Example 6.6.2 – Binomial Probability Calculations | 例6.6.2 – 二项分布概率计算

📚 Example 6.6.2 – Binomial Probability Calculations | 例6.6.2 – 二项分布概率计算

This worked example comes from the AQA A-Level Mathematics specification, Section 6.6, focusing on calculating probabilities from a binomial distribution. The problem tests your ability to apply the binomial probability formula, use cumulative tables, and interpret the conditions for a binomial model.

本例题选自 AQA A-Level 数学考纲第 6.6 节，重点训练如何从二项分布中计算概率。该问题考查你运用二项概率公式、使用累积概率表以及判断二项模型适用条件的能力。

1. Problem Statement | 题目描述

A random variable X is such that X ~ B(10, 0.3). Calculate: (a) P(X = 3), (b) P(X ≤ 2), (c) P(X ≥ 6).

设随机变量 X 服从二项分布 X ~ B(10, 0.3)。计算：(a) P(X = 3)，(b) P(X ≤ 2)，(c) P(X ≥ 6)。

This is Example 6.6.2 from the AQA textbook. The notation B(10, 0.3) tells us there are 10 independent trials, and the probability of success in each trial is 0.3.

这是 AQA 教材中的例 6.6.2。记号 B(10, 0.3) 表示共有 10 次独立试验，每次试验成功的概率为 0.3。

2. Conditions for the Binomial Model | 二项模型的条件

Before using any formula, we must verify that X actually follows a binomial distribution. The four conditions are:

在使用任何公式之前，我们必须核实 X 确实服从二项分布。四个条件如下：

There are a fixed number of trials, n = 10. | 试验次数固定，n = 10。
Each trial has exactly two outcomes: success or failure. | 每次试验只有两种结果：成功或失败。
The trials are independent of each other. | 各次试验相互独立。
The probability of success p = 0.3 is constant for every trial. | 每次试验的成功概率 p = 0.3 保持不变。

Since all four conditions hold in this example, we may proceed with the binomial probability formula.

由于本例中四个条件均满足，我们可以继续使用二项概率公式。

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

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

对于 X ~ B(n, p)，恰好 r 次成功的概率为：

P(X = r) = ⁿCᵣ × pʳ × (1 − p)ⁿ⁻ʳ

where ⁿCᵣ (also written as C(n, r) or “n choose r”) is the binomial coefficient, and (1 − p) is the probability of failure. In this question, n = 10, p = 0.3, and q = 1 − p = 0.7.

其中 ⁿCᵣ（也记作 C(n, r) 或“组合数”）是二项系数，(1 − p) 是失败概率。本题中 n = 10，p = 0.3，q = 1 − p = 0.7。

The binomial coefficient can be calculated as:

二项系数的计算方法为：

ⁿCᵣ = n! / (r! × (n − r)!)

For example, ¹⁰C₃ = 10! / (3! × 7!) = 120. Your calculator may have a dedicated nCr button; this is the quickest and most reliable method in an exam.

例如，¹⁰C₃ = 10! / (3! × 7!) = 120。计算器上通常有专用的 nCr 按键；这是考试中最快捷、最可靠的方法。

4. Part (a): Calculating P(X = 3) | 第 (a) 部分：计算 P(X = 3)

We substitute r = 3 into the formula:

我们将 r = 3 代入公式：

P(X = 3) = ¹⁰C₃ × 0.3³ × 0.7⁷

Let us compute the three components separately. First, ¹⁰C₃ = 120. Second, 0.3³ = 0.027. Third, 0.7⁷ ≈ 0.0823543. Multiplying these together gives:

我们分别计算三个部分。首先，¹⁰C₃ = 120。其次，0.3³ = 0.027。第三，0.7⁷ ≈ 0.0823543。将它们相乘得到：

P(X = 3) = 120 × 0.027 × 0.0823543 ≈ 0.2668

So P(X = 3) ≈ 0.2668, which rounds to 0.267 to three significant figures. This means that in a sample of 10 trials with success probability 0.3, we expect exactly 3 successes about 26.7% of the time.

因此 P(X = 3) ≈ 0.2668，保留三位有效数字为 0.267。这意味着在 10 次试验、成功概率为 0.3 的情况下，恰好出现 3 次成功的概率约为 26.7%。

5. Part (b): Understanding P(X ≤ 2) | 第 (b) 部分：理解 P(X ≤ 2)

The notation P(X ≤ 2) means the probability that X takes a value of 0, 1, or 2. Since these are mutually exclusive events, we add their individual probabilities:

记号 P(X ≤ 2) 表示 X 取值为 0、1 或 2 的概率。由于这些事件互斥，我们将各自的概率相加：

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

This is a cumulative probability. We could compute each term separately, but it is far more efficient to use the cumulative distribution function, often listed in statistical tables as P(X ≤ x).

这是一个累积概率。我们可以分别计算每一项，但更高效的方法是使用累积分布函数，统计表中通常以 P(X ≤ x) 的形式列出。

6. Part (b): Step-by-Step Calculation | 第 (b) 部分：分步计算

Let us compute each term individually to illustrate the process.

我们逐项计算以展示过程。

For r = 0:

对于 r = 0：

P(X = 0) = ¹⁰C₀ × 0.3⁰ × 0.7¹⁰ = 1 × 1 × 0.0282475 ≈ 0.0282

For r = 1:

对于 r = 1：

P(X = 1) = ¹⁰C₁ × 0.3¹ × 0.7⁹ = 10 × 0.3 × 0.0403536 ≈ 0.1211

For r = 2:

对于 r = 2：

P(X = 2) = ¹⁰C₂ × 0.3² × 0.7⁸ = 45 × 0.09 × 0.057648 ≈ 0.2335

Adding these together:

将它们相加：

P(X ≤ 2) = 0.0282 + 0.1211 + 0.2335 ≈ 0.3828

Therefore, P(X ≤ 2) ≈ 0.383 to three significant figures.

因此，P(X ≤ 2) ≈ 0.383，保留三位有效数字。

7. Using Cumulative Tables | 使用累积概率表

In the exam, you would normally read P(X ≤ 2) directly from a binomial cumulative table for n = 10 and p = 0.3. The table would show the value 0.3828 directly under the column p = 0.30 and the row x = 2.

在考试中，你通常可以直接从 n = 10、p = 0.3 的二项累积概率表中读取 P(X ≤ 2)。表格中，p = 0.30 列与 x = 2 行交叉处的值即为 0.3828。

x	p = 0.25	p = 0.30	p = 0.35
1	0.2440	0.1493	0.0860
2	0.5256	0.3828	0.2616
3	0.7759	0.6496	0.5138

Notice that the table gives cumulative probabilities starting from P(X = 0) up to the listed x-value. Always check whether your table gives P(X ≤ x) or P(X = x).

注意，表格给出的是从 P(X = 0) 到所列 x 值的累积概率。务必检查你的表格给出的是 P(X ≤ x) 还是 P(X = x)。

8. Part (c): Using the Complement | 第 (c) 部分：利用补集

Part (c) asks for P(X ≥ 6). The sum of all probabilities for a binomial distribution is 1, so we can rewrite this as:

第 (c) 部分要求计算 P(X ≥ 6)。二项分布所有概率之和为 1，因此我们可以改写为：

P(X ≥ 6) = 1 − P(X ≤ 5)

This is much easier to compute because P(X ≤ 5) is a direct table lookup, while P(X ≥ 6) is not directly tabulated. The complement rule avoids summing six separate probabilities.

这样计算容易得多，因为 P(X ≤ 5) 可以直接查表，而 P(X ≥ 6) 不能直接查处。补集规则避免了计算六项概率之和。

9. Part (c): Final Calculation | 第 (c) 部分：最终计算

From the cumulative table for n = 10 and p = 0.3, we find P(X ≤ 5) = 0.9527. Substituting into the complement formula:

从 n = 10、p = 0.3 的累积表中查得 P(X ≤ 5) = 0.9527。代入补集公式：

P(X ≥ 6) = 1 − 0.9527 = 0.0473

So P(X ≥ 6) ≈ 0.0473, or about 4.73%. This relatively small probability reflects that 6 or more successes is unusual when the success probability is only 0.3.

因此 P(X ≥ 6) ≈ 0.0473，约为 4.73%。这个概率较小，反映出当成功概率仅为 0.3 时，出现 6 次或更多次成功是较为罕见的事件。

As a sanity check, we could compute P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) directly and verify it equals 0.0473.

作为验算，我们可以直接计算 P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)，验证其和等于 0.0473。

10. Common Errors | 常见错误

Confusing X ≥ 6 with X > 6: Since X is discrete, P(X ≥ 6) = P(X > 5), not P(X > 6). | 混淆 X ≥ 6 与 X > 6： 由于 X 是离散变量，P(X ≥ 6) = P(X > 5)，而非 P(X > 6)。
Using the wrong complement: Some students incorrectly write P(X ≥ 6) = 1 − P(X ≤ 6). This would exclude exactly 6 from the desired region. | 用错补集： 有些学生错误地写成 P(X ≥ 6) = 1 − P(X ≤ 6)。这会把恰好等于 6 的情况从所求区域中排除。
Forgetting the binomial coefficient: The term ⁿCᵣ is essential; omitting it underestimates the probability. | 遗漏二项系数： ⁿCᵣ 这一项必不可少；省略它会低估概率。
Misreading the table: Confirm whether your table lists cumulative probabilities from 0 or from the opposite tail. | 读错表格： 确认你的表格列出的累积概率是从 0 开始还是从另一端开始。

11. Calculator Method | 计算器方法

Modern scientific calculators allow you to compute binomial probabilities directly using the distribution menu. The commands typically look like:

现代科学计算器允许你通过分布菜单直接计算二项概率。常见命令格式如下：

For P(X = r): binompdf(n, p, r) or “Binomial PD”. | 对于 P(X = r)：使用 binompdf(n, p, r) 或“Binomial PD”。
For P(X ≤ r): binomcdf(n, p, r) or “Binomial CD”. | 对于 P(X ≤ r)：使用 binomcdf(n, p, r) 或“Binomial CD”。

To find P(X ≥ 6), you would first compute binomcdf(10, 0.3, 5) and then subtract the result from 1. This matches the analytical approach above.

要计算 P(X ≥ 6)，先计算 binomcdf(10, 0.3, 5)，然后用 1 减去该结果。这与上面的解析方法一致。

12. Graph Interpretation | 图形解释

The probability mass function of X ~ B(10, 0.3) is right-skewed, with its peak at the mode. For a binomial distribution, the mode is the largest integer r such that (n + 1)p ≥ r. Here (10 + 1) × 0.3 = 3.3, so the mode is at r = 3.

X ~ B(10, 0.3) 的概率质量函数是右偏的，峰值出现在众数处。对于二项分布，众数是满足 (n + 1)p ≥ r 的最大整数 r。本例中 (10 + 1) × 0.3 = 3.3，所以众数在 r = 3。

This explains why P(X = 3) was the largest single probability among all outcomes. The tail beyond X = 5 is relatively flat and small, consistent with P(X ≥ 6) ≈ 0.0473.

这解释了为什么 P(X = 3) 是所有结果中最大的单项概率。X = 5 之后的尾部相对平坦且很小，这与 P(X ≥ 6) ≈ 0.0473 相吻合。

A visual bar chart of the distribution would show that probabilities rise from X = 0 to a maximum at X = 3, then fall steadily toward X = 10. This shape is typical for moderate p values.

从该分布的条形图可以看出，概率从 X = 0 处逐步上升，在 X = 3 处达到最大值，然后向 X = 10 稳步下降。这种形态在中等 p 值时很典型。

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Example 6.6.2 – Binomial Probability Calculations | 例6.6.2 – 二项分布概率计算