10.5 Probability Models, Relative Frequency, & Simulation
Key Takeaways
- A probability model lists every outcome in the sample space and assigns each a probability, and those probabilities must sum to exactly 1.
- A uniform model assigns equal probability to every outcome; a non-uniform model assigns probabilities from observed frequencies or from the physical setup.
- Approximating a probability by relative frequency means dividing observed successes by total trials, and the estimate improves as the number of trials grows.
- When a model and observed frequencies disagree substantially, the explanation is either an unfair setup or too few trials — both are valid answers on TABE.
Probability Models, Relative Frequency, & Simulation
Section 10.1 computed theoretical and experimental probability. This section covers the standard that sits between them: build a probability model, then check it against reality. The DRC specification asks candidates to "develop a probability model and use it to find probabilities of events" and to "compare probabilities from a model to observed frequencies; if the agreement is not good, explain possible sources of the discrepancy."
What a Probability Model Is
A probability model is a complete list of the possible outcomes together with a probability for each. Two rules govern every model:
- Each probability is between 0 and 1, inclusive.
- All the probabilities sum to exactly 1.
If a set of numbers violates either rule, it is not a probability model — and TABE writes items that ask exactly that.
Uniform models
A uniform model gives every outcome the same probability. Rolling a fair six-sided die:
| Outcome | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| Probability | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ | $\tfrac{1}{6}$ |
Sum: $6 \times \tfrac{1}{6} = 1$. ✓
Any event's probability is then the sum of its outcomes' probabilities: $P(\text{even}) = \tfrac{1}{6} + \tfrac{1}{6} + \tfrac{1}{6} = \tfrac{1}{2}$.
Non-uniform models
Most real situations are not uniform. A spinner with unequal sectors, a bag with unequal counts, or a machine with a known defect rate all need probabilities assigned individually.
A bag holds 5 red, 3 blue, and 2 green chips.
| Outcome | Red | Blue | Green |
|---|---|---|---|
| Probability | $\tfrac{5}{10} = 0.5$ | $\tfrac{3}{10} = 0.3$ | $\tfrac{2}{10} = 0.2$ |
Sum: $0.5 + 0.3 + 0.2 = 1$. ✓
Finding a missing probability. A model lists $P(A) = 0.4$, $P(B) = 0.25$, $P(C) = ?$, and those are the only outcomes. Since the total must be 1, $P(C) = 1 - 0.4 - 0.25 = \mathbf{0.35}$.
Building a Model From Observed Data
When the setup is unknown — an irregular spinner, an unfamiliar machine — you build the model from relative frequency:
A spinner is spun 200 times with these results:
| Color | Red | Blue | Yellow | Green | Total |
|---|---|---|---|---|---|
| Count | 96 | 52 | 30 | 22 | 200 |
| Relative frequency | 0.48 | 0.26 | 0.15 | 0.11 | 1.00 |
The estimated model says red is nearly half the spinner's area, and green is about a tenth. Predicting forward: in 500 more spins, expect about $0.48 \times 500 = 240$ reds.
Comparing Model to Observation
TABE's version of this standard presents a theoretical model and some observed data and asks whether they agree.
A coin is flipped 40 times and lands heads 24 times. The model says $P(\text{heads}) = 0.5$, predicting 20 heads. Observed relative frequency is $24 \div 40 = 0.6$. Is the coin unfair?
Reasonable answers, in order of likelihood:
- Ordinary random variation. Forty flips is a small sample; 24 heads is a common result for a fair coin.
- The coin or the flipping process is biased. Possible, but 40 trials is not enough evidence.
- A recording or counting error.
The same coin flipped 4,000 times landing heads 2,400 times is a very different matter. The relative frequency is identical at 0.6, but the sample is 100 times larger, and random variation is a far less plausible explanation. Sample size is what turns a discrepancy into evidence.
Sources of discrepancy to name
| Source | Description |
|---|---|
| Too few trials | small samples fluctuate widely around the true probability |
| A faulty model | the assumed probabilities do not match the real setup |
| A biased process | a weighted die, a worn spinner, a mis-set machine |
| Non-independence | outcomes influence one another, breaking the model's assumption |
| Measurement or recording error | miscounts, missing trials |
The Law of Large Numbers, Restated
As the number of trials grows, relative frequency converges toward true probability.
| Trials | Heads | Relative frequency |
|---|---|---|
| 10 | 7 | 0.70 |
| 100 | 58 | 0.58 |
| 1,000 | 517 | 0.517 |
| 10,000 | 5,032 | 0.5032 |
This is why an estimate built from 20 observations should be treated as a rough guide and one built from 2,000 as reliable — a judgment TABE asks about directly.
Simulation
A simulation models a real chance process with a simpler random device when the real situation is impractical to repeat.
Designing one takes four decisions:
- Choose a device whose probabilities match the situation — a coin for 50/50, a die for sixths, digits 0–9 for tenths, slips in a hat for anything else.
- Define what one trial represents.
- Define success.
- Run many trials and compute the relative frequency.
Problem. A store's promotion says 1 in 5 cups wins a prize. Simulate how often a customer buying 3 cups wins at least one.
- Device: digits 0–9; let 0 and 1 mean "win" (that is $\tfrac{2}{10} = \tfrac{1}{5}$) and 2–9 mean "lose."
- One trial: three digits.
- Success: at least one 0 or 1 among the three digits.
- Run 100 trials, count successes, divide by 100.
The theoretical answer is $1 - (0.8)^3 = 1 - 0.512 = 0.488$, so a well-run simulation should land near 49%. Comparing your simulated result to that theoretical value is the same model-versus-observation reasoning applied to your own experiment.
A probability model for a four-outcome process assigns P(A) = 0.32, P(B) = 0.18, and P(C) = 0.29. What must P(D) equal for this to be a valid probability model?
A machine is claimed to produce defective parts 4% of the time. In a sample of 50 parts, 6 are defective. Which is the most reasonable interpretation?