Probability Formulas: Systematic Summary & Typical Applications | 概率公式的归纳与典型应用

📚 Probability Formulas: Systematic Summary & Typical Applications | 概率公式的归纳与典型应用

Probability is the mathematical language of uncertainty. At A-Level, students are expected not only to recall individual formulas but to recognise when each rule applies, how they interconnect, and how to combine them in multi-stage problems. This article presents a structured summary of the core probability formulas, followed by typical exam-style applications with step-by-step reasoning.

概率是描述不确定性的数学语言。在 A-Level 考试中，学生不仅需要记住单个公式，更要能够判断何时使用哪一条规则、这些规则之间如何相互联系，以及在多阶段问题中如何综合运用。本文系统归纳概率的核心公式，并结合典型考试题型进行逐步剖析。

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1. Fundamental Definitions | 基本定义与记法

The probability of an event A, written P(A), is a number between 0 and 1 measuring how likely A is to occur. The sample space S contains all possible outcomes, and the empty set ∅ represents an impossible event. Two events are mutually exclusive if they cannot happen at the same time, meaning A ∩ B = ∅.

事件 A 的概率记作 P(A)，是一个介于 0 和 1 之间的数，衡量 A 发生的可能性大小。样本空间 S 包含所有可能的结果，空集 ∅ 表示不可能事件。若两个事件不能同时发生，则称它们互斥，即 A ∩ B = ∅。

• For any event A: 0 ≤ P(A) ≤ 1
• Certain event: P(S) = 1
• Impossible event: P(∅) = 0
• Complement rule: P(A′) = 1 − P(A), where A′ is “not A”

• 对任意事件 A：0 ≤ P(A) ≤ 1
• 必然事件：P(S) = 1
• 不可能事件：P(∅) = 0
• 补事件法则：P(A′) = 1 − P(A)，其中 A′ 表示“A 不发生”

P(A′) = 1 − P(A)

This complement rule is often the quickest route to a solution: when the direct probability is complicated, finding the probability of the opposite event may be far simpler.

补事件法则往往是解题的最快捷路径：当直接计算某事件的概率较复杂时，先求其对立事件的概率通常会简单得多。

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2. The Addition Rule | 加法法则

The addition rule connects the probability of a union to the probabilities of the individual events and their intersection.

加法法则将并事件（或事件）的概率与各事件自身概率及其交事件概率联系起来。

P(A ∪ B) = P(A) + P(B) − P(A ∩ B)

When A and B are mutually exclusive, P(A ∩ B) = 0, so the rule simplifies to P(A ∪ B) = P(A) + P(B).

当 A 与 B 互斥时，P(A ∩ B) = 0，此时加法法则简化为 P(A ∪ B) = P(A) + P(B)。

Typical application: A card is drawn from a standard deck of 52. Find the probability that it is either a heart or a king.

典型应用：从一副 52 张的标准扑克牌中抽取一张，求抽到红心或老 K 的概率。

Let H = heart, K = king. P(H) = 13/52, P(K) = 4/52, P(H ∩ K) = P(king of hearts) = 1/52. The events are not mutually exclusive, so P(H ∪ K) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13.

设 H = 红心，K = 老 K。P(H) = 13/52，P(K) = 4/52，P(H ∩ K) = P(红心老 K) = 1/52。这两个事件并非互斥，因此 P(H ∪ K) = 13/52 + 4/52 − 1/52 = 16/52 = 4/13。

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3. Conditional Probability | 条件概率

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

条件概率衡量的是在另一个事件已发生的条件下，某事件发生的概率，记作 P(A|B)，读作“在 B 发生的条件下 A 的概率”。

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

Rearranging this formula gives the multiplication rule, which is fundamental to tree diagrams and sequential experiments.

将该公式变形即得乘法法则，它是树形图与序贯试验的基础。

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

Typical application: A bag contains 5 red and 3 blue marbles. Two marbles are drawn without replacement. Find the probability that both are red.

典型应用：一个袋子中有 5 个红球和 3 个蓝球，不放回地依次抽取两个球，求两次都抽到红球的概率。

P(first red) = 5/8. Given that the first marble was red, only 4 red and 3 blue remain, so P(second red | first red) = 4/7. Therefore P(both red) = 5/8 × 4/7 = 20/56 = 5/14.

P(第一次为红) = 5/8。在第一次已抽到红球的条件下，袋中剩 4 红 3 蓝，所以 P(第二次为红 | 第一次为红) = 4/7。因此 P(两次都红) = 5/8 × 4/7 = 20/56 = 5/14。

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4. Independence | 事件的独立性

Two events A and B are independent if the occurrence of one does not affect the probability of the other. Mathematically, this is equivalent to three interchangeable statements.

若事件 A 的发生不影响事件 B 的概率，反之亦然，则称 A 与 B 相互独立。数学上，独立性可用以下三个等价表述来刻画。

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

Common pitfall: Independence is not the same as mutual exclusivity. In fact, two mutually exclusive events (with non-zero probabilities) are always dependent, because knowing one occurs tells you the other cannot.

常见误区：独立与互斥是两个不同的概念。事实上，两个互斥事件（概率均非零时）必定是相依的——因为知道其中一个发生，就立刻知道另一个不可能发生。

Typical application: Two fair dice are rolled. Show that the events “first die shows 3” and “sum of the two dice is 7” are independent.

典型应用：同时掷两枚均匀骰子，证明“第一枚骰子出现 3 点”与“两枚骰子点数之和为 7”这两个事件相互独立。

P(A) = 1/6. P(B): the pairs summing to 7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), so P(B) = 6/36 = 1/6. P(A ∩ B): if the first die is 3, the second must be 4, so only one favourable pair out of 36: P(A ∩ B) = 1/36. Since 1/6 × 1/6 = 1/36, the events are independent.

P(A) = 1/6。P(B)：和为 7 的组合有 (1,6)、(2,5)、(3,4)、(4,3)、(5,2)、(6,1) 共 6 种，故 P(B) = 6/36 = 1/6。P(A ∩ B)：若第一枚为 3，则第二枚必须为 4，36 种等可能结果中只有 1 种，故 P(A ∩ B) = 1/36。由于 1/6 × 1/6 = 1/36，因此两事件相互独立。

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5. The Law of Total Probability | 全概率公式

When the sample space is partitioned into mutually exclusive events B₁, B₂, …, Bₙ, the probability of any event A can be found by weighting the conditional probabilities of A given each partition.

当样本空间被划分为互斥的事件 B₁, B₂, …, Bₙ 时，任意事件 A 的概率可以通过对每个划分下的条件概率加权求和得到。

P(A) = P(B₁)P(A|B₁) + P(B₂)P(A|B₂) + … + P(Bₙ)P(A|Bₙ)

Typical application: A factory has three machines producing items. Machine I produces 50% of items with 2% defective, Machine II produces 30% with 3% defective, Machine III produces 20% with 4% defective. Find the overall defective rate.

典型应用：某工厂用三台机器生产零件。I 号机器生产 50% 的产品，次品率为 2%；II 号机器生产 30% 的产品，次品率为 3%；III 号机器生产 20% 的产品，次品率为 4%。求总次品率。

P(D) = P(M₁)P(D|M₁) + P(M₂)P(D|M₂) + P(M₃)P(D|M₃) = 0.5 × 0.02 + 0.3 × 0.03 + 0.2 × 0.04 = 0.01 + 0.009 + 0.008 = 0.027, i.e. 2.7%.

P(D) = P(M₁)P(D|M₁) + P(M₂)P(D|M₂) + P(M₃)P(D|M₃) = 0.5 × 0.02 + 0.3 × 0.03 + 0.2 × 0.04 = 0.01 + 0.009 + 0.008 = 0.027，即 2.7%。

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6. Bayes’ Theorem | 贝叶斯定理

Bayes’ theorem answers the “reverse” question: given that A has occurred, what is the probability that it came from a particular partition Bᵢ? This is one of the most powerful tools in probability and statistics.

贝叶斯定理解答的是“逆向”问题：已知 A 已发生，求它来自某个划分 Bᵢ 的概率。这是概率论与统计学中最强大的工具之一。

P(Bᵢ|A) = P(Bᵢ) × P(A|Bᵢ) / P(A)

where the denominator P(A) is computed via the law of total probability.

其中分母 P(A) 由全概率公式计算。

Typical application: Using the factory example above, if an item is found to be defective, what is the probability it was made by Machine II?

典型应用：沿用上述工厂的例子，若发现一件产品是次品，求它由 II 号机器生产的概率。

P(M₂|D) = P(M₂) × P(D|M₂) / P(D) = (0.3 × 0.03) / 0.027 = 0.009 / 0.027 = 1/3 ≈ 0.333.

P(M₂|D) = P(M₂) × P(D|M₂) / P(D) = (0.3 × 0.03) / 0.027 = 0.009 / 0.027 = 1/3 ≈ 0.333。

Thus, although Machine II produces only 30% of all items, it accounts for one third of all defective items.

由此可见，尽管 II 号机器只生产全部产品的 30%，但它贡献了全部次品的三分之一。

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7. Discrete Random Variables & Expectation | 离散型随机变量与期望

A discrete random variable X takes a finite or countable set of values x with probabilities p(x) = P(X = x). The probability distribution must satisfy two conditions.

离散型随机变量 X 取有限或可数个值 x，其概率为 p(x) = P(X = x)。概率分布必须满足两个条件。

• 0 ≤ p(x) ≤ 1 for all x
• Σ p(x) = 1 over all possible values

• 对所有 x，都有 0 ≤ p(x) ≤ 1
• 对所有可能取值求和：Σ p(x) = 1

The expected value (mean) and variance are the two most important summary measures.

期望（均值）与方差是最重要的两个概括性指标。

E(X) = Σ x · p(x)   and   Var(X) = Σ (x − μ)² · p(x) = E(X²) − [E(X)]²

Typical application: A biased die has probability distribution P(X = x) = x/21 for x = 1, 2, 3, 4, 5, 6. Find E(X).

典型应用：一枚不均匀骰子的概率分布为 P(X = x) = x/21，其中 x = 1, 2, 3, 4, 5, 6。求 E(X)。

First verify Σ p(x) = (1+2+3+4+5+6)/21 = 21/21 = 1. Then E(X) = 1×1/21 + 2×2/21 + 3×3/21 + 4×4/21 + 5×5/21 + 6×6/21 = (1+4+9+16+25+36)/21 = 91/21 ≈ 4.333.

先验证 Σ p(x) = (1+2+3+4+5+6)/21 = 21/21 = 1。然后 E(X) = 1×1/21 + 2×2/21 + 3×3/21 + 4×4/21 + 5×5/21 + 6×6/21 = (1+4+9+16+25+36)/21 = 91/21 ≈ 4.333。

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8. The Binomial Distribution | 二项分布

A binomial distribution models the number of successes in n independent trials, each with the same probability of success p. The conditions are: fixed number of trials n, two outcomes per trial (success/failure), constant p, and independence between trials.

二项分布描述的是 n 次独立试验中成功的次数，每次试验的成功概率 p 恒定。其适用条件为：试验次数 n 固定；每次试验只有成功/失败两种结果；p 保持不变；各次试验相互独立。

X ~ B(n, p):   P(X = k) = ⁿCₖ · pᵏ · (1 − p)ⁿ⁻ᵏ, for k = 0, 1, …, n

E(X) = np   and   Var(X) = np(1 − p)

Typical application: A multiple-choice test has 10 questions, each with 4 options. A student guesses every answer. Find the probability of exactly 3 correct answers.

典型应用：一份测验有 10 道选择题，每题 4 个选项。一名学生完全随机猜测，求恰好答对 3 题的概率。

Let X ~ B(10, 0.25). Then P(X = 3) = ¹⁰C₃ × (0.25)³ × (0.75)⁷ ≈ 120 × 0.015625 × 0.133484 ≈ 0.2503.

设 X ~ B(10, 0.25)，则 P(X = 3) = ¹⁰C₃ × (0.25)³ × (0.75)⁷ ≈ 120 × 0.015625 × 0.133484 ≈ 0.2503。

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9. The Poisson Distribution | 泊松分布

The Poisson distribution models the number of events occurring in a fixed interval of time or space, when events occur independently at a constant average rate λ.

泊松分布用于建模在固定时间或空间区间内事件发生的次数，其前提是事件以恒定的平均速率 λ 独立地发生。

X ~ Po(λ):   P(X = k) = e⁻ᵏ!?? — use: P(X = k) = (e^(−λ) · λᵏ) / k!, for k = 0, 1, 2, …

E(X) = Var(X) = λ

A key feature of the Poisson distribution is that its mean and variance are equal. This property is often used to check whether a Poisson model is appropriate for a data set.

泊松分布的一个重要特征是均值与方差相等。这一性质常被用来检验泊松模型是否适合某组数据。

Typical application: Calls arrive at a call centre at an average rate of 5 per minute. Find the probability that exactly 3 calls arrive in one minute.

典型应用：某呼叫中心平均每分钟接到 5 个电话。求一分钟内恰好接到 3 个电话的概率。

Let X ~ Po(5). Then P(X = 3) = (e⁻⁵ × 5³) / 3! = (0.006738 × 125) / 6 ≈ 0.1404.

设 X ~ Po(5)，则 P(X = 3) = (e⁻⁵ × 5³) / 3! = (0.006738 × 125) / 6 ≈ 0.1404。

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10. The Normal Distribution | 正态分布

The normal distribution is a continuous probability distribution described by its mean μ and standard deviation σ. It is symmetric and bell-shaped, with the total area under the curve equal to 1.

正态分布是一种连续型概率分布，由均值 μ 和标准差 σ 刻画。它呈对称的钟形曲线，曲线下方的总面积为 1。

To find probabilities, standardise X using the z-score transformation, then refer to standard normal tables.

求概率时，先将 X 通过 z 分数变换标准化，再查标准正态分布表。

Z = (X − μ) / σ   where   Z ~ N(0, 1)

Typical application: The weights of apples on a farm are normally distributed with mean 150 g and standard deviation 20 g. Find the probability that a randomly chosen apple weighs more than 180 g.

典型应用：某农场苹果的重量服从正态分布，均值 150 克，标准差 20 克。求随机选取一个苹果重量超过 180 克的概率。

z = (180 − 150)/20 = 1.5. From tables, P(Z < 1.5) = 0.9332, so P(Z > 1.5) = 1 − 0.9332 = 0.0668.

z = (180 − 150)/20 = 1.5。查表得 P(Z < 1.5) = 0.9332，故 P(Z > 1.5) = 1 − 0.9332 = 0.0668。

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11. Normal Approximation to the Binomial | 二项分布的正态近似

When n is large and p is neither too small nor too large, the binomial distribution B(n, p) can be approximated by a normal distribution with the same mean and variance. A common rule of thumb is that np > 5 and n(1 − p) > 5. A continuity correction should be applied.

当 n 较大且 p 不太小也不太大时，二项分布 B(n, p) 可用具有相同均值和方差的正态分布来近似。常用的经验法则是 np > 5 且 n(1 − p) > 5，并且需要进行连续性修正。

B(n, p) ≈ N(np, np(1 − p))

Typical application: A coin is tossed 100 times. Estimate the probability of getting between 45 and 55 heads inclusive.

典型应用：一枚硬币抛掷 100 次，估计正面出现次数在 45 到 55 次之间（含两端）的概率。

Here X ~ B(100, 0.5), with μ = 50 and σ = √(100 × 0.5 × 0.5) = 5. Apply the continuity correction: P(44.5 < X < 55.5). Lower z = (44.5 − 50)/5 = −1.1; upper z = (55.5 − 50)/5 = 1.1. From tables, P(−1.1 < Z < 1.1) = 2 × 0.8643 − 1 = 0.7286.

这里 X ~ B(100, 0.5)，μ = 50，σ = √(100 × 0.5 × 0.5) = 5。进行连续性修正：P(44.5 < X < 55.5)。下界 z = (44.5 − 50)/5 = −1.1；上界 z = (55.5 − 50)/5 = 1.1。查表得 P(−1.1 < Z < 1.1) = 2 × 0.8643 − 1 = 0.7286。

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12. Choosing the Right Formula: A Decision Guide | 公式选择指南

The table below summarises when to use each formula. This guide is equally valuable for revision and for exam strategy under time pressure.

下表总结了各公式的适用情境。这份指南对复习备考和考场上时间紧迫时的策略选择都同样有价值。

Question pattern | 题目特征	Formula | 使用公式
Probability of “A or B” | 求“A 或 B”的概率	P(A ∪ B) = P(A) + P(B) − P(A ∩ B)
Probability of “A and B” | 求“A 且 B”的概率	P(A ∩ B) = P(A) × P(B|A)
Given one event occurred, find probability of another | 已知一事件发生，求另一事件的概率	P(A|B) = P(A ∩ B) / P(B)
Reverse conditional probability | 逆向条件概率	Bayes’ theorem | 贝叶斯定理
Fixed number of independent trials | 固定次数的独立试验	Binomial B(n, p) | 二项分布
Rate of occurrence in time/space | 单位时间/空间内的发生率	Poisson Po(λ) | 泊松分布
Continuous measurements, symmetric distribution | 连续型测量值、对称分布	Normal N(μ, σ²) | 正态分布

Final tips: Always write down the defining distribution with its parameters before substituting numbers; state whether events are mutually exclusive or independent before applying a rule; and when a question says “at least one”, consider the complement. Master these formulas through consistent practice, and probability problems will become a reliable source of marks.

最后建议：在代入数值前，先写出所用分布及其参数；在套用公式前，先判断事件是否互斥或独立；当题目出现“至少一个”时，优先考虑补事件。通过持之以恒的练习掌握这些公式，概率题将成为你稳定得分的题型。

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