📚 Binomial Distribution | 二项分布
h a boolean-valued outcome: success (with probability p) or failure (with probability q = 1 − p). It is one of the cornerstones of discrete probability theory, and it appears in almost every introductory statistics course because it turns a simple repeated experiment into a compact, calculable model.
二项分布是离散概率论的基石之一。它描述了在 n 次独立试验中成功的次数，每次试验都问一个是非问题，每次结果要么成功（概率 p），要么失败（概率 q = 1 − p）。它把"重复同一个简单实验"这件事压缩成一个简洁、可计算的模型，因此几乎出现在每一门入门统计课程中。
In IB Mathematics, students encounter the binomial distribution in both the Analysis & Approaches (AA) and Applications & Interpretation (AI) courses. It forms the basis for hypothesis testing and connects to the normal distribution through approximations. In this revision guide we build the model from its four conditions, derive the probability formula, work through cumulative probabilities, means and variances, calculator skills, the normal approximation, and full hypothesis-testing procedure.
在 IB 数学中，无论是分析与方法 (AA) 还是应用与解释 (AI) 课程，学生都会接触到二项分布。它是假设检验的基础，并通过近似与正态分布建立联系。本篇复习指南将从四个前提条件出发建立模型，推导概率公式，讲解累积概率、均值与方差、计算器操作、正态近似，以及完整的假设检验流程。
1. What is a Binomial Distribution? | 什么是二项分布？
A random variable X follows a binomial distribution if it counts the number of successes in n independent Bernoulli trials, each with the same probability of success p. The possible values of X are 0, 1, 2, …, n. A single trial with only two outcomes is called a Bernoulli trial, and the binomial distribution is simply the sum of n identical, independent Bernoulli trials.
如果随机变量 X 统计的是 n 次独立伯努利试验中成功的次数，且每次成功的概率 p 相同，则 X 服从二项分布。X 的可能取值为 0, 1, 2, …, n。只有两种结果的一次试验称为伯努利试验，而二项分布本质上就是 n 次相同且独立的伯努利试验之和。
We write X ~ B(n, p) to indicate that X has a binomial distribution with parameters n and p. The number n is often called the number of trials, and p is the probability of success on each trial. Notice that the distribution is fully determined by just these two numbers: once you know n and p, every probability, the mean and the variance are fixed.
我们记作 X ~ B(n, p) 表示 X 服从参数为 n 和 p 的二项分布。n 通常称为试验次数，p 是每次试验成功的概率。注意这个分布完全由这两个数决定：只要知道 n 和 p，所有概率、均值与方差就都确定了。
al values rather than to intervals, and the sum of all probabilities P(X = 0) + P(X = 1) + … + P(X = n) equals exactly 1.
该分布是离散型的，因此概率是分配给单个取值而不是区间的，并且所有概率之和 P(X = 0) + P(X = 1) + … + P(X = n) 恰好等于 1。
2. The Four Conditions: BINS | 四个前提条件：BINS
Before you may use the binomial model, you must justify that four conditions hold. A useful memory aid is the acronym BINS: Binary, Independent, Number fixed, Same probability.
在使用二项分布模型之前，必须先说明四个条件成立。一个实用的记忆口诀是 BINS：Binary（二元）、Independent（独立）、Number fixed（次数固定）、Same probability（概率相同）。
- B – Binary: each trial has exactly two outcomes, conventionally called success and failure.
- I – Independent: the outcome of one trial does not affect any other trial.
- N – Number fixed: the number of trials n is decided in advance and does not change.
- S – Same probability: the probability of success p is constant for every trial.
- B – 二元：每次试验恰好有两个结果，习惯上称为成功与失败。
- I – 独立：一次试验的结果不影响任何其他试验。
- N – 次数固定：试验次数 n 事先确定，且不会改变。
- S – 概率相同：每次试验成功的概率 p 保持不变。
Two conditions are commonly violated in exam questions. Sampling without replacement from a small population breaks independence and the constant-p condition, so the binomial model should be applied only when the sample is small relative to the population (a common rule of thumb is less than 10%).
考试题中最常被违反的是两个条件。从小总体中进行"不放回"抽样会破坏独立性和概率恒定条件，因此只有当样本相对于总体很小时（常用经验法则是小于 10%）才可以套用二项模型。
| Condition | Question to ask | Typical failure |
| Binary | Are there only two outcomes? | Rolling a die and recording 1-6 |
| Independent | Does one result change the next? | Drawing cards without replacement |
| Number fixed | Is n decided in advance? | Counting trials until the first success |
| Same p | Is p constant every time? | Different machines with different defect rates |
3. Notation, Parameters and the Complement | 记号、参数与互补事件
lways its complement, written q = 1 − p, and this single symbol saves a great deal of writing.
两个参数定义了这个分布：n 是试验次数，p 是单次试验成功的概率。失败的概率总是它的互补概率，记作 q = 1 − p，这个符号可以省去大量书写。
X ~ B(n, p), q = 1 − p
For example, if a fair coin is tossed 20 times and X counts the number of heads, then X ~ is B(15, 0.25) with q = 0.75. Note that a single Bernoulli trial is the special case B(1, p).
例如，把一枚均匀硬币抛 20 次，X 表示正面出现的次数，则 X ~ B(20, 0.5)，其中 q = 0.5。如果一份选择题有 15 道题、每题 4 个选项，学生全部靠猜，则答对的题数服从 B(15, 0.25)，其中 q = 0.75。注意，单次伯努利试验就是 B(1, p) 这一特例。
Always state the distribution explicitly in your written answer. Examiners award a mark for correctly identifying n and p, and that mark is lost if you launch straight into a calculator calculation.
作答时一定要明确写出分布。评分标准会给"正确识别 n 与 p"的分，而如果你直接开始按计算器，这分就会丢掉。
4. The Binomial Probability Formula | 二项分布概率公式
To find the probability of exactly r successes in n trials, we must combine two ideas: the probability of one specific sequence, and the number of sequences that give r successes.
要求出 n 次试验中恰好有 r 次成功的概率，需要把两件事结合起来：某一条特定序列的概率，以及能得到 r 次成功的序列总数。
A particular sequence with r successes and (n − r) failures, for instance S S F S F …, has probability p × p × q × p × q × … = pʳ qⁿ⁻ʳ ⁿCᵣ such sequences, since we simply choose which r of the n positions hold the successes.
一条包含 r 次成功和 (n − r) 次失败的特定序列，例如 S S F S F …，其概率为 p × p × q × p × q × … = pʳ qⁿ⁻ʳ，因为试验相互独立，可以直接相乘。这样的序列共有 ⁿCᵣ 条，因为我们只需从 n 个位置中选出哪 r 个放成功。
P(X = r) = ⁿCᵣ × pʳ × qⁿ⁻ʳ, r = 0, 1, …, n
In factorial form the binomial coefficient is written out as ⁿCᵣ = n! ÷ (r! × (n − r)! ccident: the binomial theorem is precisely why the distribution carries the name binomial.
用阶乘表示时，二项系数写作 ⁿCᵣ = n! ÷ (r! × (n − r)!)，这与你在 (p + q)ⁿ 的二项展开中遇到的表达式完全相同。这并非巧合：二项式定理正是这个分布被称为"二项"分布的原因。
e probabilities over r = 0 to r = n must sum to exactly 1, which is a handy checking tool.
由于 p + q = 1，恒等式 (p + q)ⁿ = 1 保证了 r 从 0 到 n 的所有概率之和恰为 1，这也是一个方便的检验手段。
5. Worked Example: Exactly k Successes | 例题：恰好 k 次成功
A biased spinner lands on red with probability 0.3. The spinner is spun 10 times and X is the number of reds obtained. Find the probability of exactly 3 reds.
一个偏心的转盘落在红色的概率为 0.3。转动转盘 10 次，X 表示出现红色的次数。求恰好出现 3 次红色的概率。
(red or not red), spins are independent, n = 10 is fixed, and p = 0.3 is constant. Hence X ~ B(10, 0.3) and q = 0.7.
首先检查条件：每次转动是二元的（红色或非红色），各次转动相互独立，n = 10 固定，p = 0.3 不变。因此 X ~ B(10, 0.3)，q = 0.7。
P(X = 3) = ¹⁰C₃ × 0.3³ × 0.7⁷ = 120 × 0.027 × 0.0823543 ≈ 0.2668
So there is about a 26.7% chance of exactly three reds. The table below lists the whole distribution, which is a useful way of seeing the characteristic shape: it rises to a peak around the mean and then falls away.
所以恰好出现三次红色的概率约为 26.7%。下表列出了完整的分布，这有助于看出其典型形状：概率在均值附近升到峰值，然后逐渐下降。
| r | 0 | 1 | 2 | 3 | 4 | 5 |
| P(X = r) | 0.0282 | 0.1211 | 0.2335 | 0.2668 | 0.2001 | 0.1029 |
| r | 6 | 7 | 8 | 9 | 10 | Total |
| P(X = r) | 0.0368 | 0.0090 | 0.0014 | 0.0001 | 0.0000 | 1.0000 |
Notice how small the tail probabilities are. This is a general feature: for fixed n and p, the binomial distribution becomes increasingly concentrated around np as n grows.
注意尾部概率有多小。这是一个普遍特征：当 n 增大而 p 固定时，二项分布会越来越集中在 np 附近。
6. Cumulative Probabilities and Inequalities | 累积概率与不等式
s into cumulative form. The key identity is that P(X ≤ r) adds up every value from 0 to r.
大多数实际问题问的是一个范围而不是某个具体值，因此需要把不等式转换成累积形式。关键恒等式是：P(X ≤ r) 把 0 到 r 的所有取值加起来。
P(X ≤ r) = Σ P(X = k) for k = 0 to r
- P(X < r) = P(X ≤ r − 1)
- P(X ≥ r) = 1 − P(X ≤ r − 1)
- P(X > r) = 1 − P(X ≤ r)
- P(X = r) = P(X ≤ r) − P(X ≤ r − 1)
- P(X < r) = P(X ≤ r − 1)
- P(X ≥ r) = 1 − P(X ≤ r − 1)
- P(X > r) = 1 − P(X ≤ r)
- P(X = r) = P(X ≤ r) − P(X ≤ r − 1)
ecause the binomial is discrete, that expression removes the value r itself and gives the wrong answer. Always apply the "include or exclude the boundary" test before pressing the calculator.
