📚 Tree Diagrams | 树状图
Tree diagrams provide a systematic way to map out all possible outcomes of a sequence of events, making probability calculations clearer and more structured. They are especially useful when an experiment involves multiple stages, such as tossing coins, drawing items from a bag, or answering a series of true/false questions. By branching out at each stage, a tree diagram visualises the sample space and helps you apply the fundamental counting principle, the multiplication rule, and the addition rule correctly.
树状图为多个事件序列的所有可能结果提供了一种系统化的呈现方式，使概率计算更加清晰、有条理。当实验涉及多个阶段时，例如抛掷硬币、从袋中抽取物品或回答一系列判断题，树状图尤为有效。通过在每个阶段进行分支，树状图将样本空间可视化，帮助你正确应用基本计数原理、乘法法则和加法法则。
1. The Basic Structure of a Tree Diagram | 树状图的基本结构
A tree diagram is built from left to right. The starting point is a single node, representing the beginning of the experiment. From this node, branches are drawn for every possible outcome of the first event. The probability of each outcome is written along the corresponding branch. For subsequent events, new branches are drawn from the end of each previous branch, again showing all possible outcomes and their probabilities. This process continues until all stages are represented. The final outcomes are shown at the tips of the last set of branches, and each complete path from the start to an outcome represents one combined event.
树状图从左向右构建。起点是一个单独的节点，代表实验的开始。从该节点出发，为第一个事件的每种可能结果画出分支，并在分支上标注对应结果的概率。对于后续事件，从每条前一分支的末端继续画出新分支，同样展示所有可能的结果及其概率。这一过程持续进行，直到所有阶段都表示完毕。最终结果展示在最后一组分枝的末端，从起点到某个结果的每条完整路径代表一个复合事件。
2. The Multiplication Rule Along a Path | 沿路径的乘法法则
ritten on the branches of that path. This rule follows from the definition of independence or conditional probability: for two events A and B, P(A ∩ B) = P(A) × P(B | the first event, so multiplying along the path automatically accounts for any dependence.
要找出特定结果序列的概率（即沿一条路径的所有事件的联合概率），只需将该路径各分支上标注的概率相乘。这一规则源自独立性或条件概率的定义：对于两个事件 A 和 B，P(A ∩ B) = P(A) × P(B | A)。在树状图中，第二条分支上标注的概率已经以第一个事件的结果为条件，因此沿路径相乘会自动考虑任何相依关系。
3. The Addition Rule for Multiple Paths | 多路径的加法法则
through more than one path on the tree diagram. Since different paths represent mutually exclusive combined outcomes, the probability of the event is found by adding the probabilities of all paths that lead to it. This is the addition rule for mutually exclusive events: if A and B cannot happen simultaneously, P(A ∪ identify all the final outcomes that satisfy the event description and sum their path probabilities.
我们关心的事件常常可以通过树状图上的多条路径发生。由于不同路径代表互斥的复合结果，事件的概率等于所有导向该事件的路径概率之和。这就是互斥事件的加法法则：若 A 与 B 不能同时发生，则 P(A ∪ B) = P(A) + P(B)。在树状图上，你需要找出所有满足事件描述的最终结果，并将它们的路径概率相加。
4. Independent Events on a Tree Diagram | 树状图中的独立事件
e, the probability of heads on the second toss is always ½, no matter the result of the first toss. Multiplying along a path for independent events simply involves multiplying the unconditional probabilities P(A) × P(B).
当事件相互独立时，某一阶段的结果不会影响下一阶段的概率。在树状图中，这意味着无论第一阶段出现何种结果，第二组分枝上的概率都相同。例如，抛掷一枚质地均匀的硬币两次，无论第一次抛掷结果如何，第二次得到正面的概率始终是 ½。对于独立事件，沿路径相乘仅需将无条件概率 P(A) × P(B) 相乘即可。
5. Dependent Events and Conditional Probability | 相依事件与条件概率
ch probabilities differ depending on which first-stage branch is taken. The probabilities on these second branches are conditional probabilities, written as P(B | A). When multiplying along a path, you are effectively computing P(A) × P(B | A), which gives the joint probability P(A ∩ g conditional probability problems and avoiding confusion when events are not independent.
如果第一个事件的结果改变了第二个事件的概率，这些事件就是相依的。在树状图中，第二阶段分支的概率会随着所选的第一阶段分支而变化。这些第二阶段分支上的概率就是条件概率，记作 P(B | A)。沿路径相乘时，实际上是在计算 P(A) × P(B | A)，从而得到联合概率 P(A ∩ B)。树状图特别适合可视化条件概率问题，避免在事件不独立时产生混淆。
6. Sampling With and Without Replacement | 有放回与无放回抽样
the events dependent. For instance, drawing two counters from a bag containing 5 red and 3 blue counters: with replacement, P(red) = 5/8 both times; without replacement, if a red is taken first, the second P(red) becomes 4/7, and the tree branches reflect this change.
树状图的一个经典应用场景是从集合中抽取物品，可以分为有放回或无放回两种情形。有放回抽样使各阶段概率保持不变，因此事件独立。无放回抽样在每次抽取后改变了集合的构成，使得事件相依。例如，从一个装有 5 个红色和 3 个蓝色筹码的袋中抽取两次：若有放回，每次 P(红) = 5/8；若无放回，若第一次抽到红色，第二次 P(红) 变为 4/7，树状图的分支概率会反映这一变化。
7. Labelling Probabilities Accurately | 准确标注概率
ent all possible outcomes at that stage. For dependent events, check that the conditional probabilities for a given node sum to 1. Always write probabilities as fractions, decimals, or percentages consistently. When the problem gives information in words, translate it carefully: for example, 'the probability that B occurs given that A has occurred is 0.3' becomes P(B | A) = 0.3 on the branch following A.
准确标注分支至关重要。从同一个节点发出的每一组分枝的概率之和必须为 1，因为它们代表了该阶段所有可能的结果。对于相依事件，要确保给定节点下的条件概率之和为 1。始终以分数、小数或百分比一致地书写概率。当题目以文字给出信息时，需仔细转换：例如，“已知 A 发生的情况下 B 发生的概率为 0.3” 应在 A 之后的分支上标注 P(B | A) = 0.3。
8. Using Tree Diagrams to Solve Reverse Conditional Probability | 用树状图解决逆向条件概率问题
Tree diagrams are not only for finding forward probabilities (P(A ∩ B)) but are also immensely helpful when you need to find a conditional probability 'in reverse', such as P(A | B), using Bayes' theorem. First, construct the tree with the natural chronological order of events. Then identify the path(s) that lead to the given condition B. The required probability is the probability of the path that includes A and B, divided by the total probability of B (sum of all paths ending in B). This is simply the joint probability from the tree path divided by the sum of relevant joint probabilities.
树状图不仅能求正向概率（P(A ∩ B)），在需要求“逆向”条件概率（如 P(A | B)）时也极有帮助，这实际上是在使用贝叶斯定理。首先，按事件发生的自然时间顺序构建树状图。然后，找到导向已知条件 B 的路径。所求概率等于同时包含 A 和 B 的路径概率除以 B 的总概率（所有以 B 结束的路径概率之和）。这其实就是树状图路径上的联合概率除以相关的联合概率之和。
9. Worked Example 1: Independent Events | 例题 1：独立事件
A fair coin is tossed three times. Find the probability of getting exactly two heads.
一枚质地均匀的硬币抛掷三次。求恰好得到两次正面的概率。
½ × ½ = ⅛. Since these paths are mutually exclusive, P(exactly two heads) = ⅛ + ⅛ + ⅛ = ⅜.
树状图有三个层级，每层有两条分支（正和反），每条标注概率 ½。共有 2³ = 8 条等可能的路径。恰好出现两个正面的路径为：正正反、正反正、反正正。每条路径的概率为 ½ × ½ × ½ = ⅛。因为各路径互斥，P(恰好两个正面) = ⅛ + ⅛ + ⅛ = ⅜。
Although this example can be solved with binomial coefficients, the tree diagram reinforces the underlying logic of independent probability multiplication.
虽然这个例子可以用二项式系数求解，但树状图强化了独立概率相乘的基本逻辑。
10. Worked Example 2: Conditional Probability | 例题 2：条件概率
A bag contains 4 red discs and 6 blue discs. Two discs are drawn at random without replacement. Find the probability that the second disc is red given that the first disc was blue. Also find the probability that the first disc was blue given that the second disc is red.
一个袋子里装有 4 个红色圆片和 6 个蓝色圆片。随机无放回地抽取两次。求在已知第一个是蓝色的条件下第二个是红色的概率。并求在已知第二个是红色的条件下第一个是蓝色的概率。
ree: first draw, P(R₁) = 4/10 = 2/5, P(B₁) = 6/10 = 3/5. If R₁ occurred, second draw: P(R₂ | R₁) = 3/9 = 1/3, P(B₂ | R₁) = 6/9 = 2/3. If B₁ occurred, second draw: P(R₂ | B₁) = 4/9, P(B₂ | B₁) = 5/9.
首先画出树状图：第一次抽取，P(R₁) = 4/10 = 2/5, P(B₁) = 6/10 = 3/5。若第一次为红，第二次：P(R₂ | R₁) = 3/9 = 1/3, P(B₂ | R₁) = 6/9 = 2/3。若第一次为蓝，第二次：P(R₂ | B₁) = 4/9, P(B₂ | B₁) = 5/9。
(a) The required P(R₂ | B₁) is directly read from the tree: 4/9. (b) For P(B₁ | = 2/15 + 4/15 = 6/15 = 2/5. The path with B₁ and R₂ has probability 4/15. Therefore, P(B₁ | R₂) = (4/15) / (2/5) = (4/15) × (5/2) = 20/30 = ⅔.
(a) 所需的 P(R₂ | B₁) 可直接从树状图中读取：4/9。(b) 对于 P(B₁ | R₂)，需要求逆向条件概率。导致第二个为红的路径有：(R₁ → R₂) 概率为 (2/5) × (1/3) = 2/15，以及 (B₁ → R₂) 概率为 (3/5) × (4/9) = 12/45 = 4/15。总 P(R₂) = 2/15 + 4/15 = 6/15 = 2/5。包含 B₁ 和 R₂ 的路径概率为 4/15。因此，P(B₁ | R₂) = (4/15) / (2/5) = (4/15) × (5/2) = 20/30 = ⅔。
to extract all joint probabilities and then apply the conditional probability formula P(A | B) = P(A ∩ B) / P(B).
注意树状图如何使提取所有联合概率变得简单，然后应用条件概率公式 P(A | B) = P(A ∩ B) / P(B)。
11. Double-Checking Your Tree Diagram | 检查树状图的完整性
After constructing a tree diagram, always check two things: first, that the probabilities on branches emanating from a single node add up to 1; second, that the sum of the probabilities of all final outcomes (the tips of the tree) also totals 1. This serves as a quick validation. If the problem includes a condition like 'at least one of ...', ensure you have identified all the relevant paths correctly and that no paths have been overlooked or double-counted.
构建树状图后，务必检查两点：第一，从同一节点发出的分支概率之和为 1；第二，所有最终结果（树梢）的概率之和也等于 1。这可以作为快速验证。如果题目中包含“至少一个…”之类的条件，请确保正确识别了所有相关路径，且没有遗漏或重复计算任何路径。
12. Common Pitfalls to Avoid | 常见误区
One frequent mistake is confusing the multiplication and addition rules — remember, multiply along a path, add across paths. Another is mislabelling conditional probabilities, especially when the problem statement uses everyday language like 'of those who...' or 'if it is known that...'. Always translate such phrases into formal notation. Also, when sampling without replacement, forgetting to adjust the denominator and numerator for the remaining items is a typical error. Lastly, ensure fractions are simplified only at the final answer stage to avoid rounding errors in intermediate steps.
一个常见错误是混淆乘法法则和加法法则——请记住，沿路径相乘，跨路径相加。另一个误区是错误标注条件概率，尤其当题目使用“在……的人中”或“已知……时”等日常用语时。务必将这些表述转化为规范的数学符号。此外，在进行无放回抽样时，遗忘根据剩余物品调整分母和分子也是典型错误。最后，确保分数只在最终答案阶段化简，以避免中间步骤的舍入误差。
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