| English | 意味 |
|---|---|
| independent/ˌɪndɪˈpendənt/ | 独立 |
| mutually exclusive/ˈmjuːtʃuːəli eksˈkluːsɪv/ | 互斥 |
| conditional probability/kənˈdɪʃənl ˌprɒbəˈbɪlɪti/ | 条件確率 |
| tree diagram/triː ˈdaɪəɡræm/ | 木线图 |
| without replacement/wɪˈðaʊt rɪˈpleɪsmənt/ | 放回なし |
| Venn diagram/ven ˈdaɪəɡræm/ | ベン図 |
| two-way table/tuː weɪ ˈteɪbl/ | 両方向表 |
The birthday paradox
- In a room of just 23 people, there's a greater than 50% chance that two share a birthday.
- Probability is counterintuitive. The rules are simple, but the results can surprise you.
Combining probabilities
- Addition ("or"): $P(A \cup B) = P(A) + P(B) - P(A \cap B)$.
- Multiplication ("and", if independent 独立): $P(A \cap B) = P(A)\,P(B)$.
- Mutually exclusive 互斥 events can't both happen; independent events don't affect each other.
Worked example. P(rain) = 0.3, P(bus late) = 0.2, independent. P(rain or bus late) = 0.3 + 0.2 − 0.3 × 0.2 = 0.44.
Mutually exclusive ≠ independent. Mutually exclusive means they can't both happen (so $P(A \cap B) = 0$). Independent means one doesn't affect the other (so $P(A \cap B) = P(A) \times P(B)$). These are completely different concepts.
Dice: a starting point for probability
Conditional probability
P(A ∩ B) = P(A)·P(B|A)
The area model shows how conditional probabilities combine — and how Bayes finds P(A | B).
P(A) = 0.5, P(B) = 0.4 and P(A∩B) = 0.2. Are A and B 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
P(A)×P(B) = 0.5 × 0.4 = 0.2, which equals P(A∩B), so they are independent.
P(A) = 0.5, P(B) = 0.4, P(A∩B) = 0.2. What is P(A∪B)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
P(A∪B) = 0.5 + 0.4 − 0.2 = 0.7.
If A and B are mutually exclusive, then P(A∩B) = 0.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Mutually exclusive events cannot both happen, so their intersection has probability 0.
Mutually exclusive events are always 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
If A and B are mutually exclusive (and both have P > 0), then P(A∩B) = 0 ≠ P(A)×P(B), so they are NOT independent.
Conditional probability 条件概率
- Conditional: the chance of $A$ given $B$ has happened:
- Independence test: check whether $P(A \cap B) = P(A)\times P(B)$.
The normal curve: a probability is the area under it, centred on the mean
P(A∩B) = 0.2 and P(B) = 0.4. What is P(A | B)?
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
P(A|B) = P(A∩B)/P(B) = 0.2/0.4 = 0.5.
Worked example — tree diagrams 树状图
- A bag has 3 red and 2 blue balls. Two are drawn without replacement 不放回.
- P(both red) = $\dfrac{3}{5} \times \dfrac{2}{4} = \dfrac{6}{20} = 0.3$.
- P(one of each) = $\dfrac{3}{5} \times \dfrac{2}{4} + \dfrac{2}{5} \times \dfrac{3}{4} = \dfrac{6}{20} + \dfrac{6}{20} = 0.6$.
On a tree diagram, multiply probabilities along the branches, then add the paths you want.
A bag has 3 red and 2 blue. Two drawn without replacement. P(both red) = (3/5)(2/4). Find it.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
(3/5) × (2/4) = 6/20 = 0.3.
A bag has 3 red and 2 blue, drawn without replacement. Put the steps for P(both red) in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
On a tree, multiply the probabilities along the chosen branch, then simplify.
Venn diagrams 韦恩图 and two-way tables 双向表
- Venn diagrams show overlapping events visually — the intersection is $A \cap B$.
- Two-way tables organise data by two categories — read conditional probabilities from the relevant row or column.
- Count arrangements with permutations (order matters) and combinations (order does not).
You've got it
- addition (or): $P(A\cup B) = P(A) + P(B) - P(A\cap B)$
- multiplication (and, independent): $P(A\cap B) = P(A)P(B)$
- conditional: $P(A\mid B) = \dfrac{P(A\cap B)}{P(B)}$; test independence with $P(A\cap B) = P(A)P(B)$