📚 Exercise 20B.1: Binomial Distribution | 习题20B.1：二项分布
In IB Mathematics, the binomial distribution is a cornerstone of probability theory, allowing us to model the number of successes in a fixed number of independent trials. Exercise 20B.1 challenges you to apply this model to a classic scenario — perhaps tossing a fair coin multiple times or counting defective items in a batch. This article walks you through the essential concepts, the step-by-step solution to a typical Exercise 20B.1 problem, and the deeper insights you need to master the topic. By the end, you will confidently handle any binomial question on your IB exam.
在IB数学中，二项分布是概率论的基石，它帮助我们模拟在固定次数的独立试验中成功的次数。习题20B.1要求你将这一模型应用于经典场景——例如多次抛掷一枚公平硬币，或统计一批产品中的次品数量。本文将带你梳理核心概念，逐步解答一道典型的20B.1习题，并深入讲解掌握这一主题所需的洞见。读完本文，你将能自信地应对IB考试中任何二项分布问题。
1. Understanding the Binomial Setting | 理解二项分布的条件
A binomial experiment arises when four strict conditions are all satisfied: the process consists of a fixed number of trials, n; each trial is independent of the others; every trial results in only two possible outcomes — typically labelled 'success' and 'failure'; and the probability of success, p, remains constant from trial to trial. If any one of these is violated — for instance, if trials are not independent or p changes — the binomial model no longer applies. Recognising these conditions in a word problem is the first critical skill.
当一个随机试验同时满足四个严格条件时，它就服从二项分布：试验由固定次数 n 次试验组成；每次试验相互独立；每次试验只有两种可能结果——通常标记为“成功”和“失败”；且成功的概率 p 在每次试验中保持不变。如果其中任何一个条件被破坏——比如试验不独立，或者 p 发生变化——二项模型就不再适用。在文字题中识别这些条件，是第一项关键技能。
2. The Binomial Probability Formula | 二项概率公式
Once you confirm the binomial setting, the probability of obtaining exactly k successes in n trials is given by the probability mass function:
P(X = k) = ₙCₖ · pᵏ · qⁿ⁻ᵏ
where q = 1 − p is the probability of failure, and ₙCₖ (read as 'n choose k' the successes with the probability of a specific sequence of k successes and n−k failures.一旦确认了二项分布的条件，得到 n 次试验中恰好有 k 次成功的概率由概率质量函数给出：
P(X = k) = ₙCₖ · pᵏ · qⁿ⁻ᵏ
其中 q = 1 − p 是失败的概率，ₙCₖ（读作“从 n 中选 k”）计数了从 n 次试验中选择 k 次成功位置的方式数目。这个公式将成功位置的选择组合数与一个特定序列（k 次成功和 n−k 次失败）的乘机概率结合在了一起。3. Identifying n, p, and q in Problems | 在问题中识别 n, p, q
om the question statement is half the battle. Look for the total number of repetitions (n) — e.g. '10 tosses', '8 attempts' 5). The failure probability q is simply 1 − p. In Exercise 20B.1, a typical problem reads: 'A fair coin is tossed 8 times. Find the probability of getting exactly 5 heads.' Here, n = 8, success = heads, p = 0.5, and q = 0.5.
从题目叙述中提取正确的参数是成功的一半。找到总的重复次数（n）——例如“抛掷10次”、“尝试8次”。明确题目将什么定义为成功，并写出其概率（p）。通常 p 直接以小数或百分比形式给出，或者必须根据对称性推导（例如，公平硬币意味着 p = 0.5）。失败概率 q 即 1 − p。在习题20B.1中，一道典型的题目是：“抛掷一枚公平硬币8次。求恰好得到5次正面的概率。”这里，n = 8，成功 = 正面，p = 0.5，q = 0.5。
4. Calculating Combinations | 计算组合数
The term ₙCₖ, also written as C(n, k), is fundamental. It is computed using factorials:
ₙCₖ = n! / (k!(n−k)!)
For our example, ₈C₅ = 8!/(5!3! Cr function on your GDC. Getting comfortable with both methods ensures accuracy and speed.ₙCₖ 项，也写作 C(n, k)，是基础。它通过阶乘计算：
ₙCₖ = n! / (k!(n−k)!)
以我们的例子，₈C₅ = 8!/(5!3!) = (8×7×6)/(3×2×1) = 56。当 n 较小时手动计算是可行的，但在考试中你也可以使用图形计算器上的 nCr 功能。熟悉两种方法可以确保准确性和速度。5. Computing Exact Probabilities | 计算精确概率
7/32 ≈ 0.21875. Thus, the probability of exactly five heads in eight tosses is about 21.9%. Leaving the answer as an exact fraction or a rounded decimal is both acceptable — check your IB mark scheme for preferred forms.
将数字代入习题20B.1的公式：P(X = 5) = ₈C₅ × (0.5)⁵ × (0.5)³。由于 (0.5)⁵ × (0.5)³ = (0.5)⁸ = 1/256，计算变为 56 × 1/256 = 56/256 = 7/32 ≈ 0.21875。因此，8次抛掷中恰好出现5次正面的概率约为21.9%。答案可以保留精确分数或四舍五入后的小数——具体形式请参照IB的评分标准。
6. Cumulative Probabilities | 累积概率
Often, IB questions ask for probabilities such as 'at most k successes', 'less than k', or 'at least k' e k'. These require summing individual binomial probabilities. For example, P(X ≤ 5) = P(0) + P(1) + … + P(5). While you can add them manually, a GDC's binomcdf function greatly simplifies this task. The phrasing 'at least 5' means P(X ≥ 5) = 1 − P(X ≤ 4), a common trick to reduce computation.
IB题目经常要求计算诸如“最多k次成功”、“少于k次”或“至少k次”的概率。这需要对单个二项概率求和。例如，P(X ≤ 5) = P(0) + P(1) + … + P(5)。虽然可以手动相加，但图形计算器的 binomcdf 功能大大简化了这一任务。表述“至少5次”意味着 P(X ≥ 5) = 1 − P(X ≤ 4)，这是一个常用的减少计算量的技巧。
7. Using a Graphics Display Calculator (GDC) | 使用图形计算器
nomcdf(n, p, 4). Always double-check that you have selected the correct function and entered n and p in the right order. Writing down the GDC command on your exam paper, followed by the result, demonstrates clear method.
IB试卷的设计考虑了图形计算器的使用。binompdf(n, p, k) 可以得到 P(X = k)；binomcdf(n, p, k) 得到 P(X ≤ k)。要求 P(X > 4) 时，输入 1 − binomcdf(n, p, 4)。务必反复检查你是否选择了正确的函数，并以正确的顺序输入了 n 和 p。在试卷上写下GDC命令，然后写出结果，可以清晰地展示解题方法。
8. Mean and Variance of a Binomial Distribution | 二项分布的期望和方差
viation ≈ 1.41). These formulas often appear in IB Paper 2 questions requiring interpretation in context.
二项随机变量 X 有着简洁的概括：其期望值（均值）为 μ = np，方差为 σ² = npq。以抛硬币为例，μ = 8 × 0.5 = 4，σ² = 8 × 0.5 × 0.5 = 2。这告诉我们，平均来看我们期望得到4次正面，而分布的离散程度由方差2（标准差约为1.41）衡量。这些公式经常出现在IB试卷二的题目中，要求结合情境进行解释。
9. Worked Solution for Exercise 20B.1 | 习题20B.1的解答过程
Let us solve the stated problem in full: A fair coin is tossed 8 times. Find the probability of obtaining exactly 5 heads.
Step 1 – omial conditions: fixed n=8, independent tosses, two outcomes (head/tail), constant p=0.5.
Step 2 – Identify parameters: n=8, p=0.5, q=0.5, k=5.
Step 3 – Compute the combination: ₈C₅ = 56.
Step 4 – Apply formula: P(X=5) = 56 × (0.5)⁵ × (0.5)³ = 56 × (0.5)⁸ = 56/256 = 7/32.
Step 5 – Decimal approximation: 0.21875 ≈ 0.219 (3 s.f.).
Step 6 – GDC check: binompdf(8,0.5,5) yields 0.21875.
Step 7 – Interpretation: There is about a 21.9% chance of getting exactly five heads in eight tosses of a fair coin.
让我们完整解答这道题目：抛掷一枚公平硬币8次。求恰好得到5次正面的概率。
步骤1 – 验证二项条件：固定 n=8，各次抛掷独立，两种结果（正面/反面），恒定 p=0.5。
步骤2 – 确定参数：n=8, p=0.5, q=0.5, k=5。
步骤3 – 计算组合数：₈C₅ = 56。
步骤4 – 应用公式：P(X=5) = 56 × (0.5)⁵ × (0.5)³ = 56 × (0.5)⁸ = 56/256 = 7/32。
步骤5 – 小数近似：0.21875 ≈ 0.219（三位有效数字）。
步骤6 – GDC核验：binompdf(8,0.5,5) 得到0.21875。
步骤7 – 解读：抛掷公平硬币8次恰好得到5次正面的概率约为21.9%。
10. Common Mistakes to Avoid | 常见错误
Beware of several traps. First, confusing exact probability with cumulative: if the question says 'exactly 5', do not use binomcdf. Second, misinterpreting 'at least' — students often forget to subtract from 1 or include the boundary incorrectly. Third, ignoring the binomial conditions; for example, sampling without replacement from a small population changes p and invalidates the model. Finally, rounding prematurely in multi-step calculations can lead to inaccurate final answers; work with fractions or store GDC values to full precision.
要警惕几个陷阱。第一，混淆精确概率与累积概率：如果题目说“恰好5次”，不要使用 binomcdf。第二，错误理解“至少”——学生经常忘记从1中减去，或错误地包含边界值。第三，忽略二项分布的条件；例如，从小总体中不放回抽样会改变 p，使模型失效。最后，在多步计算中过早四舍五入会导致最终答案不准确；应使用分数，或在GDC中存储完整精度的数值。
11. Real-life Applications of Binomial Distribution | 二项分布的实际应用
. In medicine, a doctor predicts how many patients out of 50 will respond positively to a treatment with a known success rate. In sports analytics, a basketball player' exts strengthens your ability to formulate and interpret binomial problems on the IB exam.
二项模型远远超出抛硬币的范畴。在质量控制中，制造商可能从生产线上抽取20件产品，记录次品数量（p = 次品率）。在医学中，医生预测50名患者中有多少人会对已知成功率的治疗产生积极反应。在体育分析中，篮球运动员在一场比赛中罚球出手次数也可以用二项分布建模。识别这些背景能增强你在IB考试中构建和解读二项分布问题的能力。
12. Summary and Key Takeaways | 总结与要点
= npq, which often accompany probability calculations. Practice with a variety of wordings — 'exactly', 'at most', 'more than', 'between' — ls, binomial distribution questions become predictable marks on your IB Mathematics paper.
掌握习题20B.1意味着你已经内化了二项分布的结构：检查条件，确定 n 和 p，应用公式 P(X=k) = ₙCₖ pᵏ qⁿ⁻ᵏ，并在处理累积情形时高效使用图形计算器。记住简洁的公式 E(X) = np 和 Var(X) = npq，它们经常伴随概率计算出现。针对各种措辞——“恰好”、“最多”、“超过”、“介于”进行练习，并始终将数值答案与情境联系起来。有了这些工具，二项分布题目将成为你IB数学试卷上可以稳定拿分的部分。
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