More Conditional Probability | 进阶条件概率

📚 More Conditional Probability | 进阶条件概率

In IGCSE mathematics, you already know that conditional probability measures the chance of an event occurring given that another event has already happened. The key phrase is ‘given that’, and the notation is \( \mathrm{P}(A|B) \).

在 IGCSE 数学中，你已经学到了条件概率度量的是：在另一个事件已经发生的前提下，某事件发生的可能性。关键短语是“已知（given that）”，记作 \( \mathrm{P}(A|B) \)。

1. The Conditional Probability Formula | 条件概率公式

The most important formula is \( \mathrm{P}(A|B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \). This works whenever \( \mathrm{P}(B) > 0 \). You can rearrange it to \( \mathrm{P}(A \cap B) = \mathrm{P}(A|B) \times \mathrm{P}(B) \).

最重要的公式是 \( \mathrm{P}(A|B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \)，该公式在 \( \mathrm{P}(B) > 0 \) 时都成立。你也可以将其变形为 \( \mathrm{P}(A \cap B) = \mathrm{P}(A|B) \times \mathrm{P}(B) \)。

\( \mathrm{P}(A|B) = \frac{\mathrm{P}(A \cap B)}{\mathrm{P}(B)} \)

Always identify the ‘given’ event first. That event goes in the denominator.

一定要先找出“已知”的事件，该事件作为分母。

2. Using Two‑Way Tables | 使用双向列联表

Two-way tables give frequencies for two categories. To find a conditional probability, use the column or row total that matches the ‘given’ condition.

双向列联表给出两个分类的频数。要计算条件概率，需使用与“已知”条件对应的列合计或行合计。

	Boys	Girls	Total
Plays chess	18	12	30
Does not play chess	7	13	20
Total	25	25	50

For example, \( \mathrm{P}(\text{plays chess} | \text{girl}) = \frac{12}{25} \). The denominator is the total number of girls, not the overall total.

例如，\( \mathrm{P}(\text{下棋} | \text{女生}) = \frac{12}{25} \)。分母是女生总人数，而不是全班总人数。

3. Venn Diagrams and Conditional Probability | 韦恩图与条件概率

In a Venn diagram, \( \mathrm{P}(A|B) \) is the probability that an element is in \( A \) given that it is already in \( B \). This is the fraction of region \( B \) that lies inside \( A \).

在韦恩图中，\( \mathrm{P}(A|B) \) 表示在已知元素属于 \( B \) 的条件下，它也属于 \( A \) 的概率。即区域 \( B \) 中同时位于 \( A \) 内的部分所占比例。

Suppose \( \mathrm{P}(A \cap B) = 0.2 \) and \( \mathrm{P}(B) = 0.5 \). Then \( \mathrm{P}(A|B) = \frac{0.2}{0.5} = 0.4 \).

设 \( \mathrm{P}(A \cap B) = 0.2 \)，\( \mathrm{P}(B) = 0.5 \)。则 \( \mathrm{P}(A|B) = \frac{0.2}{0.5} = 0.4 \)。

\( \mathrm{P}(A|B) = \frac{\text{overlap region}}{\text{region B}} \)

Be careful: \( \mathrm{P}(A|B) \) is usually not equal to \( \mathrm{P}(B|A) \). Compare 0.4 with \( \mathrm{P}(B|A) = \frac{0.2}{0.3} = 0.667 \) if \( \mathrm{P}(A) = 0.3 \).

注意：\( \mathrm{P}(A|B) \) 通常不等于 \( \mathrm{P}(B|A) \)。若 \( \mathrm{P}(A) = 0.3 \)，则 \( \mathrm{P}(B|A) = \frac{0.2}{0.3} = 0.667 \)，与 0.4 不同。

4. Tree Diagrams with Conditional Probabilities | 带条件概率的树形图

Tree diagrams are excellent for multi‑stage experiments. Each branch shows a conditional probability, because the second stage depends on what happened first.

树形图非常适合多步试验。每条分支代表一个条件概率，因为第二步的概率取决于第一步的结果。

Example: A bag contains 5 red and 3 blue balls. Two balls are drawn without replacement.

例：袋中有 5 个红球和 3 个蓝球，不放回地抽取两个球。

First draw: \( \mathrm{P}(R) = \frac{5}{8} \), \( \mathrm{P}(B) = \frac{3}{8} \)

Second draw given first was R: \( \mathrm{P}(R|R) = \frac{4}{7} \), \( \mathrm{P}(B|R) = \frac{3}{7} \)

To find \( \mathrm{P}(R \text{ then } B) \), multiply along the branches: \( \frac{5}{8} \times \frac{3}{7} = \frac{15}{56} \).

求 \( \mathrm{P}(\text{先红后蓝}) \) 时，沿分支相乘：\( \frac{5}{8} \times \frac{3}{7} = \frac{15}{56} \)。

5. Without Replacement — Dependence | 无放回抽取——事件相依

When objects are drawn without replacement, the probabilities on the second branch change because the total number of objects decreases.

当物体被无放回地抽取时，第二次抽取的概率会改变，因为物体总数减少了。

First draw changes the composition.
Second draw probabilities are conditional on the first result.
Third draw probabilities are conditional on both previous results.

第一次抽取改变了袋中的组成。
第二次抽取的概率依赖于第一次的结果。
第三次抽取的概率依赖于前两次的结果。

For example, if 2 red and 3 blue are drawn without replacement from a bag containing 5 red and 5 blue, the probability of drawing two reds is \( \frac{5}{10} \times \frac{4}{9} = \frac{2}{9} \).

例如，从装有 5 红 5 蓝的袋中无放回地抽 2 个球，两球都是红的概率为 \( \frac{5}{10} \times \frac{4}{9} = \frac{2}{9} \)。

6. Independent Events | 独立事件

Two events \( A \) and \( B \) are independent if knowing that \( B \) happens does not change the probability of \( A \). That is, \( \mathrm{P}(A|B) = \mathrm{P}(A) \).

如果知道 \( B \) 发生不会改变 \( A \) 发生的概率，则事件 \( A \) 与 \( B \) 独立。即 \( \mathrm{P}(A|B) = \mathrm{P}(A) \)。

For independent events, the multiplication rule becomes \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) \).

对于独立事件，乘法法则变为 \( \mathrm{P}(A \cap B) = \mathrm{P}(A) \times \mathrm{P}(B) \)。

\( A \text{ and } B \text{ independent } \Leftrightarrow \mathrm{P}(A|B) = \mathrm{P}(A) \)

Rolling a die and tossing a coin are independent: \( \mathrm{P}(\text{head} | \text{roll a 6}) = \frac{1}{2} = \mathrm{P}(\text{head}) \).

掷骰子和抛硬币是独立事件：\( \mathrm{P}(\text{正面} | \text{掷出6}) = \frac{1}{2} = \mathrm{P}(\text{正面}) \)。

7. ‘Given That’ Word Problems | 含“已知”的解题

Exam questions use phrases like ‘given that’, ‘if it is known that’, or ‘find the probability that … given …’. Identify the condition and use the formula.

考试题常使用“已知”“如果知道”“求在……条件下的概率”等表述。先确定条件，再套用公式。

Example: In a class, 40% of students wear glasses. 15% of students are both female and wear glasses. Find \( \mathrm{P}(\text{female} | \text{wears glasses}) \).

例：某班 40% 的学生戴眼镜，15% 的学生是女生且戴眼镜。求 \( \mathrm{P}(\text{女生} | \text{戴眼镜}) \)。

\( \mathrm{P}(F|G) = \frac{0.15}{0.40} = 0.375 \)

The condition ‘wears glasses’ goes in the denominator.

条件“戴眼镜”放在分母位置。

8. Combining with Set Notation | 结合集合记号

Conditional probability often appears together with union and intersection. Recall that \( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) – \mathrm{P}(A \cap B) \).

条件概率常与并集、交集记号一起出现。回顾公式 \( \mathrm{P}(A \cup B) = \mathrm{P}(A) + \mathrm{P}(B) – \mathrm{P}(A \cap B) \)。

Suppose \( \mathrm{P}(A) = 0.6 \), \( \mathrm{P}(B) = 0.5 \), \( \mathrm{P}(A \cup B) = 0.8 \). Then \( \mathrm{P}(A \cap B) = 0.6 + 0.5 – 0.8 = 0.3 \). So \( \mathrm{P}(A|B) = \frac{0.3}{0.5} = 0.6 \).

设 \( \mathrm{P}(A) = 0.6 \)，\( \mathrm{P}(B) = 0.5 \)，\( \mathrm{P}(A \cup B) = 0.8 \)。则 \( \mathrm{P}(A \cap B) = 0.6 + 0.5 – 0.8 = 0.3 \)。因此 \( \mathrm{P}(A|B) = \frac{0.3}{0.5} = 0.6 \)。

This shows \( A \) and \( B \) are independent here because 0.6 equals \( \mathrm{P}(A) \).

这里 \( A \) 与 \( B \) 是独立的，因为 0.6 等于 \( \mathrm{P}(A) \)。

9. Real‑Life Application | 实际应用

Conditional probability is used in medical testing, weather forecasting, sports analytics and quality control. Understanding it helps you interpret statistics correctly.

条件概率广泛应用于医学检测、天气预报、体育统计和质量控制。理解它有助于你正确解读统计数据。

For example, a test for a disease has a 95% true‑positive rate. If 1% of the population has the disease, the probability that a person is actually ill given a positive test is much lower than 95% because false positives also occur.

例如，某种疾病检测的“真阳性率”为 95%。如果人群患病率为 1%，那么即使检测结果为阳性，真正患病的概率也会远低于 95%，因为还存在假阳性。

Always think about the base rate — the overall proportion of the event in the population.

一定要考虑基础率，即该事件在整体人群中的比例。

10. Common Mistakes and Exam Tips | 常见错误与考试技巧

Students often confuse the order of events or use the wrong denominator. Check which event is ‘given’.

学生经常搞混事件顺序或用错分母。请检查哪个事件是“已知”的。

Do not assume \( \mathrm{P}(A|B) = \mathrm{P}(B|A) \).
For without replacement, always adjust the total and the favourable count.
Draw a tree diagram for multi‑step problems.
Write fractions in lowest terms where possible.
Read the question: ‘given that’ tells you the condition.

不要假设 \( \mathrm{P}(A|B) = \mathrm{P}(B|A) \)。
无放回时，记得调整总数和有利数量。
多步问题画树形图。
尽量把分数化为最简。
仔细读题：“已知”指明条件。

Always show your working clearly, especially the denominator. In Edexcel IGCSE, method marks are awarded for setting up the correct conditional probability ratio.

务必清楚展示计算步骤，尤其是分母。Edexcel IGCSE 考试中，正确写出条件概率的比例会获得方法分。

Published by TutorHao | Mathematics Revision Series | aleveler.com

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More Conditional Probability | 进阶条件概率