📚 Exercise 20D: Probability Trees & Conditional Probability | 练习20D：概率树与条件概率
Exercise 20D in many IB Mathematics textbooks focuses on the use of probability tree diagrams to model sequential events, combined with the application of conditional probability and Bayes' theorem. These problems often involve drawing a tree, labelling branches with probabilities, and then calculating the likelihood of compound events or reversing the condition. In this article, we break down the key concepts, step‑by‑step methods, and common pitfalls so you can tackle every question in Exercise 20D with confidence.
在IB数学教材中，练习20D通常聚焦于使用概率树状图建立连续事件的模型，并结合条件概率和贝叶斯定理的应用。此类问题需要绘制树状图、在分支上标注概率，然后计算复合事件的可能性，或对条件进行逆向求解。本文将逐一拆解关键概念、步骤方法和常见陷阱，帮助你自信地应对练习20D的每一道题。
1. What Is Exercise 20D About? | 练习20D概述
Exercise 20D is designed to strengthen your ability to handle multi‑stage random processes. You will encounter scenarios such as drawing balls from a bag without replacement, testing for a disease with imperfect tests, or flipping coins and rolling dice in sequence. The unifying theme is that the probability of a later event depends on what happened earlier, making tree diagrams the ideal tool for visualising these dependencies.
练习20D旨在加深你对多阶段随机过程的处理能力。你将遇到诸如不放回摸球、不完美检测下的疾病筛查、或依次掷硬币与骰子等情景。共同的主题是：后续事件的概率依赖于先前发生的结果，这使得树状图成为直观呈现这些依赖关系的理想工具。
Most questions require you to calculate joint probabilities using the multiplication rule, find marginal probabilities by summing appropriate branches, and solve reverse conditional probabilities with the formula P(A|B) = P(A ∩ B) / P(B). Some exercises also test independence.
多数题目要求你运用乘法法则计算联合概率，通过将合适分支相加求出边际概率，并利用公式 P(A|B) = P(A ∩ B) / P(B) 求解逆向条件概率。部分练习还考查事件的独立性。
2. Drawing a Probability Tree | 绘制概率树
Start by identifying the stages of the experiment. For the first stage, draw branches representing all possible mutually exclusive outcomes and label them with their probabilities. The probabilities from a single point must sum to 1. For the second stage, from each outcome of the first stage, draw a new set of branches showing the subsequent outcomes. Label these with the conditional probabilities given the first‑stage outcome.
首先确定实验的阶段。针对第一阶段，画出代表所有可能互斥结果的分支，并标记相应概率。从同一点出发的各分支概率之和必须等于1。对于第二阶段，从第一阶段的每个结果出发，画出一组新的分支，显示随后的结果，并用给定第一阶段结果下的条件概率进行标记。
Always write probabilities as fractions rather than decimals to maintain precision, especially when the tree involves several steps. A typical IB tree will have two or three stages, but always confirm the exact number of trials from the question.
始终用分数而非小数来标注概率，以保持精确性，尤其是当树状图涉及多个步骤时。典型的IB题目通常有两个或三个阶段，但务必根据题意确定试验的准确次数。
3. Multiplication Rule | 乘法法则
To find the probability of a sequence of events, multiply the probabilities along the corresponding branches: P(A ∩ B) = P(A) × P(B|A). This extends naturally to three events: P(A ∩ B ∩ C) = P(A) × P(B|A) × P(C|A ∩ B).
要计算一连串事件的概率，须沿着对应分支将概率相乘：P(A ∩ B) = P(A) × P(B|A)。对于三个事件，此规则自然延伸为：P(A ∩ B ∩ C) = P(A) × P(B|A) × P(C|A ∩ B)。
If the events are independent, the conditional probabilities simplify: P(B|A) = P(B), so the joint probability becomes P(A) × P(B). Exercise 20D often mixes independent stages with dependent ones, so check carefully whether the probabilities stay the same across stages.
若事件相互独立，则条件概率简化为P(B|A) = P(B)，联合概率便成为P(A) × P(B)。练习20D经常混合独立与相依阶段，因此需仔细检查各阶段的概率是否保持不变。
4. Conditional Probability Formula | 条件概率公式
The core formula for conditional probability is P(A|B) = P(A ∩ B) / P(B), provided P(B) > 0. In tree diagrams, P(A ∩ B) is found by multiplying along the branch that includes both A and B, while P(B) is the sum of the probabilities of all branches that end in B.
条件概率的核心公式为 P(A|B) = P(A ∩ B) / P(B)，前提是 P(B) > 0。在树状图中，P(A ∩ B) 通过沿包含 A 和 B 的分支相乘得到，而 P(B) 则是所有以 B 结尾的分支概率之和。
Many students confuse P(A|B) with P(A ∩ B). Remember, the conditional probability restricts the sample space to only those outcomes where B has occurred. The tree diagram makes this restriction visible: you only consider paths that pass through B.
许多学生容易混淆 P(A|B) 与 P(A ∩ B)。请注意，条件概率将样本空间限制在 B 已发生的那些结果上。树状图使得这种限制肉眼可见：你只考虑通过 B 的路径。
5. Working Backwards with Bayes' Theorem | 逆向概率与贝叶斯定理
Exercise 20D frequently asks questions where you know the final outcome but must infer the likelihood of an earlier cause. For example, "Given that a person tests positive, what is the probability they actually have the disease?" This requires Bayes' theorem in its simplest form: P(Cause|Effect) = P(Cause ∩ Effect) / P(Effect). The numerator comes from a single branch, while the denominator is the total probability of the effect.
练习20D经常提出已知最终结果却要推断较早起因概率的问题。例如，“已知某人检测呈阳性，他真正患病的概率是多少？”这需要用到贝叶斯定理的最简形式：P(起因|效果) = P(起因 ∩ 效果) / P(效果)。分子来自单条分支，而分母是效果发生的总概率。
In a two‑stage tree with two choices in stage one (A and A') and two outcomes in stage two (B and B'), the reverse probability P(A|B) is [P(A) × P(B|A)] / [P(A) × P(B|A) + P(A') × P(B|A')]. You can derive this directly from the tree without memorising a separate formula by always applying the fraction definition.
在阶段一有两个选项（A 与 A'）且阶段二有两个结果（B 与 B'）的两阶段树状图中，逆向概率 P(A|B) 等于 [P(A) × P(B|A)] / [P(A) × P(B|A) + P(A') × P(B|A')]。你无需额外背诵公式，直接从树状图出发，始终运用分数定义即可推导得出。
6. Independent vs Dependent Events | 独立事件与相关事件
Two events A and B are independent if P(A ∩ B) = P(A) × P(B) or, equivalently, P(A|B) = P(A). In a tree, independence is visually clear when the second‑stage probabilities are identical regardless of which first‑stage outcome occurred. If the branch probabilities differ, the events are dependent.
若满足 P(A ∩ B) = P(A) × P(B) 或等价地 P(A|B) = P(A)，则事件 A 与 B 独立。在树状图中，如果无论第一阶段发生何种结果，第二阶段的概率都相同，独立关系便一目了然。如果分支概率不同，则事件相依。
Exercise 20D often tests your understanding of independence by asking you to verify whether two events from a tree are independent. Always compute both P(A ∩ B) and P(A) × P(B) to justify your answer.
练习20D通常要求你验证树状图中两事件是否独立，以检验你对独立性的理解。务必同时计算 P(A ∩ B) 与 P(A) × P(B)，并据此给出判断依据。
7. Typical Problem: Coin and Dice | 典型例题：硬币与骰子
A fair coin is tossed. If it shows heads, a fair six‑sided die is rolled; if tails, a biased four‑sided spinner with outcomes 1,2,3,4 and probabilities 0.1, 0.2, 0.3, 0.4 is used. Find the probability of obtaining an even number.
抛掷一枚均匀硬币。若出现正面，则投掷一颗均匀的六面骰子；若为反面，则使用一个偏斜的四面转盘，其结果为1、2、3、4，概率分别为0.1、0.2、0.3、0.4。求得到偶数的概率。
3/5) = 1/4 + 3/10 = 5/20 + 6/20 = 11/20. This shows how tree components are added for the 'effect'.
画出树状图：第一阶段分支 P(正面)=1/2，P(反面)=1/2。在正面之下，P(偶数) = P(2,4,6) = 3/6 = 1/2。在反面之下，偶数结果为2和4，概率为0.2+0.4=0.6 = 3/5。偶数的总概率 = (1/2 × 1/2) + (1/2 × 3/5) = 1/4 + 3/10 = 5/20 + 6/20 = 11/20。这展示了如何将树状图中的各组分相加以求得“效果”的概率。
8. Typical Problem: Defective Items | 典型例题：次品检测
A factory has two machines. Machine X produces 60% of items, with a 2% defect rate. Machine Y produces 40% of items, with a 5% defect rate. An item is chosen at random and found to be defective. What is the probability it came from machine X?
某工厂有两台机器。机器X生产60%的产品，次品率为2%。机器Y生产40%的产品，次品率为5%。随机抽检一件产品，发现其为次品。该次品来自机器X的概率是多少？
Set up the tree: first stage: P(X)=0.6, P(Y)=0.4. Second stage: P(Defect|X)=0.02, P(OK|X)=0.98; P(Defect|Y)=0.05, P(OK|Y)=0.95. We want P(X| tor: 0.6 × 0.02 = 0.012. Denominator: 0.6×0.02 + 0.4×0.05 = 0.012 + 0.02 = 0.032. Thus P(X|Defect) = 0.012/0.032 = 3/8 = 0.375. This reverse conditional is a classic Bayes' theorem problem.
建立树状图：第一阶段：P(X)=0.6，P(Y)=0.4。第二阶段：P(次品|X)=0.02，P(合格|X)=0.98；P(次品|Y)=0.05，P(合格|Y)=0.95。需求 P(X|次品)。分子：0.6 × 0.02 = 0.012。分母：0.6×0.02 + 0.4×0.05 = 0.012 + 0.02 = 0.032。因此 P(X|次品) = 0.012/0.032 = 3/8 = 0.375。这种逆向条件概率是经典的贝叶斯定理问题。
9. Using Tree Diagrams for 'At Least One' | 利用树状图求“至少一个”问题
To find the probability of at least one success in two trials, it is often quicker to calculate the complement – zero successes. In a tree, the 'zero‑success' path is simply all failures multiplied. Then subtract from 1.
求两次试验中至少成功一次的概率，通常计算其补集——零成功——更为快捷。在树状图中，“零成功”路径即全部失败分支的乘积，然后用1减去该值。
y of at least one head in two tosses of a biased coin with P(H)=0.3: P(at least one H) = 1 – P(TT) = 1 – (0.7 × 0.7) = 1 – hows exactly two paths that include heads, but the complement method is more efficient for 'at least one' problems.
例如，一枚 P(正面)=0.3 的偏斜硬币抛掷两次，求至少一次正面的概率：P(至少一次正面) = 1 – P(反面,反面) = 1 – (0.7 × 0.7) = 1 – 0.49 = 0.51。树状图显示有两条包含正面的路径，但使用补集法处理“至少一个”问题更高效。
10. Common Mistakes to Avoid | 常见错误
Mistake 1: Forgetting that probabilities on the second‑stage branches must be conditional on the first‑stage outcome. If you label them as raw probabilities, the tree becomes inconsistent and joint probabilities will be wrong.
错误1：忘记第二阶段分支上的概率必须以第一阶段结果为条件。如果将其标注为原始概率，则树状图将不再一致，联合概率也将出错。
Mistake 2: Multiplying along unconnected branches. Only multiply along a continuous path that represents a specific outcome sequence. Do not multiply across different first‑stage outcomes.
错误2：沿不连接的分支相乘。只应沿代表特定结果序列的连续路径相乘，切不可跨不同第一阶段的结果相乘。
Mistake 3: Misapplying the conditional probability formula, especially confusing P(A|B) with P(B|A). Always check which event is given and which we are seeking.
错误3：误用条件概率公式，尤其混淆 P(A|B) 与 P(B|A)。务必检查哪个事件是已知条件，哪个是我们要求的目标。
11. Practice Tips for Exercise 20D | 练习20D的解题技巧
Always draw a tree diagram for any multi‑stage probability problem, even if the question does not explicitly ask for one. Label all branches with fractions, and write the joint probabilities at the end of each path. This turns a confusing word problem into a clear visual checklist.
对于任何多阶段概率问题，即使题目没有明确要求，也一定要画出树状图。用分数标注所有分支，并在每条路径末端写出联合概率。这样一来，令人困惑的文字题便转化为清晰的视觉清单。
When facing a reverse conditional problem, immediately draft the tree, highlight the branch(es) that correspond to the given condition, and apply the fraction formula. Show your working for the numerator and denominator separately to earn method marks in the exam.
面对逆向条件概率问题时，立即画出树状图，标出与已知条件对应的分支，并运用分数公式。分别写出分子与分母的计算过程，以便在考试中获取方法分。
12. Summary: Key Formulae and Steps | 总结：关键公式与步骤
Master these core ideas for Exercise 20D:
掌握以下核心要点以应对练习20D：
Joint probability: P(A ∩ B) = P(A) × P(B|A)
联合概率：P(A ∩ B) = P(A) × P(B|A)
Total probability: P(B) = Σ P(Aᵢ) × P(B|Aᵢ)
全概率：P(B) = Σ P(Aᵢ) × P(B|Aᵢ)
Conditional probability: P(A|B) = P(A ∩ B) / P(B)
条件概率：P(A|B) = P(A ∩ B) / P(B)
Independence check: P(A ∩ B) = P(A) × P(B)
独立性检验：P(A ∩ B) = P(A) × P(B)
Follow the step‑by‑step approach: draw the tree, label probabilities, compute joint probabilities for each path, sum for marginal probabilities, and finally apply the conditional definition for reverse calculations. With this strategy, Exercise 20D becomes a consistent exercise in organised reasoning rather than a source of confusion.
遵循分步法：绘制树状图、标注概率、计算每条路径的联合概率、对边际概率求和，最后运用条件定义进行逆向计算。有了这套策略，练习20D便成为条理清楚的逻辑推演，而非混乱之源。
Published by TutorHao | Mathematics Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
或截图后用「扫一扫 → 相册」选择本图