Probability of Combined Events | 复合事件的概率求解

📚 Probability of Combined Events | 复合事件的概率求解

In probability, combined events occur when we consider two or more events happening together, either simultaneously or in sequence. Mastering the rules for unions, intersections, and conditional probability is essential for IB Mathematics.

在概率中，复合事件是指两个或多个事件同时发生或依次发生的情况。掌握并集、交集以及条件概率的运算法则，是IB数学的核心要求。

1. Sample Space and Events | 样本空间与事件

The sample space S is the set of all possible outcomes of an experiment. An event A is a subset of S. If all outcomes are equally likely, the probability of A is P(A) = number of outcomes in A divided by number in S.

样本空间 S 是试验所有可能结果的集合。事件 A 是 S 的一个子集。若所有结果等可能，则事件 A 的概率为 P(A) = A 中结果个数 ÷ S 中结果个数。

P(A) = n(A) / n(S)

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}, so P(A) = 3/6 = 1/2.

例如，掷一枚均匀骰子时，S = {1,2,3,4,5,6}。事件“掷出偶数”为 A = {2,4,6}，因此 P(A) = 3/6 = 1/2。

2. Union and Intersection | 并集与交集

The union of two events A and B, written A ∪ B, is the event that at least one of A or B occurs. The intersection A ∩ B is the event that both A and B occur.

两个事件 A 与 B 的并集记作 A ∪ B，表示 A 或 B 至少有一个发生。交集 A ∩ B 表示 A 与 B 同时发生。

The general addition rule connects these two concepts:

一般的加法法则将这两个概念联系起来：

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

This formula accounts for the double-counting of the intersection when both A and B are possible. If A and B cannot both happen, the last term disappears.

这个公式避免了 A 与 B 同时发生部分被重复计算。若 A 与 B 不可能同时发生，则最后一项为零。

3. Mutually Exclusive Events | 互斥事件

Two events are mutually exclusive if they cannot occur at the same time. In set notation, A ∩ B = ∅, so P(A ∩ B) = 0.

两个事件互斥是指它们不可能同时发生。用集合语言表示，A ∩ B = ∅，所以 P(A ∩ B) = 0。

For mutually exclusive events, the addition rule simplifies to:

对于互斥事件，加法法则简化为：

P(A ∪ B) = P(A) + P(B)

For example, when drawing one card from a deck, the events “draw a heart” and “draw a spade” are mutually exclusive because a single card cannot be both suits.

例如，从一副牌中抽一张时，“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。

A common mistake is to assume that mutually exclusive events are independent. They are actually the opposite: if A and B are mutually exclusive and both have positive probability, then P(A ∩ B) = 0, which is not equal to P(A)P(B) unless one of them is impossible.

常见错误是认为互斥事件就是独立事件。事实上它们相反：若 A 与 B 互斥且概率都为正，则 P(A ∩ B) = 0，不等于 P(A)P(B)，除非其中一个概率为零。

4. Independent Events | 独立事件

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 the multiplication rule:

两个事件 A 与 B 独立是指其中一个发生不影响另一个发生的概率。数学上，独立性由乘法法则定义：

P(A ∩ B) = P(A) × P(B)

For instance, rolling a die and tossing a coin are independent experiments. The probability of rolling a 6 and getting heads is (1/6) × (1/2) = 1/12.

例如，掷骰子和抛硬币是独立试验。掷出 6 且硬币为正面的概率为 (1/6) × (1/2) = 1/12。

Do not confuse independent with mutually exclusive. Independent events usually have non-empty intersections, whereas mutually exclusive events cannot overlap.

不要将独立与互斥混淆。独立事件通常有非空交集，而互斥事件没有交集。

5. Conditional Probability | 条件概率

Conditional probability measures the probability of event A given that event B has already occurred. It is denoted P(A|B), read as “the probability of A given B”.

条件概率表示在事件 B 已经发生的条件下，事件 A 发生的概率。记作 P(A|B)，读作“在 B 条件下 A 的概率”。

P(A|B) = P(A ∩ B) / P(B), where P(B) > 0

For example, if a class has 30 students and 18 have studied calculus, then P(studied calculus) = 18/30. If we also know that 12 of the calculus students are female, then P(female | studied calculus) = 12/18 = 2/3.

例如，一个班有30名学生，其中18人学过微积分，则 P(学过微积分) = 18/30。如果学过微积分的学生中有12名女生，则 P(女生 | 学过微积分) = 12/18 = 2/3。

Conditional probability is the foundation for many IB exam problems, especially those involving two-stage experiments.

条件概率是许多IB考题的基础，尤其是涉及两阶段试验的问题。

6. Multiplication Rule | 乘法法则

Rearranging the conditional probability formula gives the general multiplication rule:

重排条件概率公式，得到一般乘法法则：

P(A ∩ B) = P(A|B) × P(B) = P(B|A) × P(A)

This rule is useful when a problem gives conditional probabilities directly, or when we need to compute the probability of a sequence of events.

该法则在题目直接给出条件概率，或需要计算一串事件发生的概率时非常有用。

For example, drawing two cards without replacement: P(first is Ace) = 4/52. P(second is Ace | first was Ace) = 3/51. Hence P(both are Aces) = (4/52) × (3/51).

例如，从一副牌中不放回地连续抽两张：P(第一张是Ace) = 4/52。P(第二张是Ace | 第一张是Ace) = 3/51。因此 P(两张都是Ace) = (4/52) × (3/51)。

7. Tree Diagrams | 树形图

Tree diagrams are powerful tools for solving multi-stage probability problems. Each branch represents a possible outcome, and the probability written on a branch is usually conditional on the outcomes already chosen.

树形图是解决多阶段概率问题的有力工具。每条分支代表一个可能结果，分支上标注的概率通常是基于已发生结果的条件概率。

To find the probability of a single path, multiply the probabilities along that path. To find the probability of several outcomes, add the probabilities of the separate paths.

求一条路径的概率时，将路径上各概率相乘。求若干结果的总概率时，将各条路径的概率相加。

Example: A bag contains 3 red and 5 blue marbles. Two marbles are drawn without replacement. The probability of drawing one red and one blue in any order is:

例：一个袋子中有3个红球和5个蓝球。不放回地连续取出两个球。取出一红一蓝（任意顺序）的概率为：

P(R then B) + P(B then R) = (3/8 × 5/7) + (5/8 × 3/7) = 30/56 = 15/28

Notice that the denominator of the second draw decreases from 8 to 7 because one marble has been removed.

注意第二次抽取的分母从8变为7，因为已经取走了一个球。

8. Venn Diagrams and Two-Way Tables | 文氏图与双向表

Venn diagrams visually represent unions, intersections, and complements. They are especially helpful for conditional probability, because we can focus on only the region holding the given condition.

文氏图直观地表示并集、交集和补集。它对于条件概率尤其有用，因为我们只需关注满足条件的区域。

Two-way tables (contingency tables) organize data by two categorical variables. They make it easy to compute marginal, joint, and conditional probabilities.

双向表（列联表）按两个分类变量组织数据，便于计算边缘概率、联合概率和条件概率。

For example, a survey of 100 students: 40 study mathematics, 30 study physics, and 15 study both. Then by the addition rule, the number studying at least one is 40+30−15=55, so P(math ∪ physics) = 55/100.

例如，调查100名学生：40人学数学，30人学物理，15人两科都学。根据加法法则，至少学一科的人数为40+30−15=55，因此 P(数学∪物理) = 55/100。

When using a Venn diagram, always label the intersection first, then work outward to avoid double-counting.

使用文氏图时，务必先标出交集部分，再由内向外填充，避免重复计算。

9. Common Pitfalls and Exam Strategies | 常见陷阱与解题策略

Pitfall 1: Confusing “and” with “or”. “And” means intersection, while “or” means union. A question asking for P(A ∪ B) must not be answered by simply multiplying P(A) and P(B).

陷阱1：混淆“且”与“或”。“且”表示交集，“或”表示并集。求 P(A ∪ B) 时不能简单地将 P(A) 与 P(B) 相乘。

Pitfall 2: Forgetting the subtraction term in the addition rule. Always ask whether A and B can overlap.

陷阱2：在加法法则中忘记减去交集项。一定要先判断 A 与 B 是否可能重叠。

Pitfall 3: Using independence when the events involve sampling without replacement. In that case, probabilities change after the first selection.

陷阱3：在不放回抽样时误用独立性。这种情况下，第一次选择后概率会改变。

Strategy: Write down all given probabilities first, then draw a Venn diagram or tree diagram if the problem is multi-stage. Finally, identify whether the question asks for conditional, joint, or union probability.

策略：先写出所有已知概率，如果是多阶段问题就画文氏图或树形图，最后判断问题是要求条件概率、联合概率还是并集概率。

10. Worked Example | 综合例题

A box contains 4 red chips and 6 blue chips. Two chips are drawn at random without replacement. Find the probability that the second chip is red given that the first chip was blue.

一个盒子中有4个红色筹码和6个蓝色筹码。不放回地随机抽取两个筹码。已知第一个筹码是蓝色，求第二个筹码是红色的概率。

Given the first chip is blue, there are now 9 chips left: 4 red and 5 blue. Therefore:

已知第一个筹码是蓝色后，盒子中还剩9个筹码：4红5蓝。因此：

P(second red | first blue) = 4/9

Let us also compute the probability that both chips are blue:

我们再计算两个筹码都是蓝色的概率：

P(first blue) = 6/10, P(second blue | first blue) = 5/9

P(both blue) = 6/10 × 5/9 = 30/90 = 1/3

This example illustrates how the multiplication rule and conditional probability work together in combined-event problems.

这个例题说明了乘法法则与条件概率在复合事件问题中如何联合运用。

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Probability of Combined Events | 复合事件的概率求解