📚 Laws of Probability Operations in IB Mathematics | IB数学：概率运算律详解
Probability is a core topic in IB Mathematics, appearing in both Analysis and Approaches (AA) and Applications and Interpretation (AI). Mastering the laws of probability operations is essential for solving exam questions on random events, conditional probability, and statistical inference.
概率是IB数学的核心内容，出现在分析与方法（AA）以及应用与解释（AI）两门课程中。掌握概率运算律对于解决随机事件、条件概率和统计推断等考试题目至关重要。
1. Sample Space and Events | 样本空间与事件
Before applying any probability law, you must clearly define the sample space, denoted S, which is the set of all possible outcomes of an experiment. An event A is a subset of S.
在应用任何概率运算律之前，必须先明确定义样本空间（记作 S），即一次试验所有可能结果构成的集合。事件 A 是 S 的子集。
For example, when rolling a fair six-sided die, S = {1, 2, 3, 4, 5, 6}. The event “rolling an even number” is A = {2, 4, 6}.
例如，掷一枚均匀的六面骰子时，S = {1, 2, 3, 4, 5, 6}。事件“掷出偶数”为 A = {2, 4, 6}。
In IB problems, always state S and the event(s) symbolically first. This helps you apply the correct formula and avoid confusion.
在IB题目中，务必先用符号表示 S 和各个事件。这有助于你应用正确的公式并避免混淆。
2. The Complement Rule | 对立事件法则
The complement of an event A, written A′ (or A^c), contains all outcomes in S that are not in A. The probability of the complement is given by:
事件 A 的对立事件记作 A′（或 A^c），包含 S 中所有不属于 A 的结果。对立事件的概率为：
P(A′) = 1 − P(A)
This rule is especially useful when the probability of an event itself is difficult to calculate directly, but the probability of its complement is easier.
该法则在直接计算某事件概率较困难、而计算其对立事件概率更为容易时特别有用。
For example, the probability of getting at least one head when tossing a coin three times is easier to find as 1 − P(no heads).
例如，抛掷一枚硬币三次，至少出现一次正面的概率，可以更方便地通过 1 − P(没有正面) 来计算。
3. The Addition Law for Mutually Exclusive Events | 互斥事件的加法法则
Two events A and B are mutually exclusive (disjoint) if they cannot occur at the same time, meaning A ∩ B = ∅. In that case, the probability of A or B occurring is the sum of their individual probabilities:
如果事件 A 和 B 不能同时发生，即 A ∩ B = ∅，则称它们是互斥（不相容）的。此时，A 或 B 发生的概率等于各自概率之和：
P(A ∪ B) = P(A) + P(B)
For instance, when drawing a card from a standard deck, the events “draw a heart” and “draw a spade” are mutually exclusive, because one card cannot be both suits at once.
例如，从一副标准扑克牌中抽一张牌，事件“抽到红心”和“抽到黑桃”是互斥的，因为一张牌不可能同时属于两种花色。
Be careful: this special rule only works when the intersection is empty. If A and B can both happen, you must use the general addition law.
注意：这个特殊规则仅当交集为空时成立。若 A 和 B 可能同时发生，则必须使用一般加法法则。
4. The General Addition Law | 一般加法法则
For any two events A and B, the probability that at least one of them occurs is:
对于任意两个事件 A 和 B，至少有一个发生的概率为：
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
The term P(A ∩ B) is subtracted because it has been counted twice when adding P(A) and P(B). This formula works for all events, including mutually exclusive ones, where P(A ∩ B) = 0.
减去 P(A ∩ B) 是因为它在分别相加 P(A) 和 P(B) 时被重复计算了一次。该公式对所有事件都成立，包括互斥事件（此时 P(A ∩ B) = 0）。
Example: In a class of 30 students, 18 study physics, 15 study chemistry, and 10 study both. The probability that a randomly selected student studies physics or chemistry is P(P ∪ C) = 18/30 + 15/30 − 10/30 = 23/30.
例如：一个班有30名学生，18人学物理，15人学化学，10人两科都学。随机选一名学生，他学物理或化学的概率为 P(P ∪ C) = 18/30 + 15/30 − 10/30 = 23/30。
5. Conditional Probability | 条件概率
Conditional probability measures the probability of event A occurring given that event B has already occurred. It is denoted P(A | B) and defined as:
条件概率衡量的是在事件 B 已经发生的条件下，事件 A 发生的概率，记作 P(A | B)，定义为：
P(A | B) = P(A ∩ B) / P(B), provided P(B) > 0
Rearranging this formula gives the general multiplication law, which is useful in many contexts.
重新整理该公式可以得到一般乘法法则，它在许多情境下非常有用。
In IB exams, conditional probability often appears in table form or tree diagrams. Always identify the condition carefully: “given that” or “if … then” usually signals a condition.
在IB考试中，条件概率常以表格或树状图形式出现。务必仔细识别条件：“已知……”或“如果……那么……”通常表示条件。
6. The Multiplication Law for Independent Events | 独立事件的乘法法则
Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, independence is defined by:
如果事件 A 的发生不影响事件 B 的概率，则称 A 和 B 独立。数学上，独立性定义为：
P(A ∩ B) = P(A) × P(B)
Equivalently, for independent events, P(A | B) = P(A) and P(B | A) = P(B).
等价地，对独立事件，有 P(A | B) = P(A) 且 P(B | A) = P(B)。
Do not confuse “independent” with “mutually exclusive”. Mutually exclusive events cannot be independent (unless one has probability zero) because if A occurs, B certainly cannot.
不要将“独立”与“互斥”混淆。互斥事件不可能是独立的（除非其中一个概率为零），因为若 A 发生，B 就一定不发生。
Example: If you flip a fair coin and roll a fair die, the events “heads” and “roll a 6” are independent. Thus P(heads and 6) = (1/2) × (1/6) = 1/12.
例如：抛一枚均匀硬币并掷一枚均匀骰子，“正面”和“掷出6”是独立事件。因此 P(正面且掷出6) = (1/2) × (1/6) = 1/12。
7. The General Multiplication Law | 一般乘法法则
Without independence, we can still find the probability of both A and B occurring using conditional probability:
即使不独立，我们仍然可以用条件概率求出 A 和 B 同时发生的概率：
P(A ∩ B) = P(A) × P(B | A) = P(B) × P(A | B)
This law is essential for problems where the events are dependent, such as drawing two items without replacement from a finite set.
该法则对事件相互依赖的问题至关重要，例如从有限集合中不放回地抽取两个物品。
Example: A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement. The probability that both are red is P(R₁ ∩ R₂) = P(R₁) × P(R₂ | R₁) = (5/8) × (4/7) = 20/56 = 5/14.
例如：一个袋子中有5个红球和3个蓝球。不放回地抽取两个球。两个都是红球的概率为 P(R₁ ∩ R₂) = P(R₁) × P(R₂ | R₁) = (5/8) × (4/7) = 20/56 = 5/14。
8. Tree Diagrams and Conditional Probabilities | 树状图与条件概率
Tree diagrams are powerful visual tools for solving multi-stage probability problems. Each branch represents a possible outcome, and the probability written on each branch is often a conditional probability.
树状图是解决多阶段概率问题的强大可视化工具。每条分支代表一种可能结果，分支上标注的概率通常是条件概率。
For two-stage experiments, multiply along branches to find the probability of a particular sequence, and add the probabilities of different sequences that lead to the same final event.
对于两阶段试验，沿各分支相乘得到特定序列的概率，并将通向相同最终事件的不同序列概率相加。
Below is a typical tree diagram summary for drawing two items without replacement from a bag with 3 red and 2 blue items:
下面是一个典型的树状图总结：从装有3个红球和2个蓝球的袋子中不放回地抽取两个球：
| First draw | Second draw (given first) | Branch probability |
| R (3/5) | R (2/4) | (3/5)×(2/4)=3/10 |
| R (3/5) | B (2/4) | (3/5)×(2/4)=3/10 |
| B (2/5) | R (3/4) | (2/5)×(3/4)=3/10 |
| B (2/5) | B (1/4) | (2/5)×(1/4)=1/10 |
Notice that the second-draw probabilities are conditional on the first draw, not independent.
注意，第二次抽取的概率是以第一次抽取为条件的，因此并不独立。
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