表からの確率と制限されたグループ
導入済み| English | 意味 |
|---|---|
| independent events/ˌɪndɪˈpendənt ɪˈvents/ | 独立事象 |
| conditional probability/kənˈdɪʃənl ˌprɒbəˈbɪlɪti/ | 条件付き確率 |
A decision before an answer
- “Among bus users” changes the population in the denominator. Dividing by everybody answers a different question.
- Your goal: Calculate event and complement probabilities from counts.
Read the relationship
- A probability based on equally represented observations is favourable count divided by the relevant total. A two-way table has joint counts in cells, group totals on the margins and a grand total. Read whether the question asks about a joint event, either event or one event given another before choosing the denominator.
- Use the conditioning group as the denominator.
Of 20 cyclists, 8 wear red. Of all 60 people, 18 wear red. P(red given cyclist):
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 conditioning group is the 20 cyclists.
Use the defining rule
- For P(A given B), restrict attention to the B group and divide the A-and-B count by the B total. It is not generally equal to P(B given A), because the conditioning groups differ. A fraction with the correct numerator but the wrong population still answers the wrong question.
- Distinguish independence from disjointness.
Two disjoint positive-probability events are 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
One event occurring makes the other impossible.
Check the conditions
- The complement probability is 1-P(A). For either A or B, adding separate probabilities double-counts overlap, so subtract P(A and B). Disjoint events have no overlap; independent events satisfy P(A and B)=P(A)P(B). Two positive-probability disjoint events cannot be independent, since observing one rules out the other.
- Distinguish independence from disjointness.
Original counts: bus users—18 carry a packed lunch, 12 do not; non-bus users—12 carry lunch, 8 do not. Total 50. P(lunch)=30/50=0.6; P(lunch given bus)=18/30=0.6; P(bus given lunch)=18/30=0.6 in this particular balanced table. Change one cell and these need not stay equal. P(bus and lunch)=18/50=0.36, equal here to 0.6·0.6, so the table shows independence of these categories.
P(A)=0.35. P(not A)=____.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Complement probabilities sum to 1.
Apply the task format
- For without-replacement draws, update both the remaining favourable count and total at each step. For independent repeated choices the probabilities multiply without that update. State the sampling assumptions before calculating. A table of observed relative frequencies supports empirical probabilities for that population, not a guarantee about every future draw.
- Distinguish independence from disjointness.
Equal conditional answers in one table are coincidence or structure to check, not a universal rule. Condition first, then count.
このケースに当てはまる答えはどれか？
数に基づいて事象とその補事象の確率を計算する
P(A given B) must equal P(B given A).
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 denominator changes with the conditioning group.
Keep the distinctions
- conditional probability 条件概率 — Probability calculated within a stated conditioning event.
- independent events 独立事件 — Events for which knowing one does not change the probability of the other.
- Calculate event and complement probabilities from counts.
- Use the conditioning group as the denominator.
- Distinguish independence from disjointness.
Match each term with its precise meaning in this lesson.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence 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
The order follows the stated process; check each stage before the next.