📚 Tree Diagrams for Combined Probabilities | 树形图求解复合概率
Tree diagrams provide a visual method for solving problems involving combined probabilities, especially when two or more events occur in sequence. They allow us to map every possible outcome and its associated probability, making the structure of the problem transparent and the calculations intuitive.
树形图为解决复合概率问题提供了一种直观的可视化方法，尤其适用于两个或多个事件依次发生的情形。它能够帮助我们列出所有可能的结果及其对应的概率，使问题结构一目了然，计算过程也更符合直觉。
1. The Structure of a Tree Diagram | 树形图的基本结构
Each branch from a node represents a possible outcome, and the probability of that outcome is written alongside the branch. The sum of probabilities from every node must be exactly 1, because the outcomes at that stage are mutually exclusive and exhaustive.
从节点出发的每条分支代表一种可能结果，该结果的概率标注在分支旁。每个节点上所有分支的概率之和必须恰好为 1，因为这些结果在这一阶段是互斥且穷尽的。
For a two-stage experiment, the first set of branches shows the outcomes of the first event, and the second set shows the outcomes of the second event given the first. The final endpoint of each path represents one combined outcome of the whole experiment.
对于两阶段试验，第一组分支表示第一个事件的结果，第二组分支表示在已知第一个结果的前提下第二个事件的结果。每条路径的末端代表整个试验的一个组合结果。
2. The Multiplication Rule | 乘法法则
To find the probability that two events occur in sequence, multiply the probabilities along the branches. For events A and B, the probability of A occurring first and then B is written as:
要求两个事件依次发生的概率，需要沿相应分支将概率相乘。对于事件 A 和 B，先发生 A 再发生 B 的概率记为：
P(A then B) = P(A) × P(B|A)
Here P(B| l probability of B given that A has already occurred. If A and B are independent, then P(B|A) = P(B), so the rule simplifies to P(A then B) = P(A) × P(B).
其中 P(B|A) 表示在 A 已发生的条件下 B 的条件概率。如果 A 和 B 相互独立，那么 P(B|A) = P(B)，因此该法则简化为 P(A then B) = P(A) × P(B)。
3. Independent vs Dependent Events | 独立事件与非独立事件
Two events are independent when the occurrence of one does not change the probability of the other. For example, tossing a coin twice produces independent events, because the result of the first toss has no effect on the second.
两个事件彼此独立，意味着其中一个事件的发生不会改变另一个事件发生的概率。例如，抛两次硬币就是独立事件，因为第一次抛掷的结果不会影响第二次。
Events are dependent when the outcome of the first event changes the probability of the second. In a tree diagram, dependent events show different probabilities on the second-level branches depending on which first-level branch was taken.
当第一个事件的结果会改变第二个事件的概率时，这些事件就是非独立的。在树形图中，非独立事件在第二层的分支概率会因第一层所选的分支不同而发生变化。
4. Example 1: Tossing a Fair Coin Twice | 示例 1：抛两次均匀硬币
Toss a fair coin twice. The first toss can be Heads (H) or Tails (T), each with probability 0.5. For the second toss, the probabilities remain 0.5 each because the events are independent.
抛两次均匀硬币。第一次抛掷可能为正面 H 或反面 T，每种概率均为 0.5。由于事件相互独立，第二次抛掷时每种结果的概率仍然是 0.5。
The probability of getting two heads is found by multiplying along the branch H then H:
得到两次正面的概率需要沿“H 然后 H”的分支进行相乘：
P(HH) = 0.5 × 0.5 = 0.25
Similarly, the probability of exactly one head can be found by adding the probabilities of the two paths H then T, and T then H:
类似地，恰好得到一次正面的概率，可以分别求出“H 然后 T”与“T 然后 H”两条路径的概率，再相加：
P(exactly one H) = 0.25 + 0.25 = 0.5
5. Example 2: Drawing Balls with Replacement | 示例 2：有放回地取球
A bag contains 3 red balls and 2 blue balls. A ball is drawn, its colour is recorded, and then it is replaced before a second draw. Because the ball is replaced, the composition of the bag is the same for the second draw, so the events are independent.
一个袋子中有 3 个红球和 2 个蓝球。先取出一球并记录颜色，然后放回，再进行第二次抽取。由于取出的球已被放回，第二次抽取时袋中的组成不变，因此事件相互独立。
The probability of drawing a red ball first is 3/5, and the probability of drawing a red ball second is also 3/5. Therefore:
第一次取出红球的概率是 3/5，第二次取出红球的概率仍为 3/5。因此：
P(RR) = (3/5) × (3/5) = 9/25
The probability of getting one red and one blue is found by adding the two possible orders:
得到“一红一蓝”的概率需要将两种可能顺序的概率相加：
P(RB or BR) = (3/5 × 2/5) + (2/5 × 3/5) = 12/25
6. Example 3: Drawing Balls without Replacement | 示例 3：无放回地取球
Using the same bag of 3 red and 2 blue balls, now suppose the first ball is not replaced. This makes the events dependent because the composition of the bag changes after the first draw.
仍然使用装有 3 个红球和 2 个蓝球的袋子，但现在第一次取出的球不放回。由于第一次抽取后袋中的组成发生变化，事件变为非独立的。
If a red ball is drawn first, there remain 2 red and 2 blue balls. Hence the probability of a second red is 2/4, not 2/5. The probability of two reds becomes:
如果第一次取出的是红球，袋中还剩 2 个红球和 2 个蓝球。因此第二次取出红球的概率是 2/4，而不是 2/5。两次都取出红球的概率变为：
P(RR) = (3/5) × (2/4) = 3/10
use removing a red ball reduces the proportion of red balls available for the second draw.
注意，这个结果小于有放回时的 9/25 ≈ 0.36。无放回时两次都是红球的概率为 3/10 = 0.3，因为取走一个红球后，第二次抽取时红球所占比例下降了。
7. Finding "At Least One" or "Neither" | 求“至少一次”或“均不发生”的概率
outcome. A powerful approach is to use the complement rule: P(at least one) = 1 − P(none).
许多问题要求计算“至少发生一次”的概率。一个有效的方法是使用对立事件法则：P(至少一次) = 1 − P(一次也不发生)。
For example, in the without-replacement problem above, the probability of drawing no red balls is:
例如，在上面无放回的问题中，没有取出任何红球的概率为：
P(BB) = (2/5) × (1/4) = 1/10
Therefore, the probability of getting at least one red ball is:
因此，至少取到一个红球的概率为：
P(at least one red) = 1 − 1/10 = 9/10
This method frequently saves time because it avoids adding several branch probabilities.
这种方法通常能节省时间，因为它避免了逐一相加多条分支概率的繁琐计算。
8. Conditional Probability from a Tree Diagram | 从树形图中读取条件概率
Tree diagrams are also useful for finding conditional probabilities. The conditional probability P(B|A) is defined by the formula:
树形图同样适用于求解条件概率。条件概率 P(B|A) 的定义公式为：
P(B|A) = P(A and B) / P(A)
From the tree, P(A and B) is the product of the probabilities along the path A then B, and P(A) is the sum of all branch probabilities that begin with A.
在树形图中，P(A and B) 等于沿“A 然后 B”路径相乘的积，而 P(A) 是所有以 A 开头的分支概率之和。
der the without-replacement example. We know P(RR) = 3/10 and P(first R) = 3/5. Therefore:
以无放回取球为例。已知 P(RR) = 3/10，且 P(第一次为红球) = 3/5。因此：
P(second R | first R) = (3/10) / (3/5) = 1/2
This matches the branch probability directly, confirming that the tree already contains the conditional probabilities on its second-level branches.
该结果与第二层分支上的概率一致，这也说明树形图第二层的分支上已经直接标注了条件概率。
9. Common Mistakes | 常见错误
Forgetting to add the probabilities of two different orders when asked for "one red and one blue". These are two mutually exclusive paths, so their probabilities must be summed.
在求“一红一蓝”时，忘记将两种不同顺序对应的概率相加。这两条路径互斥，因此必须将它们的概率求和。
Using the same branch probabilities for dependent events. When sampling without replacement, the second-level probabilities must be adjusted after the first draw.
对非独立事件使用了相同的分支概率。在无放回抽样时，第二层概率必须根据第一次抽取后的结果进行调整。
Adding probabilities along a path instead of multiplying them. The multiplication rule applies when moving along a single branch sequence; the addition rule applies when combining different paths.
在同一路径中使用了加法，而正确做法是乘法。沿同一条分支序列推进时应使用乘法法则；只有合并不同路径时才使用加法。
Not checking that the probabilities from a node sum to 1. This simple check can reveal arithmetic errors early.
没有检查同一节点上概率之和是否为 1。这一简单检查可以提早发现计算错误。
10. Worked Exam-Style Example | 考试风格例题
A student travels to school either by bus or by bicycle. The probability that he uses the bus is 0.7, and the probability that he cycles is 0.3. If he takes the bus, the probability that he is late is 0.2. If he cycles, the probability that he is late is 0.1.
一名学生上学时要么乘公交车，要么骑自行车。他乘坐公交车的概率为 0.7，骑自行车的概率为 0.3。若乘坐公交车，他迟到的概率为 0.2；若骑自行车，他迟到的概率为 0.1。
Find the probability that he is late on any given day.
求他在任意一天迟到的概率。
The two mutually exclusive paths that result in being late are: bus then late, and bicycle then late. Therefore:
导致迟到的两条互斥路径分别是：乘公交然后迟到，以及骑车然后迟到。因此：
P(late) = 0.7 × 0.2 + 0.3 × 0.1 = 0.14 + 0.03 = 0.17
A second typical question asks: given that the student is late, what is the probability that he took the bus? This is a conditional probability:
第二个典型问题是：已知该学生迟到了，他乘坐公交车的概率是多少？这是一个条件概率问题：
P(bus | late) = 0.14 / 0.17 = 14/17
This shows how tree diagrams connect directly to the conditional probability formula tested in A-Level exams.
这表明树形图如何与 A-Level 考试中经常考查的条件概率公式直接联系起来。
11. Practice Questions | 练习问题
A fair six-sided die is rolled twice. Find the probability that both rolls show an even number.
一枚均匀六面骰子连续掷两次。求两次结果都是偶数的概率。
Answer: (3/6) × (3/6) = 1/4
A bag contains 4 red and 6 blue balls. Two balls are drawn without replacement. Find the probability that both balls are of the same colour.
一个袋子中有 4 个红球和 6 个蓝球。不放回地连续取出两个球。求两个球颜色相同的概率。
Answer: (4/10 × 3/9) + (6/10 × 5/9) = 12/90 + 30/90 = 42/90 = 7/15
On a farm, 60% of the animals are sheep and 40% are cows. If an animal is a sheep, the probability that it is white is 0.8; if a cow, the probability that it is white is 0.3. Find the probability that a randomly chosen animal is white.
一个农场中，60% 的动物是羊，40% 是牛。若为羊，其为白色的概率为 0.8；若为牛，其为白色的概率为 0.3。求随机选择一只动物为白色的概率。
Answer: 0.6 × 0.8 + 0.4 × 0.3 = 0.48 + 0.12 = 0.6
12. Summary | 总结
When solving combined probability problems with a tree diagram, always follow the same core steps: draw the branches for each event, write each probability on the appropriate branch, multiply along a path to find the probability of a single combined outcome, and add the probabilities of mutually exclusive paths when the question asks for more than one acceptable outcome.
使用树形图求解复合概率问题时，应始终遵循以下核心步骤：为每个事件画出分支，在对应分支上标注概率，沿单一组合结果所在路径进行乘法运算，当问题涉及多个可接受结果时，将互斥路径的概率相加。
Remember to check whether events are independent or dependent, and adjust the second-level probabilities accordingly. With regular practice, tree diagrams become one of the most reliable tools in the A-Level probability toolkit.
务必判断事件是独立还是非独立，并据此调整第二层分支的概率。通过经常练习，树形图会成为 A-Level 概率部分最可靠的工具之一。
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