📚 Probability of Combined Events | 复合事件的概率
Probability is the mathematical language of uncertainty. It measures how likely an event is to occur on a scale from 0 (impossible) to 1 (certain). When two or more events are combined, we need clear rules to calculate the chance that both happen, or that at least one happens. This article introduces the essential methods for solving combined-event probability problems in the IGCSE Mathematics syllabus.
概率是描述不确定性的数学语言，它用 0 到 1 之间的数值衡量事件发生的可能性：0 表示不可能发生，1 表示必然发生。当两个或多个事件结合在一起时，我们需要明确的规则来计算”两者都发生”或”至少一个发生”的概率。本文将系统介绍 IGCSE 数学考纲中求解复合事件概率的必备方法。
1. Understanding Probability Basics | 理解概率基础
Every probability problem begins with identifying the sample space, which is the set of all possible outcomes. For a single event A, the probability is written as P(A) and calculated by dividing the number of favourable outcomes by the total number of equally likely outcomes.
每个概率问题都始于确定样本空间，即所有可能结果构成的集合。对于单一事件 A，其概率记为 P(A)，其数值等于有利结果数除以等可能结果总数。
P(A) = 有利结果数 ÷ 总结果数 = n(A) ÷ n(S)
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The probability of any event is always between 0 and 1 inclusive.
任何事件的概率始终介于 0 和 1 之间（含端点）。
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P(A) = 0 means the event is impossible; P(A) = 1 means the event is certain.
P(A) = 0 表示事件不可能发生；P(A) = 1 表示事件必然发生。
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The sum of the probabilities of all possible outcomes of an experiment equals 1.
一次试验中所有可能结果的概率之和等于 1。
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The complement rule states that P(A′) = 1 − P(A), where A′ means “not A”.
补事件法则指出：P(A′) = 1 − P(A)，其中 A′ 表示”非 A”。
For example, when rolling a fair six-sided die, the probability of rolling a 4 is 1 ÷ 6 = 1/6, and the probability of not rolling a 4 is 1 − 1/6 = 5/6.
例如，掷一枚均匀的六面骰子时，掷出 4 的概率为 1 ÷ 6 = 1/6，而掷不出 4 的概率为 1 − 1/6 = 5/6。
2. The Addition Rule: P(A or B) | 加法法则：P(A 或 B)
For mutually exclusive events — events that cannot happen at the same time — the probability that either A or B occurs is the sum of their individual probabilities.
对于互斥事件，即不能同时发生的事件，A 与 B 中至少一个发生的概率等于它们各自概率之和。
P(A ∪ B) = P(A) + P(B)
Here the symbol ∪ means “union” or “OR”. For example, when drawing one card from a standard deck of 52 cards, the events “draw a heart” and “draw a spade” are mutually exclusive, so the probability of drawing a heart or a spade is 13/52 + 13/52 = 26/52 = 1/2.
这里符号 ∪ 表示并集，即”或”。例如，从一副 52 张的标准扑克牌中抽取一张，”抽到红心”与”抽到黑桃”是互斥事件，因此抽到红心或黑桃的概率为 13/52 + 13/52 = 26/52 = 1/2。
For events that are not mutually exclusive, we must subtract the probability that both occur, to avoid double-counting the overlap.
对于并非互斥的事件，为避免重复计算重叠部分，我们必须减去两者同时发生的概率。
P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Here ∩ means “intersection” or “AND”. For example, the events “draw a red card” and “draw a king” are not mutually exclusive, because the king of hearts and the king of diamonds belong to both. Thus P(red or king) = 26/52 + 4/52 − 2/52 = 28/52 = 7/13.
这里 ∩ 表示交集，即”且”。例如，”抽到红色牌”与”抽到 K”并非互斥事件，因为红心 K 和方块 K 同时属于两者。因此 P(红色或 K) = 26/52 + 4/52 − 2/52 = 28/52 = 7/13。
3. The Multiplication Rule: P(A and B) | 乘法法则：P(A 且 B)
For independent events, where the outcome of one event does not affect the outcome of the other, the probability that both A and B occur is the product of their individual probabilities.
对于相互独立的事件，即一个事件的结果不影响另一个事件的结果时，A 与 B 同时发生的概率等于它们各自概率的乘积。
P(A ∩ B) = P(A) × P(B)
For example, rolling a 4 on a fair die and flipping heads on a fair coin are independent events, so P(4 and heads) = 1/6 × 1/2 = 1/12.
例如，掷骰子得到 4 与抛硬币得到正面是相互独立的事件，因此 P(4 且正面) = 1/6 × 1/2 = 1/12。
For dependent events, the probability of the second event depends on the outcome of the first. In this case we use conditional probability.
对于相关事件，第二个事件的概率取决于第一个事件的结果。此时需要运用条件概率。
P(A ∩ B) = P(A) × P(B | A)
The expression P(B | A) means “the probability of B given that A has occurred”. For example, drawing two cards without replacement: the probability of drawing two aces in a row is 4/52 × 3/51 = 12/2652 = 1/221.
表达式 P(B | A) 表示”在 A 已发生的条件下 B 发生的概率”。例如，不放回地连续抽取两张牌：连续抽到两张 A 的概率为 4/52 × 3/51 = 12/2652 = 1/221。
4. Drawing Tree Diagrams | 绘制树状图
Tree diagrams are a visual method for representing successive events. Each branch is labelled with a probability, and the probability of reaching the end of a path is the product of the probabilities along that path.
树状图是表示连续事件的直观工具。每条分支上标注对应概率，到达某条路径终点的概率等于该路径上所有分支概率的乘积。
Consider tossing a fair coin twice. The tree diagram has two stages; at each stage the branches “heads” (H) and “tails” (T) each have probability 1/2. The four possible paths and their probabilities are shown below.
考虑抛一枚均匀硬币两次。树状图包含两个阶段；每个阶段”正面”(H) 与”反面”(T) 分支的概率均为 1/2。四条可能路径及其概率如下表所示。
| 路径 | 计算结果 | 概率 |
| H 然后 H | 1/2 × 1/2 | 1/4 |
| H 然后 T | 1/2 × 1/2 | 1/4 |
| T 然后 H | 1/2 × 1/2 | 1/4 |
| T 然后 T | 1/2 × 1/2 | 1/4 |
To find the probability of an event described by several paths, simply add the probabilities of those paths. Therefore the probability of getting exactly one head is 1/4 + 1/4 = 1/2.
若某事件对应多条路径，只需将这些路径的概率相加。因此恰好得到一个正面的概率为 1/4 + 1/4 = 1/2。
5. Independent vs Dependent Events | 独立事件与相关事件
Distinguishing between independent and dependent events is crucial when applying the multiplication rule. Independent events do not influence each other, while dependent events do.
区分独立事件与相关事件是正确运用乘法法则的关键。独立事件之间互不影响，而相关事件之间会相互影响。
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Sampling with replacement: after each draw, the item is returned, so the sample space stays the same and the events are independent.
有放回抽样：每次抽取后将物品放回，样本空间保持不变，各事件相互独立。
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Sampling without replacement: the item is not returned, so the sample space changes and the events are dependent.
无放回抽样：每次抽取后不放回，样本空间发生改变，各事件相互关联。
The table below compares the two situations for selecting two objects from a bag containing 3 red and 2 blue marbles.
下表比较了从装有 3 个红球和 2 个蓝球的袋子中选取两个球的两种情形。
| 情形 | P(先红后红) | 说明 |
| 有放回 | 3/5 × 3/5 = 9/25 | 第二次仍是 5 个球 |
| 无放回 | 3/5 × 2/4 = 3/10 | 第二次只剩 4 个球 |
Always read the question carefully to determine whether replacement happens. This detail changes the denominator in the second probability.
解题时务必仔细审题，判断是否存在放回。这一细节会改变第二个概率的分母。
6. Venn Diagrams in Probability | 概率中的韦恩图
Venn diagrams provide a clear way to visualise the relationships between events. In a typical diagram, the rectangle represents the entire sample space S, and circles represent events A and B. The regions show the intersection, the union, and the complements.
韦恩图能够清晰地展示事件之间的关系。在典型示意图中，矩形表示整个样本空间 S，圆形表示事件 A 与 B。图中的各个区域分别表示交集、并集和补集。
For two overlapping events A and B, the four regions correspond to: (1) A only; (2) B only; (3) both A and B; (4) neither A nor B. Reading probabilities directly from a Venn diagram can save time and reduce errors.
对于两个有重叠的事件 A 与 B，图中四个区域分别对应：(1) 仅 A；(2) 仅 B；(3) A 与 B 同时成立；(4) 两者均不成立。直接从韦恩图中读取概率既省时又能减少错误。
For example, suppose in a class of 30 students, 18 study physics, 14 study chemistry, and 8 study both. Then the number studying only physics is 18 − 8 = 10, the number studying only chemistry is 14 − 8 = 6, and the number studying neither is 30 − (10 + 8 + 6) = 6. The probability that a randomly chosen student studies physics or chemistry is (10 + 8 + 6) ÷ 30 = 24/30 = 4/5.
例如，某班有 30 名学生，其中 18 人学习物理，14 人学习化学，8 人两科都学。则仅学物理的人数为 18 − 8 = 10，仅学化学的人数为 14 − 8 = 6，两科均不学的人数为 30 − (10 + 8 + 6) = 6。随机选择一名学生，其学习物理或化学的概率为 (10 + 8 + 6) ÷ 30 = 24/30 = 4/5。
P(A ∪ B) = P(A) + P(B) − P(A ∩ B) 与韦恩图完全一致
This formula corresponds exactly to the overlap structure of the Venn diagram: adding the two circles counts the overlap twice, so we subtract it once.
该公式与韦恩图的重叠结构完全对应：将两个圆形区域相加时，重叠部分被重复计算了两次，因此需要减去一次。
7. Worked Example: Two Dice | 实例解析：掷两颗骰子
Two fair six-sided dice are rolled together. Each die has faces numbered 1 to 6, so the total number of equally likely outcomes is 6 × 6 = 36.
同时掷两枚均匀的六面骰子。每枚骰子有 1 到 6 共 6 个面，因此等可能结果总数为 6 × 6 = 36。
We first calculate the probability that the sum of the two scores is 7. The favourable pairs are (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), giving 6 outcomes.
首先计算两枚骰子点数之和为 7 的概率。有利的点数对为 (1, 6)、(2, 5)、(3, 4)、(4, 3)、(5, 2)、(6, 1)，共 6 种结果。
P(和为 7) = 6 ÷ 36 = 1/6
Next we calculate the probability that the sum is 11. The favourable pairs are (5, 6) and (6, 5), giving 2 outcomes.
然后计算点数之和为 11 的概率。有利的点数对为 (5, 6) 和 (6, 5)，共 2 种结果。
P(和为 11) = 2 ÷ 36 = 1/18
Because the events “sum = 7” and “sum = 11” cannot occur at the same time, they are mutually exclusive. Therefore the probability that the sum is 7 or 11 is found by addition.
由于”和为 7″与”和为 11″不可能同时发生，所以它们是互斥事件。因此点数之和为 7 或 11 的概率通过加法求得。
P(和为 7 或 11) = 1/6 + 1/18 = 3/18 + 1/18 = 4/18 = 2/9
The full frequency distribution of sums is shown below; it confirms that the sum 7 is the most likely outcome.
点数之和的完整频数分布如下表所示；可以看出和为 7 是最可能出现的结果。
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