What Does a Fixed Probability Claim?
In Matching a Model to a Situation Using Variable Type, we asked whether the variable and the process support a particular model. For a binomial model, the process includes a claim that is easy to overlook: every trial has the same probability of success. This tutorial focuses on whether that fixed probability is realistic in a particular setting.
A fixed probability does not mean that the same number of successes occurs in every group of trials. It means that the chance of success on each individual trial is modeled by the same value of \(p\). The observed proportion of successes can vary from one stretch of trials to another, even if that assumption is true. Conversely, two stretches with similar observed proportions do not prove that their underlying probabilities are equal.
Consider a basketball player taking free throws. A model with \(p=0.75\) says that each shot has a 0.75 chance of being made, whether it is the first shot or the last. The value need not be exactly the player’s long-run real-world chance; it is a simplification used to calculate probabilities. Whether that simplification is useful depends on the situation and the question.
A player’s chance might plausibly change with fatigue, an injury, pressure, a change in shooting routine, or the circumstances around each attempt. Those are reasons to question a constant \(p\), not proof that \(p\) actually changes. The same idea applies outside sports: a production process may warm up, weather may shift, or the people responding to a survey may differ over time.
Inspect the Process, Not Just the Pooled Rate
A sensible assessment begins with how the trials are produced. Ask whether important conditions remain similar across the trials. Are the equipment, environment, participants, and procedures stable? Does the order of trials matter? Is there a known reason for success to become more or less likely? These questions help decide whether one shared probability is a reasonable approximation.
Then look for patterns in the data, if data are available. Divide trials into meaningful groups, such as early and late attempts or different operating conditions, and compare their success proportions. A pattern may flag a concern about constant \(p\). But observed proportions fluctuate by chance, especially in small groups, so a difference by itself does not establish that the underlying probabilities differ.
The overall success proportion can conceal variation between groups. If one group has a high success probability and another has a low one, the pooled proportion averages across them. That average can describe the combined data, but it does not make the probability on every individual trial equal to the average. A binomial model with that pooled value may therefore misrepresent the process.
The constant-probability condition is separate from independence. A process could have the same success probability on every trial while outcomes still depend on one another; it could also have independent outcomes with probabilities that change from trial to trial. The next tutorial considers independence assumptions in real settings. For a binomial model, both the common-probability claim and the independence claim need to be plausible.
A Practical Check for Constant \(p\)
Use the following questions to organize your reasoning. They do not provide a mechanical test that proves a probability is fixed. Instead, they help identify when the assumption is well supported, questionable, or simply unknown.
Be precise about what counts as one trial and what outcome counts as success. A probability can only be assessed for a clearly described process.
Consider time, conditions, participants, equipment, or other factors that could change the chance of success from one trial to the next.
If data are available, compare success proportions across meaningful groups or time periods. Treat differences as evidence to consider, not automatic proof of changing probabilities.
Explain whether one probability seems reasonable for the question, whether it is a rough approximation, or whether the information is not enough to decide.
A good answer is appropriately cautious. If the scenario says conditions are stable and supplies a probability for each trial, you can use that model as stated. If it describes changing conditions, explain why a single probability may be unrealistic. If it gives only a pooled success rate, do not claim that the rate has been shown to apply equally to every trial.
Worked Example: Free Throws Across a Game
Worked Example: Free Throws Across a Game
A fictional coach uses \(p=0.75\) as a player’s free-throw success probability. In a game, the player makes 10 of 12 free throws in the first half and 5 of 12 in the second half. Assess whether the constant-probability assumption is reasonable, and calculate the probability of making at most 5 of 12 shots in the second half if the stated model applies.
State. Let \(X\) be the number of made free throws in the second half. The proposed model treats the 12 attempts as trials with success probability \(p=0.75\) on each attempt.
Plan. First compare the observed proportions, while remembering that sample proportions vary by chance. Then, conditionally, calculate a binomial probability under the proposed fixed-\(p\) model. For that calculation, assume a fixed 12 attempts, two outcomes per attempt, the same \(p=0.75\) on every attempt, and independent attempts. The calculation describes what the model predicts; it cannot by itself establish why the observed result occurred.
Do. The first-half success proportion is \(10/12\approx0.833\), and the second-half proportion is \(5/12\approx0.417\). Across the full game, the pooled proportion is:
The second-half proportion is noticeably lower than the first-half proportion, so the data give a reason to ask whether fatigue or another game condition affected the chance of success. With just 12 attempts in each half, however, the difference does not prove that the underlying probabilities changed.
Under the stated binomial model, \(X\sim B(12,0.75)\). “At most 5” means \(X\leq5\). As in Using binomcdf for At Most Probabilities:
A formula check gives the same result by adding the probabilities for 0 through 5 makes:
Thus, if the fixed-\(p\) model and independence assumption apply, making 5 or fewer of 12 attempts has a probability of about 0.0143, rounded to four decimal places. It would be an unusual result under that model. It still does not identify the cause: chance variation, an inaccurate value of \(p\), changing conditions, or a failure of another model assumption could be relevant.
Conclude. The difference between the half-game proportions makes constant \(p=0.75\) worth questioning, but these data alone do not prove the probability changed. Under the proposed binomial model, the probability of at most 5 makes in the second half is about 0.0143.
This example also shows why it matters to say “under the model.” A small model probability means the observed event would be unusual if the model’s assumptions were correct. It is a reason to examine the assumptions, not a direct measurement of the player’s actual probability on every shot.
Worked Example: A Pooled Rate Hides Different Conditions
Worked Example: A Pooled Rate Hides Different Conditions
A fictional greenhouse tests seeds in two conditions. The model assigns a sprouting probability of 0.90 to each of 10 seeds in a warm, steady section and 0.50 to each of 10 seeds in a cooler section. The recorded results are 9 sprouts in the warm section and 5 in the cooler section. Can the 20 trials reasonably be modeled as a binomial count with one common probability?
State. Success is a seed sprouting, and the proposed count is the total number of sprouts among 20 seeds.
Plan. A binomial model would require one common success probability across all 20 trials, in addition to the other binomial conditions discussed in Choosing Between Binomial and Other Models. Compare the probabilities specified for the two conditions before using the pooled observed proportion.
Do. The success probabilities differ by condition: 0.90 for the warm-section seeds and 0.50 for the cooler-section seeds. The observed proportions are \(9/10=0.90\) and \(5/10=0.50\); the pooled observed proportion is:
The pooled rate of 0.70 summarizes the combined results, but it does not replace the two stated trial probabilities. A binomial model \(B(20,0.70)\) would claim that each seed has the same 0.70 chance of sprouting, which conflicts with the stated probabilities for the two sections. The pooled proportion can be reported descriptively, but it does not justify that common-probability model.
Conclude. The 20-seed count should not be modeled as binomial with one shared \(p\) under the stated conditions. The two groups have different success probabilities, and pooling them conceals that difference.
The lesson is not that a binomial model is impossible whenever conditions vary slightly. Models are simplifications, and a common probability may be a useful approximation when differences are small relative to the purpose of the analysis. But if the probabilities are known to differ substantially, one pooled rate can give a misleading picture of the trial process.
Worked Example: Looking for a Pattern Across Batches
Worked Example: Looking for a Pattern Across Batches
A fictional packaging line records whether each item passes a quality check. The line checks four consecutive batches of 10 items. The numbers that pass are 8, 7, 4, and 5. Assess what these data suggest about using a constant success probability for all 40 items.
State. A success is an item passing the quality check. The question is whether one probability seems reasonable across the four consecutive batches.
Plan. Compare the batch proportions and the pooled proportion. A sequence of noticeably different proportions may flag changing conditions, but small groups can show variation by chance. Without more information, do not treat these observed proportions as exact values of the underlying probabilities.
Do. The batch proportions are:
| Batch | Items checked | Passes | Observed proportion passing |
|---|---|---|---|
| 1 | 10 | 8 | 0.80 |
| 2 | 10 | 7 | 0.70 |
| 3 | 10 | 4 | 0.40 |
| 4 | 10 | 5 | 0.50 |
The first two batches have higher observed proportions than the last two. Across all batches, \(8+7+4+5=24\) of 40 items pass, so the pooled proportion is \(24/40=0.60\). That pooled number does not show whether the process stayed stable; the batch-by-batch pattern raises a question about whether the line’s conditions changed.
The proportions are based on only 10 items per batch, so they can vary even if a common probability is a reasonable model. It would help to check whether the machine setting, materials, or inspection process changed between batches, and to collect more data in a planned way. The observed pattern is a useful warning to investigate, not proof of a change in probability.
Conclude. The batch proportions and their order make a constant success probability worth investigating. The pooled proportion of 0.60 alone is not enough to establish that the same probability applied to every item, nor do these small batches prove that it changed.
Common Mistakes and AP Exam Tips
- Equating equal probabilities with equal observed proportions. A constant \(p\) does not require every batch to have the same fraction of successes. Full-credit wording distinguishes the modeled chance on each trial from the proportions observed in finite groups.
- Treating the pooled proportion as proof of a common \(p\). A pooled rate averages outcomes across the trials. If conditions differ, it may hide important differences. State what the pooled rate describes and avoid claiming it applies to each trial without support.
- Claiming that a difference proves probabilities changed. Observed differences can result from chance variation. Explain what the pattern suggests and what additional information would help assess it; do not turn a descriptive comparison into certainty.
- Checking only the probability assumption. A common \(p\) is one feature of a binomial model, not the only one. Keep it distinct from independence, and remember that both assumptions need attention.
- Interpreting an unusual model probability as a cause. A small probability is calculated under the stated model. It does not tell you which assumption, if any, failed, or prove that conditions changed.
Key Takeaway
A fixed success probability is a claim about the chance assigned to each trial, not a requirement that observed success proportions match exactly. Stable conditions can make one probability a reasonable model; changing conditions or patterns across meaningful groups can make the assumption questionable. A pooled rate may summarize combined results while concealing different trial probabilities.
Check Your Understanding
Answer each question using the distinction between a trial probability and an observed success proportion.
- A player makes 7 of 10 shots early in practice and 4 of 10 later. What does this comparison suggest about a constant probability, and why does it not prove that the probability changed?
- A machine has a stated pass probability of 0.8 when cold and 0.6 after it has been running for an hour. Why would a single binomial model with one common \(p\) be questionable for a count spanning both conditions?
- Two batches have different observed success proportions. Does that alone establish that their underlying probabilities differ? Explain.
- Can a constant-probability model produce different observed proportions in two groups? Explain why.
- In one or two sentences, describe how a pooled success proportion could conceal changing probabilities across time.