📚 Using Tree Diagrams to Analyse Probabilities of Multiple Events | 用树状图分析多重事件概率
A tree diagram is a powerful visual tool in probability that represents all possible outcomes of a sequence of events. It helps students and exam candidates systematically multiply and add probabilities without losing track of branches. In IB mathematics, tree diagrams are especially useful for solving problems involving independent and dependent events, conditional probability, and Bayes’ theorem.
树状图是概率中一种强大的可视化工具，它能表示一系列事件的所有可能结果。它帮助考生系统地进行概率的乘法和加法，不会遗漏分支。在 IB 数学中，树状图特别适用于解决涉及独立事件、依赖事件、条件概率和贝叶斯定理的问题。
1. Basic Probability Rules | 基本概率规则
When two events occur one after another, the probability that both occur is the product of the probability of the first event and the conditional probability of the second event given the first. This is known as the multiplication rule.
当两个事件先后发生时，两者同时发生的概率等于第一个事件发生的概率乘以在第一个事件发生的条件下第二个事件发生的概率，这就是乘法法则。
Pr(A∩B) = Pr(A) × Pr(B|A)
If the events are independent, the condition is irrelevant, so Pr(B|A) = Pr(B), giving Pr(A∩B) = Pr(A) × Pr(B).
如果事件相互独立，条件就不起作用，因此 Pr(B|A) = Pr(B)，即 Pr(A∩B) = Pr(A) × Pr(B)。
The addition rule is used when adding probabilities of disjoint pathways. A tree diagram naturally shows which paths can be added.
加法法则用于计算互不相交路径的概率之和。树状图能够自然地显示哪些路径可以进行加法运算。
2. Constructing a Tree Diagram | 构建树状图
Start with a root node. For each trial, draw a branch for every possible outcome and label each branch with its probability. Every path from the root to a terminal leaf represents a combined outcome, and its probability is the product of the probabilities written along that path.
从一个根节点开始。对每一次试验，为每一个可能的结果画一条分支，并在分支上标注概率。从根到末端叶子的每一条路径代表一个组合结果，其概率等于路径上所有概率的乘积。
Always check that the probabilities on branches leaving the same node add to 1. This verifies that no outcomes are missing.
始终检查从同一个节点出发的所有分支概率之和是否为 1。这样可以验证是否漏掉了某些结果。
Labels on later branches may be conditional probabilities, especially when the experiment is done without replacement.
后面分支上的标签可能是条件概率，尤其是在不放回抽取的试验中。
3. Independent Events: Tossing a Coin Twice | 独立事件：抛硬币两次
Consider tossing a fair coin twice. The first toss has branches H and T, each with probability 0.5. From each of those branches, the second toss again has H and T, each with probability 0.5.
考虑抛两次均匀硬币。第一次抛掷有 H 和 T 两个分支，概率均为 0.5。从每个分支出发，第二次抛掷仍然有 H 和 T，概率均为 0.5。
The four complete paths are HH, HT, TH and TT. Each path has probability 0.5 × 0.5 = 0.25, so every outcome is equally likely.
四条完整路径是 HH、HT、TH 和 TT。每条路径的概率为 0.5 × 0.5 = 0.25，因此每个结果等可能发生。
Tree diagrams can also compute more complex events. The probability of at least one head is 1 − 0.25 = 0.75, since only TT contains no head.
树状图还能用于计算更复杂的事件。至少出现一次正面的概率为 1 − 0.25 = 0.75，因为只有 TT 没有正面。
4. Dependent Events: Drawing Without Replacement | 依赖事件：不放回抽取
Suppose a bag contains 3 red and 2 blue counters. Two counters are drawn without replacement. The first draw has P(R₁) = 3/5 and P(B₁) = 2/5.
假设一个袋子中有 3 个红色和 2 个蓝色计数器。不放回抽取两个计数器。第一次抽取的概率为 P(R₁) = 3/5，P(B₁) = 2/5。
If the first counter is red, the second draw has P(R₂|R₁) = 2/4 and P(B₂|R₁) = 2/4. If the first counter is blue, the second draw has P(R₂|B₁) = 3/4 and P(B₂|B₁) = 1/4.
如果第一次抽到红色，第二次抽取的概率为 P(R₂|R₁) = 2/4，P(B₂|R₁) = 2/4。如果第一次抽到蓝色，第二次抽取的概率为 P(R₂|B₁) = 3/4，P(B₂|B₁) = 1/4。
These conditional probabilities differ from the first draw because the total number of counters has decreased.
这些条件概率与第一次抽取不同，因为计数器总数已经减少。
The probability of drawing two red counters is 3/5 × 2/4 = 3/10. Similarly, the probability of one blue then one red is 2/5 × 3/4 = 3/10.
抽取两个红色计数器的概率为 3/5 × 2/4 = 3/10。类似地，先蓝后红的概率为 2/5 × 3/4 = 3/10。
5. Identifying Conditional Probabilities | 识别条件概率
In a tree diagram, the probability on a second or later branch is a conditional probability. For example, P(R₂|R₁) = 2/4 means “the probability of red on the second draw given that the first draw was red”.
在树状图中，第二个或之后分支上的概率是条件概率。例如 P(R₂|R₁) = 2/4 表示“在第一次已经抽到红色的条件下，第二次抽到红色的概率”。
Pr(A|B) = Pr(A∩B) / Pr(B)
A tree diagram provides the numerator Pr(A∩B) directly as the probability of the relevant path. The denominator Pr(B) is the sum of all paths that end in B.
树状图直接提供了分子 Pr(A∩B)，即相应路径的概率。分母 Pr(B) 是所有以 B 为终点的路径概率之和。
This relationship is essential for solving conditional probability questions in IB exams.
这种关系对于求解 IB 考试中的条件概率问题至关重要。
6. Finding Total Probability Using a Tree | 用树状图求事件总概率
To find the probability of an event that can occur through multiple pathways, add the probabilities of all paths that end in that event. This is the law of total probability.
要求一个可以通过多条路径发生的事件概率，只需将末端为该事件的所有路径概率相加。这就是全概率公式。
In the counter example, the probability that the second counter is red is:
在计数器例子中，第二个计数器是红色的概率为：
Pr(R₂) = (3/5 × 2/4) + (2/5 × 3/4) = 3/10 + 3/10 = 3/5
This is more intuitive than memorising the formula, because the tree shows exactly which combinations lead to R₂.
这比死记公式更直观，因为树状图明确显示了哪些组合会导致 R₂。
7. Probability of “At Least One” | 计算“至少一个”的概率
A common exam question asks for the probability that an event occurs at least once in several trials. Instead of adding many paths, subtract the probability of the single “none” path from 1.
常见考题要求重复若干次试验后事件至少发生一次的概率。与其累加多条路径，不如用 1 减去唯一的“不发生”路径概率。
For example, if a biased coin shows heads with probability 0.6 and is tossed three times independently, the probability of at least one head is 1 − (0.4)³ = 0.936.
例如，一枚偏向正面的硬币正面概率为 0.6，独立抛掷三次，至少出现一次正面的概率为 1 − (0.4)³ = 0.936。
The “none” path is easy to identify because it consists entirely of failures. This method works for dependent trials as well, as long as the branch probabilities are multiplied correctly.
“不发生”路径很容易识别，因为它完全由失败组成。只要分支概率相乘正确，这种方法同样适用于依赖试验。
8. Tree Diagrams and Bayes’ Theorem | 树状图与贝叶斯定理
Sometimes we know the final result and need to work backwards to find the probability of an earlier event. For example, given that the second counter drawn is red, what is the probability that the first counter was blue?
有时我们知道最终结果，需要反推某个先前事件发生的概率。例如，已知第二次抽到的是红色计数器，求第一次抽到蓝色的概率。
Pr(B₁|R₂) = Pr(B₁∩R₂) / Pr(R₂)
From the tree, Pr(B₁∩R₂) = 3/10 and Pr(R₂) = 3/5, so Pr(B₁|R₂) = (3/10) / (3/5) = 1/2.
从树状图中可知 Pr(B₁∩R₂) = 3/10，Pr(R₂) = 3/5，因此 Pr(B₁|R₂) = (3/10) / (3/5) = 1/2。
This reverse conditional probability is exactly Bayes’ theorem. The tree diagram avoids memorisation by showing the numerator and denominator clearly.
这种反向条件概率正是贝叶斯定理。树状图通过清晰地展示分子和分母，避免了死记硬背。
9. Common Mistakes | 常见错误
Treating dependent events as independent. When drawing without replacement, the probabilities must change after each draw.
把依赖事件当作独立事件。 在不放回抽取时，每次抽取后的概率必须更新。
Adding probabilities without multiplying along branches. Each path requires multiplication before any addition.
不对分支概率相乘就直接相加。 每条路径必须先相乘，然后才能相加。
Using decimals and fractions inconsistently. Choose one format and keep it throughout the problem.
小数和分数混用。 应当选择一种格式并贯穿始终。
Forgetting that all branch probabilities from a node sum to 1. Check this to ensure no path is missing.
忘记从同一节点出发的所有分支概率之和必须为 1。 检查这一点可以避免路径遗漏。
10. Exam Tips for IB Mathematics | IB 数学考试技巧
Read the question carefully to decide whether the events are independent or dependent. Words like “with replacement” signal independence, while “without replacement” signal dependence.
仔细审题，判断事件是独立还是依赖。出现“有放回”时表示独立，“无放回”时表示依赖。
Always draw a tree diagram unless the question tells you otherwise. A clear diagram earns method marks even if the final answer is wrong.
除非题目另有要求，始终画出树状图。即使最终答案错误，清晰的图也能帮助获得方法分。
Show every multiplication and addition step. This helps you check calculations and gain partial credit.
写出每一步乘法和加法过程。这有助于检查计算，并获得部分分数。
Finally, verify that every probability lies between 0 and 1, and that all final probabilities add to 1.
最后，检验每个概率都在 0 到 1 之间，并且所有最终概率之和为 1。
11. Practice Problems with Solutions | 练习与答案
Problem 1: A factory machine produces a defective item with probability 0.1. Two items are selected independently. Find the probability that exactly one is defective.
问题 1： 某工厂机器生产次品的概率为 0.1。独立抽取两个产品，求恰好有一个是次品的概率。
Solution: Exactly one defective occurs through the paths D (defective) then N (non-defective), or N then D.
解答： 恰好一个次品可以通过次品-合格或合格-次品两条路径发生。
Pr = (0.1 × 0.9) + (0.9 × 0.1) = 0.18
Problem 2: A bag contains 5 white and 3 black counters. Two counters are drawn without replacement. Find the probability that both counters have the same colour.
问题 2： 袋子中有 5 个白色和 3 个黑色计数器。不放回抽取两个，求两个计数器颜色相同的概率。
Solution: The required probability is the sum of the white-white path and the black-black path.
解答： 所求概率等于白-白路径与黑-黑路径之和。
Pr = (5/8 × 4/7) + (3/8 × 2/7) = 20/56 + 6/56 = 26/56 = 13/28
12. Conclusion | 总结
Tree diagrams convert complex probability problems into manageable visual steps. They clarify the difference between independent and dependent events, support conditional probability, and make Bayes’ theorem intuitive.
树状图将复杂的概率问题转化为易于处理的可视化步骤。它理清了独立事件与依赖事件的区别，支持条件概率计算，并让贝叶斯定理变得直观。
By mastering the construction and interpretation of tree diagrams, you will gain confidence and accuracy in solving multi-event probability questions in IB mathematics.
通过掌握树状图的构建与解读，你将在 IB 数学中更有信心、更准确地解决多重事件概率问题。
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