📚 Mastering Binomial Distribution: Exercise 18E.1 | 掌握二项分布：练习18E.1
Exercise 18E.1 in the IB Mathematics curriculum introduces students to calculating probabilities using the binomial distribution. This set of problems typically focuses on identifying binomial experiments, applying the exact probability formula, and interpreting results in context. Mastering these fundamentals is essential for success in both the analysis and application of statistical models.
在 IB 数学课程中，练习 18E.1 引导学生利用二项分布计算概率。这组题目通常侧重于识别二项试验、应用精确概率公式并在实际情境中解读结果。掌握这些基础知识对于统计模型的学习与应用至关重要。
1. Conditions for a Binomial Experiment | 二项试验的条件
Before applying any formula, you must verify that the scenario meets the four key conditions: a fixed number of trials (n), each trial is independent, there are only two possible outcomes (success/failure), and the probability of success (p) remains constant. In Exercise 18E.1, most problems explicitly state or imply these conditions. For instance, repeatedly tossing a coin or randomly selecting items with replacement ensures independence. If sampling is without replacement but the population is large enough (usually more than 10 times the sample size), the binomial model can still be used as a reasonable approximation.
在应用任何公式之前，你需要验证问题情境是否满足四个关键条件：试验次数固定 (n)、每次试验独立、每次试验只有两种可能结果（成功/失败），且成功的概率 (p) 在每次试验中保持不变。在练习 18E.1 中，大多数题目都会明确或暗示这些条件。例如，反复抛掷硬币或有放回地随机抽取物品，都能确保独立性。如果是无放回抽样，但总体足够大（通常超过样本量的 10 倍），仍然可以用二项分布作为合理的近似。
2. The Binomial Probability Formula | 二项概率公式
If a random variable X follows a binomial distribution, written as X ~ B(n, p), then the probability of obtaining exactly r successes is P(X = r) = ⁿCᵣ × pʳ × (1 – p)ⁿ⁻ʳ (also called the binomial coefficient, sometimes written as C(n, r) or nCr) is given by n! / (r!(n–r)!). It counts the number of combinations of n items taken r at a time. The factor pʳ is the probability of r independent successes, and (1 – p)ⁿ⁻ʳ is the probability of the remaining n– cise 18E.1, you will often evaluate this formula manually for small n, or use a calculator's nCr function to speed up the computation.
如果随机变量 X 服从二项分布，记作 X ~ B(n, p)，那么恰好得到 r 次成功的概率为 P(X = r) = ⁿCᵣ × pʳ × (1 – p)ⁿ⁻ʳ。其中，ⁿCᵣ（也称为二项系数，有时写作 C(n, r) 或 nCr）的计算公式为 n! / (r!(n–r)!)。它表示从 n 次试验中选择 r 次成功的组合数。pʳ 是 r 次独立成功的概率，而 (1 – p)ⁿ⁻ʳ 是其余 n–r 次失败的概率。在练习 18E.1 中，对于较小的 n，你通常需要手动计算该公式，或者使用计算器的 nCr 功能以加快运算。
3. Example 1: Coin Tosses (Exact Probability) | 例1：抛硬币（精确概率）
the probability of getting exactly 4 heads. Here we identify n = 10, p = 0.5, r = 4, so X ~ B(10, 0.5). Using the formula: P(X = 4) = ¹⁰C₄ × (0.5)⁴ × (0.5)⁶. First, compute ¹⁰C₄ = 10!/(4!6! = 4) = 210 × 1/1024 = 210/1024, which simplifies to 105/512, approximately 0.2051 or 20.51% probability of any single specific count is often lower than students intuitively expect.
一枚均匀硬币抛掷 10 次。求恰好出现 4 次正面的概率。这里 n = 10, p = 0.5, r = 4，因此 X ~ B(10, 0.5)。代入公式：P(X = 4) = ¹⁰C₄ × (0.5)⁴ × (0.5)⁶。首先计算 ¹⁰C₄ = 10!/(4!6!) = 210。接着，(0.5)⁴ × (0.5)⁶ = (0.5)¹⁰ = 1/1024。因此 P(X = 4) = 210 × 1/1024 = 210/1024，化简为 105/512，约等于 0.2051，即 20.51%。这个例子说明，即使是一枚完全均匀的硬币，任何一个特定正面次数出现的概率往往比直觉预期的要低。
4. Example 2: Defective Items in a Batch | 例2：一批产品中的次品
A factory produces components with a known defect rate of 5% If we randomly select 20 components, what is the probability that exactly 2 are defective? Let X be the number of defective items, so X ~ 5. (0.95)¹⁸ is approximately 0.3972. Multiplying: 190 × 0.0025 × 0.3972 ≈ 0.1887, or 18.87% ability, while modest, is the most likely individual outcome; yet it is still less than 19% Such insights are crucial in quality control and often appear in IB examination questions.
某工厂生产的产品次品率为 5%。如果随机抽取 20 个产品，求恰好有 2 个次品的概率。设 X 为次品件数，则 X ~ B(20, 0.05)，r = 2。代入公式：P(X = 2) = ²⁰C₂ × (0.05)² × (0.95)¹⁸。使用计算器，²⁰C₂ = 190，(0.05)² = 0.0025，(0.95)¹⁸ 约等于 0.3972。相乘得 190 × 0.0025 × 0.3972 ≈ 0.1887，即 18.87%。这个概率虽然不大，却是所有可能结果中概率最高的；但即便如此仍低于 19%。这类洞察在质量控制中非常关键，也是 IB 考试中常见的题型。
5. Using Your GDC for Binomial Probabilities | 使用图形计算器计算二项概率
e coin example. For cumulative probabilities, such as P(X ≤ 4), use binomcdf (or binomCdf) — the 'c' stands for cumulative — ith these functions, the mark scheme usually requires you to write down the distribution X ~ B(n, p) and the formula being used, so never skip the reasoning steps.
IB 考试期望你能高效地使用图形计算器 (GDC)。对于精确概率，通常使用 binompdf （或 binomPdf）函数，语法为 binompdf(n, p, r)。例如，binompdf(10, 0.5, 4) 返回的结果正与硬币例子中我们计算的一致。对于累积概率，比如 P(X ≤ 4)，则使用 binomcdf （或 binomCdf），其中 'c' 代表累积，语法为 binomcdf(n, p, 4)。虽然练习 18E.1 可以借助这些功能快速完成，但评分标准通常要求你写出分布 X ~ B(n, p) 以及所使用的公式，因此切勿省略推理步骤。
6. Cumulative Probabilities: 'At Most' and 'At Least' | 累积概率：“最多”和“至少”
Exercise 18E.1 frequently extends to questions involving phrases like 'at most 3 successes' or 'at least 5 successes'. For 'at most r' you compute P(X ≤ r) directly using binomcdf(n, p, r). For 'at least r' you use the complement rule: P(X ≥ r) = 1 – P(X ≤ r – 1). For example, in 10 coin tosses, the probability of at least 8 heads is P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – his technique saves time by avoiding the manual addition of many individual probabilities.
练习 18E.1 经常会扩展到包含“最多 3 次成功”或“至少 5 次成功”这类表述的题目。对于“最多 r 次”，直接使用 binomcdf(n, p, r) 计算 P(X ≤ r)。对于“至少 r 次”，利用补集规则：P(X ≥ r) = 1 – P(X ≤ r – 1)。例如，在 10 次抛硬币中，至少 8 次正面的概率为 P(X ≥ 8) = 1 – P(X ≤ 7) = 1 – binomcdf(10, 0.5, 7)。这种方法避免了手动加总大量单个概率，能节省不少时间。
Be careful with inclusive and exclusive boundaries: 'more than r' is P(X > r) = 1 – P(X ≤ r). 'Fewer than r' is P(X < r) = P(X ≤ r – re-read the wording and, if in doubt, sketch a quick number line to confirm the interval.
务必注意包含与不包含边界的情况：“大于 r 次” 对应 P(X > r) = 1 – P(X ≤ r)。“少于 r 次” 对应 P(X < r) = P(X ≤ r – 1)。仔细审题，如有疑问，可以快速画一条数轴来确认区间。
7. Expected Value and Variance | 期望值和方差
For a binomial random variable X ~ number of successes (the mean) is simply E(X) = n × p. The variance is Var(X) = n × p × (1 – p) = n p q, where q = 1 – p. The standard deviation is σ = √n p q. These summary statistics often appear in Exercise 18E.1 as a follow‑up to probability calculations. For example, if a large batch of 200 components has a 5% nderstanding expectation and variance lays the groundwork for normal approximations and hypothesis testing later in the syllabus.
对于二项随机变量
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