More Tree Diagrams | 进阶树状图

📚 More Tree Diagrams | 进阶树状图

Tree diagrams are not just for tossing coins. They become far more powerful when we need to model a sequence of events where later probabilities depend on earlier results. In this revision article you will learn how to extend the basic tree-diagram method to conditional probability, sampling without replacement and multi-stage experiments.

树状图不只是用来解决掷硬币问题。当我们需要刻画一系列事件、且后续事件发生的概率取决于前面结果时，树状图会变得非常强大。在这篇复习文章中，你将学会如何把基本树状图法推广到条件概率、无放回抽样和多阶段试验。

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1. The Basic Rules You Already Know | 你已经掌握的基本规则

For any two-step tree diagram, the probability of a particular path is found by multiplying the probabilities along the branches. If two different paths represent mutually exclusive outcomes, their probabilities are added when the question asks for the probability of ‘either one’.

对任何两步树状图，某一条路径的概率都是把这条路径上所有分支的概率相乘。如果两条不同路径代表互斥结果，那么问题问到“两者之一”的概率时，就要把对应路径的概率相加。

P(path) = P(first branch) × P(second branch)

P(A or B) = P(A) + P(B) for mutually exclusive outcomes

Always check that the probabilities at the ends of all branches add to 1. This is a quick way to spot arithmetic mistakes.

一定要检查所有末端分支的概率之和是否为 1。这是快速发现计算错误的好方法。

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2. Independent Events and Replacement | 独立事件与有放回

When a trial is repeated and the outcome of the first trial has no effect on the second, the events are independent. In a tree diagram, the probabilities on the second set of branches are identical to those on the first set.

当一次试验被重复进行，且第一次的结果不影响第二次时，事件就相互独立。在树状图中，第二层的分支概率与第一层完全相同。

For example, a fair coin is tossed twice. The probability of heads remains 1/2 on the second toss, no matter what happened on the first toss.

例如，将一枚均匀硬币抛掷两次。无论第一次结果如何，第二次抛出正面的概率仍然是 1/2。

P(exactly one head) = P(H then T) + P(T then H) = (1/2 × 1/2) + (1/2 × 1/2) = 1/4 + 1/4 = 1/2

This is the ‘with replacement’ situation: the original state is restored before the next event.

这就是“有放回”的情形：下一次事件发生前，原来的状态已经恢复。

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3. Dependent Events and Without Replacement | 相关事件与无放回

When you take items from a set without replacing them, the probabilities change after each pick. The second branch must use conditional probabilities.

当你从一个集合中不放回地取出物品时，每取一次后概率都会改变。第二次分支必须使用条件概率。

Example: a bag contains 4 red and 6 green counters. Two counters are drawn without replacement.

例：一个袋中有 4 个红色计数器和 6 个绿色计数器。不放回地取出两个计数器。

• The first draw: P(R) = 4/10 and P(G) = 6/10. | 第一次抽取：P(R) = 4/10，P(G) = 6/10。
• If the first is red: P(R|R) = 3/9 and P(G|R) = 6/9. | 如果第一次是红色：P(R|R) = 3/9，P(G|R) = 6/9。
• If the first is green: P(R|G) = 4/9 and P(G|G) = 5/9. | 如果第一次是绿色：P(R|G) = 4/9，P(G|G) = 5/9。

Path	Probability
R then R	(4/10) × (3/9) = 12/90 = 2/15
R then G	(4/10) × (6/9) = 24/90 = 4/15
G then R	(6/10) × (4/9) = 24/90 = 4/15
G then G	(6/10) × (5/9) = 30/90 = 1/3

Notice that the denominators change after the first draw because one counter has been removed.

注意，第一次抽取后分母会改变，因为已经取出一个计数器。

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4. Conditional Probability Notation | 条件概率记号

The symbol P(A|B) is read as ‘the probability that A occurs given that B has already occurred’. On a tree diagram, the numbers written on the second layer of branches are usually conditional probabilities.

符号 P(A|B) 读作“在 B 已发生的条件下 A 发生的概率”。在树状图中，第二层分支上写的通常就是条件概率。

In the example above, P(R|G) means the probability of getting a red counter on the second draw, given that the first draw was green.

在上面的例子中，P(R|G) 表示已知第一次抽到绿色，第二次抽到红色的概率。

P(R|G) = 4/9

The vertical bar is not a division sign; it means ‘given that’.

这里竖线不是除号，而是表示“在……条件下”。

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5. Finding Conditional Probabilities from a Tree | 用树状图求条件概率

To calculate P(A|B) from a tree, first identify P(A∩B) and P(B), then divide. The denominator is the total probability of all paths that include B, and the numerator is the probability of the path or paths that include both A and B.

要从树状图计算 P(A|B)，首先确定 P(A∩B) 和 P(B)，再相除。分母是包含 B 的所有路径的概率总和，分子是同时包含 A 和 B 的路径的概率。

Example: a bag contains 3 red and 5 blue balls. Two balls are drawn without replacement. Find P(first is red | second is red).

例：一个袋中有 3 个红球和 5 个蓝球。不放回地取出两个球。求“在第二次是红球的条件下，第一次是红球”的概率。

P(R₁ ∩ R₂) = (3/8) × (2/7) = 6/56

Published by TutorHao | IGCSE Mathematics Revision Series | aleveler.com

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