Sampling inference, expected value and counting
| English | Français |
|---|---|
| expected value/ekˈspektɪd ˈvæljuː/ | valeur attendue |
| combination/ˌkɒmbɪˈneɪʃn/ | combinaison |
A decision before an answer
- Choosing a captain and deputy is different from choosing two unranked representatives. Roles make order matter.
- Your goal: Interpret a population estimate and simulation evidence without claiming proof.
Read the relationship
- A random sample supports inference to its source population; a margin of error gives a sampling-uncertainty interval around an estimate. Random assignment instead supports a causal treatment comparison. In a simulation, a result rarely produced under a claim can be evidence against that claim; failure to find an unusual result does not prove the claim true.
- Calculate conditional, compound and expected-value quantities.
Choose two unranked representatives from four people. Number of choices:
4·3/2=6; order does not matter.
Use the defining rule
- Conditional probability restricts the denominator to the conditioning group. For either A or B, subtract their overlap after adding separate probabilities. Independence permits multiplication of unchanged probabilities; without-replacement draws require updated counts. Expected value is the sum of each possible outcome times its probability, not necessarily an outcome achieved in one trial.
- Choose permutations or combinations according to whether order matters.
A random award is 10 with probability 0.2 and 0 otherwise. Expected award:
10·0.2+0·0.8=2.
Check the conditions
- Use the multiplication principle for successive choices. For n distinct objects, choosing r in order gives n!/(n-r)!; choosing an unranked set gives n!/[r!(n-r)!]. Dividing by r! removes the multiple orderings of the same set. These formulas assume distinct objects and no replacement; different conditions need a corresponding count.
- Choose permutations or combinations according to whether order matters.
From five students, captain/deputy pairs number 5·4=20; unranked pairs number 5·4/2=10. A game awards 4 points with probability 1/4 and 0 otherwise, so expected award is 1 point per play. A sample estimate 52% ±3 percentage points gives 49%–55%; it is not a guarantee about every sample or individual.
An estimate 64% ±4 points has lower endpoint ____%.
Subtract four percentage points.
Apply the task format
- Translate the event to a count before calculating its probability. If a favourable event can occur in several disjoint ways, add those ways; if paths overlap, avoid double-counting. A numerical answer must be between zero and one for a probability, while an expected cost or score has the units of the outcomes.
- Choose permutations or combinations according to whether order matters.
Order, replacement and conditioning change the calculation. Failing to reject a statistical claim is not proving it true.
Which answer fits this case?
Interpret a population estimate and simulation evidence without claiming proof
Failing to reject a claim establishes that it is certainly true.
It means the evidence did not justify rejection under the procedure.
Keep the distinctions
- expected value 期望值 — A probability-weighted average of possible outcomes.
- combination 组合 — An unordered selection of distinct objects.
- Interpret a population estimate and simulation evidence without claiming proof.
- Calculate conditional, compound and expected-value quantities.
- Choose permutations or combinations according to whether order matters.
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.