| English | 意味 |
|---|---|
| probability/ˌprɒbəˈbɪlɪti/ | 確率 |
| mutually exclusive/ˈmjuːtʃuːəli eksˈkluːsɪv/ | 排反 |
| independent/ˌɪndɪˈpendənt/ | 独立 |
| complement/ˈkɒmplɪmənt/ | 対立事象 |
| permutation/ˌpɜːmjuːˈteɪʃn/ | 順列 |
| combination/ˌkɒmbɪˈneɪʃn/ | 組合せ |
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
See why "and" multiplies while "or" adds — and when each rule applies.
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
There is usually one way to get none and many ways to get at least one.
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)?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
P(no red) = 6/10 × 5/9 = 1/3, so the complement is 2/3.
Drawing without replacement leaves the two draws independent.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The first draw changes what is left, so the second probability depends on it.
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The order of a committee does not change who is on it, so order does not matter.
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.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
It means the calculation is wrong. Probabilities never exceed 1 — usually you added instead of multiplying.
Using the class table, calculate P(bus | late). Give a decimal to three places.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Among the six late students, four are bus users: 4/6≈0.667.
Match each given condition to the correct denominator.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
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.