Probability · 概率
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| probability/ˌprɒbəˈbɪlɪti/ | 概率 | gài lǜ |
| mutually exclusive/ˈmjuːtʃuːəli eksˈkluːsɪv/ | 互斥 | hù chì |
| independent/ˌɪndɪˈpendənt/ | 独立 | dú lì |
| complement/ˈkɒmplɪmənt/ | 对立事件 | duì lì shì jiàn |
| permutation/ˌpɜːmjuːˈteɪʃn/ | 排列 | pái liè |
| combination/ˌkɒmbɪˈneɪʃn/ | 组合 | zǔ hé |
Changing the denominator changes the conditional question
- In a class of 20, 10 use a bus. Four bus users and two non-bus users are late.
- P(late | bus)=4/10, while P(bus | late)=4/6. The same overlap count has a different denominator because the given group changes.
Count a defined outcome space
- Probability 概率 lies between 0 and 1. For equally likely outcomes it is favourable count divided by total count.
- Define what counts as one outcome. An ordered sequence of draws is different from an unordered set of selected people.
Separate independence from mutual exclusion
- Mutually exclusive 互斥 events cannot both happen, so their union adds probabilities. In general subtract the overlap: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$.
- Independent 独立 events satisfy $P(A\cap B)=P(A)P(B)$. More generally use $P(A\cap B)=P(A)P(B\mid A)$; condition on the information actually given.
Independent events, combined · 独立事件的组合
See why "and" multiplies while "or" adds — and when each rule applies. · 理解为何“和”相乘、“或”相加，以及各自适用情形。
Use a complement when it simplifies the event
- The complement 对立事件 has probability $P(A^c)=1-P(A)$. For at least one red draw, its complement is no red draws.
- Without replacement, the remaining composition depends on earlier draws. A tree can keep each conditional branch denominator and numerator visible.
A question asks for P(at least one). What is usually the fastest route? · 题目要求 P(至少一次)。通常最快的路径是什么？
There is usually one way to get none and many ways to get at least one. · 通常只有一种全不的情况，却有多种至少一次的情况。
At least one red, without replacement. A bag contains four red and six blue counters. For two draws, $P(BB)=(6/10)(5/9)=1/3$. Therefore $P(\text{at least one red})=1-P(BB)=2/3$. The probability changes on the second draw because a counter has been removed.
4 red and 6 blue balls, two drawn without replacement. What is P(at least one red)? · 4 红球 6 蓝球，不放回抽取两个。P(至少一红) 是多少？
P(no red) = 6/10 × 5/9 = 1/3, so the complement is 2/3. · P(无红) = 6/10 × 5/9 = 1/3，故补集为 2/3。
Drawing without replacement leaves the two draws independent. · 不放回抽取使两次抽取相互独立。
The first draw changes what is left, so the second probability depends on it. · 第一次抽取改变了剩余内容，因此第二次概率依赖于前者。
Count ordered and unordered choices differently
- A permutation 排列 counts ordered selections. A podium of first, second and third has roles that make order matter.
- A combination 组合 counts unordered selections. A three-person committee with no assigned roles is unchanged when its members are listed in another order.
Choosing 3 students for a committee from 10. Permutation or combination? · 从 3 名学生中选出委员会成员， 10. 排列还是组合？
The order of a committee does not change who is on it, so order does not matter. · 委员会成员顺序不影响实际人选，故顺序无关紧要。
Conditional probabilities have different denominators. The fraction of bus users who are late differs from the fraction of late students who use the bus. Sheet 2.4 gives the whole table so each conditional group can be counted.
Carry the reasoning to a new case
- Compare P(late | bus)=0.4 with P(late)=6/20=0.3; the events are not independent in this table.
- The association does not prove the bus caused lateness.
A calculated probability of 1.2 means the event is very likely. · 计算出的概率为 1.2，意味着事件极有可能发生。
It means the calculation is wrong. Probabilities never exceed 1 — usually you added instead of multiplying. · 这意味着计算有误。概率不可能超过 1——通常是误加而非误乘。
Using the class table, calculate P(bus | late). Give a decimal to three places.
Among the six late students, four are bus users: 4/6≈0.667.
Match each given condition to the correct denominator.
The denominator counts the group conditioned on.
A probability outside 0 to 1 is invalid. A result inside that interval can still be wrong: check the outcome definition, overlap and replacement rule. An association in one class table does not establish a causal transport effect.