📚 An Important Result: The Addition Rule for Probability | 重要结果：概率加法法则
evel Mathematics, many probability questions depend on one deceptively simple formula: P(A ∪ B) = P(A) + P(B) − P(A ∩ ty, and it is an important result because it prevents double-counting when events overlap.
在 AQA A-Level 数学中，许多概率题都依赖一个看似简单却非常重要的公式：P(A ∪ B) = P(A) + P(B) − P(A ∩ B)。这就是概率加法法则，它能避免在事件重叠时重复计数。
1. What the Result Says | 结果的内容
nts A and B from the same sample space, the probability that A or B occurs is given by P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The symbol ∪ means union, read as 'A or B or both', while ∩ means intersection, read as 'both A and B'.
对于同一样本空间中的任意两个事件 A 和 B，A 或 B 发生的概率为 P(A ∪ B) = P(A) + P(B) − P(A ∩ B)。符号 ∪ 表示并集，读作“A 或 B 或二者同时发生”；符号 ∩ 表示交集，读作“A 和 B 同时发生”。
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
This formula is given in the AQA formulae booklet, but you must know when and how to use it without hesitation.
该公式在 AQA 公式手册中给出，但你必须能够毫不犹豫地知道何时以及如何使用它。
2. Why the Intersection Is Subtracted | 为什么要减去交集
If you simply add P(A) and P(B), any outcome that belongs to both A and B is counted twice. Subtracting P(A ∩ B) removes one of those duplicate counts, leaving the true probability of the union.
如果只是把 P(A) 和 P(B) 相加，同时属于 A 和 B 的结果会被计算两次。减去 P(A ∩ B) 可以消除重复计数，从而得到并集的真实概率。
In a Venn diagram, the overlap region A ∩ B is included in both circle A and circle B. The addition rule is the algebraic way of correcting this double-counting.
在维恩图中，重叠区域 A ∩ B 既包含在圆 A 中，也包含在圆 B 中。加法法则正是用代数方法修正这种重复计数。
3. Mutually Exclusive Events | 互斥事件
Two events A and B are mutually exclusive if they cannot happen at the same time, so A ∩ B = ∅ and P(A ∩ B) = 0. In this case the addition rule simplifies to P(A ∪ B) = P(A) + P(B).
如果两个事件 A 和 B 不能同时发生，则它们是互斥事件，因此 A ∩ B = ∅ 且 P(A ∩ B) = 0。此时加法法则简化为 P(A ∪ B) = P(A) + P(B)。
Example: When rolling a fair six-sided die, let A be the event 'the score is odd' and B be the event 'the score is 6'. Then A = {1, 3, 5} and B = {6}, so A ∩ B = ∅. Therefore P(A ∪ B) = 3/6 + 1/6 = 4/6 = 2/3.
例如：掷一枚均匀六面骰子，设 A 为“点数为奇数”，B 为“点数为 6”。则 A = {1, 3, 5}，B = {6}，所以 A ∩ B = ∅。因此 P(A ∪ B) = 3/6 + 1/6 = 4/6 = 2/3。
4. The Three-Event Extension | 三事件推广
The addition rule can be extended to three events A, B and C. The formula is P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C).
加法法则可以推广到三个事件 A、B 和 C。公式为 P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A ∩ B) − P(A ∩ C) − P(B ∩ C) + P(A ∩ B ∩ C)。
P(A ∪ B ∪ C) = P(A) + P(B) + P(C) − P(A∩B) − P(A∩C) − P(B∩C) + P(A∩B∩C)
This extension is less common in AS questions but appears in A2 statistics and problem-solving questions. The triple intersection is added back because it has been subtracted too many times.
这种推广在 AS 题目中较少见，但会出现在 A2 统计和问题解决题中。三重交集之所以要加回去，是因为它之前被减去了太多次。
5. Linking to Conditional Probability | 与条件概率的联系
The intersection term P(A ∩ B) can be rewritten using conditional probability: P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B). Substituting this into the addition rule gives P(A ∪ B) = P(A) + P(B) − P(A) × P(B | A).
交集项 P(A ∩ B) 可以用条件概率改写：P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B)。将其代入加法法则可得 P(A ∪ B) = P(A) + P(B) − P(A) × P(B | A)。
This version is useful when a question gives a conditional probability such as P(B | A) instead of the intersection P(A ∩ B) directly.
当题目给出的是条件概率（如 P(B | A)）而不是直接给出交集 P(A ∩ B) 时，这个版本非常有用。
6. Independent Events | 独立事件
If events A and B are independent, then P(A ∩ B) = P(A) × P(B). The addition rule then becomes P(A ∪ B) = P(A) + P(B) − P(A) × P(B).
如果事件 A 和 B 相互独立，则 P(A ∩ B) = P(A) × P(B)。此时加法法则变为 P(A ∪ B) = P(A) + P(B) − P(A) × P(B)。
Do not use P(A ∩ ates that A and B are independent or the context clearly implies it. Independence must be established or given.
除非题目说明 A 和 B 相互独立，或者上下文清楚表明二者独立，否则不要使用 P(A ∩ B) = P(A) × P(B)。独立性必须是被建立或给定的。
7. Complement and 'At Least One' Problems | 补集与“至少一个”问题
is to work with the complement. The probability that at least one of A or B occurs is P(A ∪ B), and the probability that neither occurs is P(A′ ∩ B′) = 1 − P(A ∪ B).
一个非常有用的替代方法是利用补集。A 或 B 至少一个发生的概率是 P(A ∪ B)，而二者都不发生的概率是 P(A′ ∩ B′) = 1 − P(A ∪ B)。
(A′) × P(B′) = (1 − P(A))(1 − P(B)). Therefore P(at least one) = 1 − (1 − P(A))(1 − P(B)).
如果 A 和 B 相互独立，那么 A′ 和 B′ 也相互独立，因此 P(二者都不发生) = P(A′) × P(B′) = (1 − P(A))(1 − P(B))。所以 P(至少一个发生) = 1 − (1 − P(A))(1 − P(B))。
8. Worked Example 1: Union of Two Events | 例题1：两事件并集
AQA-style question: Given P(A) = 0.35, P(B) = 0.45 and P(A ∩ B) = 0.15, find P(A ∪ B) and P(neither A nor B).
AQA 风格例题：已知 P(A) = 0.35，P(B) = 0.45，P(A ∩ B) = 0.15，求 P(A ∪ B) 和 P(A 与 B 都不发生)。
Using the addition rule: P(A ∪ B) = 0.35 + 0.45 − 0.15 = 0.65. Then P(neither) = 1 − P(A ∪ B) = 1 − 0.65 = 0.35.
使用加法法则：P(A ∪ B) = 0.35 + 0.45 − 0.15 = 0.65。然后 P(二者都不发生) = 1 − P(A ∪ B) = 1 − 0.65 = 0.35。
Always check that your final probabilities lie between 0 and 1. Here 0.65 and 0.35 are both valid.
始终检查最终概率是否在 0 和 1 之间。这里的 0.65 和 0.35 都是有效的。
9. Worked Example 2: Independent Events | 例题2：独立事件
Question: A and B are independent events with P(A) = 0.25 and P(B) = 0.60. Find P(A ∪ B), and find the probability that exactly one of A or B occurs.
例题：A 和 B 是独立事件，P(A) = 0.25，P(B) = 0.60。求 P(A ∪ B)，并求 A 或 B 恰好一个发生的概率。
Since A and B are independent, P(A ∩ B) = 0.25 × 0.60 = 0.15. Therefore P(A ∪ B) = 0.25 + 0.60 − 0.15 = 0.70.
因为 A 和 B 独立，所以 P(A ∩ B) = 0.25 × 0.60 = 0.15。因此 P(A ∪ B) = 0.25 + 0.60 − 0.15 = 0.70。
The probability that exactly one occurs is P(A ∪ B) − P(A ∩ B) = 0.70 − 0.15 = 0.55. This removes the overlap where both occur.
恰好一个发生的概率是 P(A ∪ B) − P(A ∩ B) = 0.70 − 0.15 = 0.55。这样就去除了二者同时发生的重叠部分。
10. Worked Example 3: Three Events | 例题3：三事件
ve probabilities P(A) = 0.50, P(B) = 0.40, P(C) = 0.30. The pairwise intersections are P(A ∩ B) = 0.20, P(A ∩ C) = 0.10, P(B ∩ C) = 0.10, and the triple intersection is P(A ∩ B ∩ C) = 0.05. Find P(A ∪ B ∪ C).
例题：三个事件 A、B、C 的概率分别为 P(A) = 0.50，P(B) = 0.40，P(C) = 0.30。两两交集为 P(A ∩ B) = 0.20，P(A ∩ C) = 0.10，P(B ∩ C) = 0.10，三重交集为 P(A ∩ B ∩ C) = 0.05。求 P(A ∪ B ∪ C)。
Apply the three-event rule: P(A ∪ B ∪ C) = 0.50 + 0.40 + 0.30 − 0.20 − 0.10 − 0.10 + 0.05 = 1.20 − 0.40 + 0.05 = 0.85.
应用三事件法则：P(A ∪ B ∪ C) = 0.50 + 0.40 + 0.30 − 0.20 − 0.10 − 0.10 + 0.05 = 1.20 − 0.40 + 0.05 = 0.85。
11. Common Mistakes and How to Avoid Them | 常见错误及避免方法
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