2.3 Chance Models and Simulation — Answers and Rubrics
These solutions explain the new practice tasks in this lesson. For simulation parts, a valid recorded run may differ from another learner’s run; the rubric checks the model, recorded work and interpretation.
Original generated practice for the Fall 2026 course. Return to the lesson and independent checks.
A Welcome-Desk Simulation: worked answer
In a new model, each visitor to a welcome desk independently has a 30% chance of receiving a voucher. One complete trial contains four visitors. Estimate the probability that at least two of the four receive vouchers. The following 20 independent four-digit groups were newly generated for this task. Each digit is equally likely to be 0–9. For every part of this task, use the prescribed encoding: digits 0, 1 and 2 mean voucher; digits 3–9 mean no voucher.
2286 5035 3180 2443 0005 3763 4428 6784 5403 4121 1689 9403 0406 8540 5970 4853 8513 0459 9934 3078
Part a
Define one complete trial, one possible outcome, and the event of interest. Use the prescribed voucher digits 0, 1 and 2 and explain why this encoding matches 30%. Describe how independence is maintained and explain why the results are random.
A complete trial is four independent visitors. An outcome could be voucher, no voucher, voucher, no voucher; the event contains every four-visitor outcome with at least two vouchers. Digits 0, 1 and 2 represent vouchers, so 3 of 10 equally likely digits give probability 0.30. Digits 3–9 represent no voucher. Generate a fresh independent digit for each visitor; with cards, replace and mix after each draw. Chance determines the generated digits and thus the outcome.
| Evidence for credit | Points |
|---|---|
| Defines a four-visitor trial and a valid four-result outcome/event. | 1 |
| Uses the prescribed voucher digits 0, 1 and 2 and explains that 3 of 10 equally likely digits gives probability 0.30. | 1 |
| Explains chance and an independence-preserving procedure. | 1 |
Part b
Execute the encoding on every supplied group. Record each trial’s voucher count, the number of trials satisfying the event, and the trial total. Calculate and interpret the empirical probability estimate.
The voucher counts, in order, are 2, 1, 2, 1, 3, 0, 1, 0, 1, 3, 1, 1, 2, 1, 1, 0, 1, 1, 0, 1. Trials 1, 3, 5, 10 and 13 qualify: 5 of 20, so the estimate is 0.25. In these simulated groups, 25% contained at least two voucher recipients. This estimates the event probability; it is not the individual-visitor probability 0.30.
| Evidence for credit | Points |
|---|---|
| Records all 20 voucher counts consistently. | 1 |
| Counts five qualifying trials and uses 20 as the denominator. | 1 |
| Interprets 0.25 for four-visitor groups and identifies it as an estimate. | 1 |
Part c
Run 80 additional independent four-visitor trials using a ten-sided die, random-digit technology, or newly generated random digits. Use the same prescribed encoding: 0, 1 and 2 mean voucher. Keep a complete reproducible record of all 80 outcomes and voucher counts. Combine these with the supplied 20 trials to estimate the probability from 100 trials. State your event count and denominator.
Runs differ. A valid response records 80 new independent four-digit groups using the encoding in part a, counts qualifying groups, and reports , where is the new qualifying-group count. The 100-trial denominator counts groups, not 400 visitors. Credit requires actual recorded trials and the matching calculation; a statement that a simulation could be run does not complete this part.
| Evidence for credit | Points |
|---|---|
| Provides all 80 actual new four-digit groups or equivalent recorded outcomes. | 1 |
| Uses the prescribed voucher digits 0, 1 and 2 with independent outcomes and repetitions. | 1 |
| Correctly combines the qualifying count with the supplied five and divides by 100. | 1 |
Part d
Explain what the true probability of the four-visitor event means in the long run. Two classmates’ independent 100-trial estimates are 0.31 and 0.36. Explain why they may differ, what a much larger independent run is expected to show, and why a run of losses does not force a win next.
The true event probability is the relative frequency approached over many independent four-visitor trials under this fixed model. Finite random runs have sampling variation, so 0.31 and 0.36 can differ. The law of large numbers predicts stabilization toward the same event probability as independent trial counts grow, without guaranteeing improvement at every step. Losses do not change a later visitor’s 30% chance or force a voucher.
| Evidence for credit | Points |
|---|---|
| States a fixed long-run four-visitor event probability. | 1 |
| Attributes differing estimates to random finite-run variation. | 1 |
| Explains LLN and rejects forced balancing or changed next-visitor chance. | 1 |