Probability from a table and a restricted group
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| independent events/ˌɪndɪˈpendənt ɪˈvents/ | 独立事件 | dú lì shì jiàn |
| conditional probability/kənˈdɪʃənl ˌprɒbəˈbɪlɪti/ | 条件概率 | tiáo jiàn gài lǜ |
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):
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.
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)=____.
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.
Which answer fits this case?
Calculate event and complement probabilities from counts
P(A given B) must equal P(B given A).
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.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.