–数学的单元格除以数学学生的总数。如果错误地除以全体学生总数，答案就是错的，而且往往过小。
7. Venn Diagram and False Positives Puzzle | 凡恩图与假阳性谜题
Medical testing problems illustrate the ‘false positive paradox’ and challenge intuition. With a rare disease (low prevalence) and a very accurate test, the probability a positive test indicates disease can still be low because false positives from the huge healthy population dominate. Students who skip the tree diagram or the Bayes setup often overestimate P(D|+). Make a table: Disease × Test result, fill with assumed numbers, then compute conditional proportions. This turns abstract percentages into concrete frequencies, drastically reducing errors.
医学检测问题体现了’假阳性悖论’，并对直觉构成挑战。在一种罕见疾病（低患病率）且检测非常精确的情况下，检测结果呈阳性表明确实患病的概率仍然可能很低，因为来自庞大健康人群的假阳性主导了结果。那些跳过树形图或贝叶斯设定的学生往往会高估 P(D|+)。制作一个表格：疾病 × 检测结果，用假设的数字填充，然后计算条件比例。这能把抽象的百分比转化为具体的频数，从而大幅减少错误。
For instance: population 10,000, prevalence 1% (100 have disease). Test sensitivity 99% (99 of 100 test pos.), specificity 95% (5% of 9,900 = 495 false positives). Total positives = 99+495 = 594. P(D|+) = 99/594 ≈ 16.7%. Many students guess near 99%.
例如：人口 10,000，患病率 1%（100 人患病）。检测灵敏度 99%（100 人中 99 人测出阳性），特异度 95%（9,900 人中的 5% = 495 个假阳性）。总阳性数 = 99+495 = 594。P(D|+) = 99/594 ≈ 16.7%。很多学生会猜接近 99%。
8. Algebraic Errors with Conditional Probability Notation | 条件概率符号的代数错误
Manipulating expressions like P(A’ ∩ B) or P[(A ∪ B) | C] requires caution. A common error is to write P(A’ ∩ B) = 1 – P(A ∩ B) – this is false. The correct complement rule for intersections uses the whole sample space: P(A’ ∩ B) = P(B) – P(A ∩ B), not 1 minus. Similarly, students often misapply De Morgan’s laws inside conditional probability. Remember that P[(A ∪ B) | C] = P(A ∪ B ∩ C)/P(C), and you must expand the numerator carefully using set properties. Always translate to intersections before simplifying.
对诸如 P(A’ ∩ B) 或 P[(A ∪ B) | C] 这类表达式进行操作需要谨慎。一个常见错误是写 P(A’ ∩ B) = 1 – P(A ∩ B)——这是错的。交集运算正确的补集规则使用的是整个样本空间：P(A’ ∩ B) = P(B) – P(A ∩ B)，而不是 1 减。同样，学生常常在条件概率中误用德摩根定律。记住 P[(A ∪ B) | C] = P(A ∪ B ∩ C)/P(C)，你必须仔细使用集合性质展开分子。在化简之前，总是先转化为交集。
Practice with simple numeric Venn diagrams: shade the region, write the probability, then translate into symbols. This builds robust intuition.
用简单的数值凡恩图进行练习：给区域涂上阴影，写出概率，然后转化为符号。这能建立起稳健的直觉。
9. Overlooking Conditional Independence | 忽视条件独立
A-level questions sometimes ask whether A and B are conditionally independent given C, meaning P(A ∩ B | C) = P(A | C)P(B | C). The error is to test for unconditional independence and assume it carries over. Two events can be dependent unconditionally but independent conditionally, and vice versa. For example, let C be ‘using a certain study method’, A = ‘pass exam’ and B = ‘complete homework’. Overall passing and homework might be dependent, but among those using the method, they could become independent. Always check with the specific condition.
A-Level 题目有时会问在给定 C 的条件下 A 和 B 是否条件独立，即 P(A ∩ B | C) = P(A | C)P(B | C)。错误做法是检验无条件独立性并假定这可以传递。两个事件在无条件时可能相依，而在条件下可能独立；反过来也一样。例如，设 C 为’使用某种学习方法’，A = ‘考试通过’，B = ‘完成作业’。整体上通过和作业可能相依，但在使用该方法的学生中，它们可能变成独立。务必针对特定条件进行检验。
This requires calculating three probabilities all conditioned on C and comparing product vs intersection.
这需要计算三个都以 C 为条件的概率，并比较乘积和交集。
10. Numerical Slips with Fractions and Decimals | 分数与小数的数值疏忽
Conditional probability problems often involve multiple fractions: P(A) = 2/5, P(B|A) = 3/8, etc. When computing P(A ∩ B) = 2/5 × 3/8, cancel and simplify to 3/20 before proceeding. Leaving unsimplified fractions leads to errors in the denominator of Bayes’ formula. Also, decimal rounding too early can cause answers that are marked incorrect. Keep at least four significant figures during intermediate steps. Use fraction mode on your calculator whenever possible.
条件概率问题常涉及多个分数：P(A) = 2/5，P(B|A) = 3/8 等等。计算 P(A ∩ B) = 2/5 × 3/8 时，要在进入下一步之前先约分得到 3/20。保留未化简的分数会导致在贝叶斯公式的分母中出错。此外，过早对小数进行四舍五入可能导致被判定为错误的答案。在中间步骤中至少保留四位有效数字。只要可能，就使用计算器的分数模式。
Another slip: when computing P(B) using total probability, add fractions with different denominators carefully. Draw a quick table or tree to keep the parts visible.
另一个马虎点：用全概率公式计算 P(B) 时，仔细地将不同分母的分数相加。快速画一个表格或树形图，让各个部分一目了然。
11. Ignoring the ‘Without Replacement’ Condition | 忽视’不放回’条件
Many probability questions involve selecting items without replacement, making the trials dependent. If a student assumes independence, they use constant probabilities across draws. For example, when drawing cards from a deck without replacement, the probability of a second heart given first heart is 12/51, not 13/52. Treat each scenario as a tree with changing denominators. Students who set up a binomial distribution (which assumes replacement) for without–replacement questions get the entire model wrong.
许多概率题涉及不放回地选取物品，这使得各次试验彼此相依。如果学生假定独立，他们就会在各次抽取中使用恒定概率。例如，从不放回的一副牌中抽牌，已知第一张是红心，第二张也是红心的概率是 12/51，而不是 13/52。要把每种情形都当作分母不断变化的树形图来处理。对不放回问题套用二项分布（假定放回）的学生，会把整个模型搞错。
Always check: ‘replaced’ or ‘not replaced’? Underline these words in the question.
始终检查：’放回’还是’不放回’？在题目中把这些词划下来。
12. The Worksheet Approach: Self-Test Questions | 练习题方法：自我检测题
To cement these concepts, treat the following as a diagnostic worksheet. Cover the answers and attempt each; then uncover and correct with the methods above.
要巩固这些概念，请将以下内容当作诊断练习题。遮住答案逐一尝试；然后对照上面的方法进行纠正。
| Question / 问题 | Common error / 常见错误 | Correct approach / 正确方法 |
|---|---|---|
| P(A)=0.4, P(B)=0.5, P(A∩B)=0.3. Find P(A|B) / 求 P(A|B) | Dividing 0.3 by 0.4 (reversing) / 除以 0.4（方向反了） | P(A|B) = 0.3/0.5 = 0.6 |
| Bag: 4 Red, 6 Blue. Draw two without replacement. P(2nd Red | 1st Blue) / 不放回，求P(第二个红|第一个蓝) | Using 4/10 or 4/9 after 1st Red / 用了4/10或先红后4/9 | After a blue is removed, 4R,5B remain. Probability = 4/9. |
| P(Disease)=0.02, P(Pos|D)=0.98, P(Pos|¬D)=0.03. Find P(D|Pos) / 求 P(患病|阳性) | Answering 0.98, or 0.02×0.98 in numerator only / 答案0.98，或分子只用0.02×0.98 | Denom = 0.02×0.98+0.98×0.03 = 0.0196+0.0294 = 0.049; P = 0.0196/0.049 = 0.4. |
Regularly interleave such problems with theory review. The muscle memory of checking ‘given’, drawing the reduced space, and applying the definition prevents the vast majority of mistakes.
定期将此类问题与理论复习穿插进行。检查’条件’、画出缩减后的空间并应用定义的肌肉记忆，能预防绝大多数错误。
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