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Using the Binomial Distribution in Context | 二项分布在实际问题中的应用

2026-08-30 06:49 · 本文部分内容由AI辅助创作

📚 Using the Binomial Distribution in Context | 二项分布在实际问题中的应用

This guide corresponds to Exercise 6D in the AQA A-Level Mathematics Year 1 Statistics textbook. We explore how to apply the binomial distribution to real-world scenarios, how to choose the correct parameters, and how to interpret probability statements involving inequalities. Worked examples and exam-style questions are included throughout.

本指南对应 AQA A-Level 数学 Year 1 统计教材中的 Exercise 6D。我们将探讨如何将二项分布应用于现实情境,如何选取正确的参数,以及如何解读涉及不等式的概率表述。全文包含典型例题与考试风格练习。


1. Recognising the Binomial Distribution | 识别二项分布

A discrete random variable X follows a binomial distribution if it counts the number of successes in a fixed number of independent trials, where each trial has the same probability of success. We write X ~ B(n, p), where n is the number of trials and p is the probability of success on each trial.

如果一个离散随机变量 X 统计的是在固定次数的独立试验中成功的次数,且每次试验的成功概率相同,则 X 服从二项分布。记作 X ~ B(n, p),其中 n 是试验次数,p 是每次试验成功的概率。

For a situation to be modelled by a binomial distribution, all four of the following conditions must hold:

要用二项分布对某一情境建模,必须同时满足以下四个条件:

In Exercise 6D problems, you must first check these conditions before applying any formula. If a problem involves sampling without replacement from a small population, the binomial distribution is not appropriate.

在 Exercise 6D 的题目中,你必须先检验这些条件,然后才能套用公式。如果问题涉及从不大的总体中无放回抽样,则不宜使用二项分布。


2. The Binomial Probability Formula | 二项概率公式

For X ~ B(n, p), the probability of exactly r successes is given by:

若 X ~ B(n, p),则恰好 r 次成功的概率为:

P(X = r) = ⁿCᵣ × pʳ × (1 − p)ⁿ⁻ʳ

Here ⁿCᵣ is the binomial coefficient, often written as C(n, r) or ⁿCᵣ, and it counts the number of ways to choose r successes from n trials. You may compute it using your calculator's nCr function.

其中 ⁿCᵣ 是二项系数,也常写作 C(n, r),它表示从 n 次试验中选出 r 次成功的方式数。你可以用计算器上的 nCr 功能直接计算。

For example, if X ~ B(8, 0.35), then the probability of exactly 3 successes is:

例如,若 X ~ B(8, 0.35),则恰好 3 次成功的概率为:

P(X = 3) = ⁸C₃ × 0.35³ × 0.65⁵ = 56 × 0.042875 × 0.116029 ≈ 0.2786

When solving Exercise 6D problems, always write out the formula first and substitute the values clearly. This not only earns method marks but also helps you catch arithmetic errors.

在解答 Exercise 6D 题目时,务必先写出公式,再清楚代入数值。这样既能获得方法分,也有助于发现计算错误。


3. Cumulative Probabilities | 累积概率

re cumulative probabilities, which you can obtain from your calculator or from cumulative binomial tables.

许多题目会要求计算涉及不等式的概率,如 P(X ≤ r)、P(X ≥ r) 或 P(a ≤ X ≤ b)。这些需要使用累积概率,可通过计算器或累积二项分布表获得。

Your calculator has a function usually labelled 'Binomial CD' or 'binomCDF'. You input the values of n, p and the upper (and lower) limits, and it returns the cumulative probability directly.

你的计算器上通常有一个标记为 'Binomial CD' 或 'binomCDF' 的功能键。输入 n、p 以及上下限后,它会直接返回累积概率。

look up P(X ≤ r) for given values of n and p. To find other forms, apply these identities:

在考试中使用表格时,通常直接查表得到给定 n 和 p 下 P(X ≤ r) 的值。要计算其他形式,可以利用以下恒等式:

P(X ≥ r) = 1 − P(X ≤ r − 1)

P(a ≤ X ≤ b) = P(X ≤ b) − P(X ≤ a − 1)

These identities are essential because most tables and calculator functions give cumulative probabilities from zero upward only.

这些恒等式非常重要,因为大多数表格和计算器功能只给出从零开始的累积概率。


4. Choosing the Correct Parameters | 正确选取参数

The most common mistake in Exercise 6D is misidentifying n or p. Always ask yourself:

Exercise 6D 中最常见的错误是误判 n 或 p 的取值。请时刻问自己:

For instance, if a production line produces components with a 2% defect rate, and we inspect 20 components, then X ~ B(20, 0.02), where 'success' is defined as finding a defective component. Notice that 'success' does not always mean something good — it simply means the outcome of interest.

例如,若一条生产线的次品率为 2%,我们抽查 20 个零件,那么 X ~ B(20, 0.02),此处 '成功' 定义为发现次品。注意这里的 '成功' 不一定指好事——它仅指我们关心的结果发生。

given directly. You may need to deduce it from the context, such as when a question states 'one in six people prefer brand A' — then p = 1/6.

有时概率 p 不会直接给出。你需要从情境中推断,例如当题目说 '六个人中有一人偏好品牌 A' 时,p = 1/6。

Always convert percentages to decimals before substituting into the formula. For a 2% defect rate, use p = 0.02, never 2.

在代入公式前,务必将百分数转换为小数。例如 2% 的次品率应使用 p = 0.02,绝不能写成 2。


5. Worked Example: Quality Control | 例题一:质量控制

A factory produces electronic chips, and 3% of all chips are defective. A quality inspector randomly selects 12 chips from the production line. Let X be the number of defective chips in the sample.

某工厂生产电子芯片,全部芯片中有 3% 是次品。质检员从生产线随机抽取 12 枚芯片。设 X 为样本中的次品数。

(a) State the distribution of X.

(a) 写出 X 的分布。

Solution: There are 12 independent trials, each with a constant probability of 'success' (defective) equal to 0.03. Hence X ~ B(12, 0.03).

解答:共有 12 次独立试验,每次 '成功'(次品)的概率恒为 0.03。因此 X ~ B(12, 0.03)。

(b) Find the probability that exactly one chip is defective.

(b) 求恰好一枚芯片是次品的概率。

P(X = 1) = ¹²C₁ × 0.03¹ × 0.97¹¹ = 12 × 0.03 × 0.7153 ≈ 0.2575

(c) Find the probability that at most two chips are defective.

(c) 求至多两枚芯片是次品的概率。

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

stribution function on your calculator with n = 12, p = 0.03 and upper limit 2, we obtain:

使用计算器上的二项分布功能,输入 n = 12、p = 0.03、上限 2,得到:

P(X ≤ 2) ≈ 0.9926

This means there is a 99.26% chance that no more than two chips in the sample are defective.

这意味着样本中次品数不超过两枚的概率为 99.26%。


6. Worked Example: Examination Questions | 例题二:考试中的选择题

A multiple-choice test consists of 10 questions, each with 4 options. A student guesses every answer at random. Let X be the number of correct answers.

一份选择题试卷包含 10 道题目,每题有 4 个选项。一名学生完全随机猜测所有答案。设 X 为答对的题数。

(a) Find the distribution of X.

(a) 求 X 的分布。

rrect guess is p = 1/4 = 0.25. Since the student guesses independently on each question, X ~ B(10, 0.25).

解答:每道题是一次试验,n = 10。猜对的概率为 p = 1/4 = 0.25。由于每道题独立猜测,所以 X ~ B(10, 0.25)。

(b) Find the probability that the student passes, meaning at least 5 correct answers.

(b) 求该学生及格的概率,即至少答对 5 题。

We need P(X ≥ 5). Using the complement identity:

我们需要求 P(X ≥ 5)。利用补事件公式:

P(X ≥ 5) = 1 − P(X ≤ 4)

From the calculator, P(X ≤ 4) ≈ 0.9219 for B(10, 0.25). Therefore:

由计算器得 B(10, 0.25) 的 P(X ≤ 4) ≈ 0.9219。因此:

P(X ≥ 5) = 1 − 0.9219 = 0.0781

The student has only a 7.81% chance of passing by guessing alone. This illustrates how the binomial distribution quantifies intuitive ideas about chance.

该学生仅凭猜测及格的概率只有 7.81%。这说明二项分布能够将我们对概率的直觉加以精确量化。


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7. Working with 'At Least' and 'More Than' | 处理 '至少' 与 '超过'

Exam questions frequently use English phrases that map onto specific inequalities. You must translate these correctly into mathematical symbols:

考试题目经常使用某些英文短语,它们对应特定的不等式。你必须正确地将它们转换为数学符号:

Phrase | 短语 Inequality | 不等式 Calculator Input | 计算器输入
At most r | 至多 r 次 P(X ≤ r) Lower = 0, Upper = r
Fewer than r | 少于 r 次 P(X ≤ r − 1) Lower = 0, Upper = r − 1
At least r | 至少 r 次 P(X ≥ r) = 1 − P(X ≤ r − 1) Lower = r, Upper = n (or complement)
More than r | 超过 r 次 P(X > r) = 1 − P(X ≤ r) Lower = r + 1, Upper = n (or complement)
Between a and b inclusive | 从 a 到 b(含) P(a ≤ X ≤ b) Lower = a, Upper = b

Pay particular attention to whether the endpoints are included. 'At least 3' includes 3, while 'more than 3' excludes 3. This distinction changes the upper or lower limit by one.

要特别注意端点是否包含在内。'至少 3' 包含 3,而 '超过 3' 不包含 3。这一区别会使上界或下界相差 1。


8. Common Misconceptions | 常见误区

Students often confuse the binomial distribution with other distributions or misuse its assumptions. Here are the most frequent pitfalls:

学生经常将二项分布与其他分布混淆,或误用其假设条件。以下是最常见的陷阱:

In AQA exam mark schemes, recognising that a scenario is not binomial (e.g., sampling without replacement) can still earn you a mark if you justify your answer correctly.

在 AQA 的评分标准中,识别出某个情境不属于二项分布(例如无放回抽样)并给出正确理由,同样可以获得分数。


9. Using Expected Value and Variance | 期望与方差的应用

For a binomial distribution X ~ B(n, p), the expected value (mean) and variance are:

对于二项分布 X ~ B(n, p),其期望(均值)和方差分别为:

E(X) = np

Var(X) = np(1 − p)

ected number of heads is E(X) = 50 × 0.4 = 20, with variance Var(X) = 50 × 0.4 × 0.6 = 12.

这些公式在 Exercise 6D 中非常有用。例如,一枚偏倚硬币出现正面的概率为 0.4,若抛掷 50 次,则出现正面的期望次数为 E(X) = 50 × 0.4 = 20,方差为 Var(X) = 50 × 0.4 × 0.6 = 12。

deviation is the square root of the variance: σ = √[np(1 − p)]. In the example above, σ = √12 ≈ 3.464.

标准差是方差的开平方:σ = √[np(1 − p)]。上例中 σ = √12 ≈ 3.464。

Some problems ask you to estimate a probability from the expected value. For instance, 'estimate the number of defective items in a batch of 500' simply requires you to compute np.

有些题目会要求你利用期望值估算概率。例如,'估算 500 件产品中的次品数量' 只需计算 np 即可。


10. Exam-Style Questions | 考试风格练习

The following questions mirror the style of AQA examination questions on Exercise 6D. Try them before reading the solutions.

以下题目模拟 AQA 考试中 Exercise 6D 的题型。请先尝试解答,再对照答案。

Question 1. A biased die is rolled 20 times. The probability of rolling a 6 is 0.15. Let Y be the number of sixes obtained.

题目 1. 一枚偏倚骰子掷 20 次。掷出 6 点的概率为 0.15。设 Y 为掷出 6 点的次数。

(a) Write down the distribution of Y. | (a) 写出 Y 的分布。

(b) Find P(Y = 3). | (b) 求 P(Y = 3)。

(c) Find P(Y ≥ 2). | (c) 求 P(Y ≥ 2)。

Solution. (a) Y ~ B(20, 0.15).

解答。 (a) Y ~ B(20, 0.15)。

(b) P(Y = 3) = ²⁰C₃ × 0.15³ × 0.85¹⁷ ≈ 1140 × 0.003375 × 0.06311 ≈ 0.2428.

(c) P(Y ≥ 2) = 1 − P(Y ≤ 1) = 1 − [P(Y = 0) + P(Y = 1)].

P(Y = 0) = 0.85²⁰ ≈ 0.0388

P(Y = 1) = 20 × 0.15 × 0.85¹⁹ ≈ 0.1368

Therefore P(Y ≥ 2) = 1 − (0.0388 + 0.1368) = 1 − 0.1756 = 0.8244.

因此 P(Y ≥ 2) = 1 − (0.0388 + 0.1368) = 1 − 0.1756 = 0.8244。

Question 2. A medicine is effective for 80% of patients. A doctor treats 8 patients with this medicine. Find the probability that exactly 6 patients recover.

题目 2. 某药物对 80% 的患者有效。一名医生用该药治疗 8 名患者。求恰好 6 名患者康复的概率。

Solution. Let R ~ B(8, 0.8). Then:

解答。 设 R ~ B(8, 0.8),则:

P(R = 6) = ⁸C₆ × 0.8⁶ × 0.2² = 28 × 0.262144 × 0.04 ≈ 0.2936

You could also model the number of patients who do not recover using S ~ 0.2). Then P(R = 6) = P(S = 2), which gives the same answer. Choosing the more convenient 'success' can simplify calculations.

你也可以用未康复人数 S ~ B(8, 0.2) 来建模。此时 P(R = 6) = P(S = 2),结果相同。选择更方便的 '成功' 定义可以简化计算。


11. Tips for Full Marks | 满分技巧

To secure all available marks in binomial distribution questions, follow these examiner-approved steps:

要在二项分布题目中获得满分,请遵循以下考官认可的步骤:

In AQA examinations, you may use your calculator's statistical functions freely, but you must still write down the distribution and the probability statement you are evaluating.

在 AQA 考试中,你可以自由使用计算器的统计功能,但仍必须写出分布和所计算的概率表达式。


12. Practice Problems | 巩固练习

Test your understanding with these additional problems. Solutions are provided in condensed form.

请通过以下额外练习检验你的理解。答案以简略形式给出。