Probability Tree Diagrams | 概率树形图

📚 Probability Tree Diagrams | 概率树形图

Probability tree diagrams are a visual tool used to map out all the possible outcomes of two or more events. They show each event as a branch, with probabilities written along the branches, making it easy to calculate the likelihood of combined events step by step. This method is especially useful when working with both independent and dependent events in KS3 mathematics.

概率树形图是一种可视化工具，用于列出两个或多个事件的所有可能结果。每个事件被表示为一个分支，概率标注在分支上，从而可以逐步计算组合事件的可能性。在 KS3 数学中，该方法对于处理独立事件和相关事件都特别有用。

---

1. What is a Probability Tree Diagram? | 什么是概率树形图？

A probability tree diagram is a branching diagram that displays all possible outcomes of a sequence of events. Each branch represents one possible result, and the probabilities for each outcome are written along the branches. The diagram begins with a single starting point and then 'splits' at each stage.

概率树形图是一种分支图，显示一系列事件的所有可能结果。每个分支代表一种可能的结果，每个结果的概率标注在分支旁。图从单个起点开始，然后在每个阶段“分叉”。

---

2. Key Features of a Tree Diagram | 树形图的关键特征

Every tree diagram has three important features: starting point, branches (with probabilities), and final outcomes. The probabilities on branches from the same point always add up to 1. The outcomes are listed at the end of each complete path, and the total probability of all final outcomes sums to 1.

每个树形图有三个重要特征：起点、分支（带概率）和最终结果。从同一点发出的分支上的概率之和总是等于 1。最终结果列在每个完整路径的末端，并且所有最终结果的概率之和等于 1。

---

3. The Multiplication Rule (AND) | 乘法规则（与）

r along a single path, we multiply the probabilities along that path. This is known as the 'AND' as probability 1/3, then the probability of both occurring in sequence is 1/2 × 1/3 = 1/6.

当我们需要两个事件沿着同一条路径同时发生时，我们将该路径上的概率相乘。这被称为“与”规则。例如，若事件 A 的概率为 1/2，事件 B 的概率为 1/3，则两者依次发生的概率为 1/2 × 1/3 = 1/6。

---

4. The Addition Rule (OR) | 加法规则（或）

oduce the same type of final outcome, we add the probabilities of those paths. This is the 'OR' rule. For example, 'getting exactly one head in two coin tosses' can happen via HT or TH, so we calculate 1/4 + 1/4 = 1/2.

如果有多个不同的路径产生同一类型的最终结果，我们将这些路径的概率相加。这就是“或”规则。例如，“两次抛硬币中恰好出现一次正面”可以通过 HT 或 TH 发生，因此计算为 1/4 + 1/4 = 1/2。

---

5. Independent Events vs Dependent Events | 独立事件与相关事件

Independent events are those where the outcome of one event does not affect the probability of another. For example, tossing a coin and rolling a dice. Dependent events are those where the first outcome changes the probability of the next, such as drawing counters from a bag without replacement.

独立事件是指一个事件的结果不影y响另一个事件概率的事件。例如，抛硬币和掷骰子。相关事件是指第一个结果改变下一个事件概率的事件，例如在不放回的情况下从袋子中抽取计数物。

---

6. Example 1: Tossing a Fair Coin Twice | 例1：抛掷一枚均匀硬币两次

ducing four equally likely outcomes: HH, HT, TH, TT. Each has probability 1/2 × 1/2 = 1/4.

抛掷一枚均匀硬币两次。得到正面（H）的概率为 1/2，反面（T）为 1/2。树形图在第一次抛掷时分叉，然后在第二次抛掷时再次分叉，产生四个等可能的结果：HH、HT、TH、TT。每个结果的概率为 1/2 × 1/2 = 1/4。

• Start
  • First toss: H (1/2)
    • Second toss: H (1/2) → HH (1/4)
    • Second toss: T (1/2) → HT (1/4)
  • First toss: T (1/2)
    • Second toss: H (1/2) → TH (1/4)
    • Second toss: T (1/2) → TT (1/4)

P(at least one head) = P(HT) + P(TH) + P(HH) = 1/4 + 1/4 + 1/4 = 3/4

---

7. Example 2: Drawing Colored Balls Without Replacement | 例2：不放回地抽取彩球

A bag contains 5 red balls and 3 blue balls. Two balls are drawn one after the other without replacement. The probability of drawing a red first is 5/8. If a red is taken, the probability of red second becomes 4/7, and blue becomes 3/7. The tree diagram helps to compute combined probabilities such as both red.

一个袋子装有 5 个红球和 3 个蓝球。依次抽取两个球，不放回。第一次抽到红球的概率为 5/8。如果抽走一个红球，第二次抽到红球的概率变为 4/7，蓝球变为 3/7。树形图有助于计算诸如两个都是红球的组合概率。

• Start
  • First draw: R (5/8)
    • Second: R (4/7) → RR: 5/8 × 4/7 = 20/56 = 5/14
    • Second: B (3/7) → RB: 5/8 × 3/7 = 15/56
  • First draw: B (3/8)
    • Second: R (5/7) → BR: 3/8 × 5/7 = 15/56
    • Second: B (2/7) → BB: 3/8 × 2/7 = 6/56 = 3/28

P(both red) = 5/14

---

广告 / Sponsored

8. Example 3: Drawing Colored Balls With Replacement | 例3：放回地抽取彩球

If the first ball is put back before the second draw, the probabilities remain unchanged. Again starting with 5 red and 3 blue, the first draw probability is 5/8 and 3/8. The second draw also has the same probabilities. The tree diagram shows that events are independent.

如果在第二次抽取前将第一个球放回，概率保持不变。仍以 5 红 3 蓝开始，第一次抽取概率为 5/8 和 3/8。第二次抽取的概率相同。树形图显示这些事件是独立的。

• Start
  • First: R (5/8)
    • Second: R (5/8) → RR: 5/8 × 5/8 = 25/64
    • Second: B (3/8) → RB: 5/8 × 3/8 = 15/64
  • First: B (3/8)
    • Second: R (5/8) → BR: 3/8 × 5/8 = 15/64
    • Second: B (3/8) → BB: 3/8 × 3/8 = 9/64

P(at least one red) = 25/64 + 15/64 + 15/64 = 55/64

---

9. Common Mistakes and How to Avoid Them | 常见错误及避免方法

One frequent error is forgetting that the probabilities on branches from the same point must sum to 1. Another is adding probabilities when they should be multiplied, or multiplying when paths should be added. Always check whether events are independent or dependent before drawing the tree.

一个常见错误是忘记从同一点发出的分支概率之和必须为 1。另一个是该相乘的时候相加，或该相加的时候相乘。在绘制树形图前，务必检查事件是独立的还是相关的。

Also, be careful with fractions when multiplying; simplify only at the end if possible. And remember that without replacement, the denominator changes for the second event.

此外，分数相乘时要小心，尽可能在最后才约分。还要记住，不放回时，第二个事件的分母会改变。

---

10. Practice Question with Step-by-Step Solution | 练习题及分步解答

Question: A box contains 4 strawberry sweets and 6 orange sweets. Two sweets are taken at random, one after the other, without replacement. Find the probability that both sweets are the same flavour.

问题：一个盒子里有 4 颗草莓糖和 6 颗橘子糖。随机依次取两颗糖，不放回。求两颗糖口味相同的概率。

Solution: cond orange = 6/9 = 2/3. If first is orange, second strawberry = 4/9, second orange = 5/9.

解答：第一次抽取：P(草莓) = 4/10 = 2/5，P(橘子) = 6/10 = 3/5。若第一颗是草莓，第二颗草莓 = 3/9 = 1/3，第二颗橘子 = 6/9 = 2/3。若第一颗是橘子，第二颗草莓 = 4/9，第二颗橘子 = 5/9。

both same) = P(SS) + P(OO) = (2/5 × 1/3) + (3/5 × 5/9) = 2/15 + 15/45 = 2/15 + 5/15 = 7/15

---

11. Real-life Applications of Tree Diagrams | 树形图的实际应用

Tree diagrams are not just for textbook questions. They are used in business to evaluate decision paths, in medicine to assess the probability of test results, and in sports strategy. Any situation involving a sequence of uncertain events can be modelled with a probability tree.

树形图不仅出现在课本问题中。它们被商业用于评估决策路径，在医学中用于评估测试结果的概率，以及在体育策略中。任何涉及一系列不确定事件的情形都可以用概率树形图建模。

For example, a weather forecast might use a tree to show the chance of rain on two consecutive days, helping event planners make better decisions.

例如，天气预报可能使用树形图显示连续两天下雨的概率，帮助活动策划者做出更好的决策。

---

12. Summary – Key Points to Remember | 总结 – 关键点记忆

• All branches from one point must have probabilities that sum to 1. | 从同一点发出的所有分支概率之和必须为 1。
• Multiply along a branch for 'AND' events. | “与”事件沿分支相乘。
• Add probabilities from different branches for 'OR' combined outcomes. | “或”组合结果：将不同分支的概率相加。
• Check whether events are independent or dependent before assigning probabilities. | 在分配概率前检查事件是独立还是相关。
• Simplify fractions carefully and check your total probability equals 1. | 仔细约分并检查总概率是否等于 1。

---

Published by TutorHao | Mathematics Revision Series | aleveler.com

更多咨询请联系16621398022（同微信）