When a Count Is Not Binomial
In The BINS Checklist for Binomial Conditions, you learned to check for Binary outcomes, Independent trials, a fixed Number of trials, and the Same probability of success on every trial. This tutorial uses that checklist to spot situations that fail one or more conditions. The goal is not just to say “not binomial,” but to identify what feature of the process causes the problem.
Two situations deserve particular attention. Drawing cards without replacement changes what remains available after each draw. Stopping when the first success occurs makes the number of trials depend on the outcomes. Both can sound like repeated yes-or-no trials, but that alone is not enough for a binomial model.
Two Warning Signs to Look For
Changing what remains available is a warning sign for sampling without replacement. If an item is removed after selection, the next selection may have a different chance of success. The outcomes are also dependent: what happens on one draw affects the composition of the remaining group. These features commonly make both the Independent and Same-probability conditions fail.
For sampling without replacement, the 10% condition helps judge whether treating the selections as approximately independent is reasonable: the sample size should be no more than 10% of the population. Meeting that guideline can support a binomial approximation; it does not make the draws literally independent or the success probability exactly constant.
A process that waits for an outcome is a warning sign for a variable number of trials. Ask whether the number of trials is chosen before the process starts, or whether the process ends because a particular result has occurred. If the ending depends on the outcomes, the count of trials is not fixed.
- Does each trial still have two relevant outcomes?
- Can an outcome change the chances on later trials?
- Is the number of trials set in advance?
- Is the probability of success the same on every trial?
Keep the last two questions separate. A process might have independent trials but changing success probabilities, or a common success probability for each position but dependent outcomes. Either way, a failed condition rules out a binomial model as described.
Worked Example: Drawing Cards Without Replacement
Worked Example: Drawing Cards Without Replacement
A standard deck has 52 cards, including 13 hearts. Eight cards are drawn one at a time without replacement. Let \(H\) be the number of hearts drawn. Is an exact binomial model appropriate for \(H\)?
State. One trial is drawing one card. Define success as drawing a heart. The variable \(H\) counts the hearts among the eight draws.
Plan. Apply BINS. In particular, check whether each draw leaves the probability of a heart unchanged, and use the 10% condition to assess whether a binomial approximation is supported.
Do. Binary: Each card drawn is a heart or not a heart. Independent: The draws are dependent because each card is removed. For example, drawing a heart leaves fewer hearts in the deck for the next draw. Number fixed: Exactly eight cards are drawn. Same probability: The probability of a heart changes as cards are removed. At the first draw it is \(13/52=0.25\). If the first card is a heart, the probability at the second draw is \(12/51\approx0.2353\). If the first card is not a heart, it is \(13/51\approx0.2549\).
The 10% check gives \(8/52\approx0.1538\), or about 15.38% of the deck. Since \(0.1538>0.10\), the sample exceeds 10% of the population. The usual guideline does not support treating these draws as approximately independent.
Conclude. \(H\) is not exactly binomial: the draws are dependent, and the probability of success changes. Also, the 10% condition does not support a binomial approximation for these eight draws from a 52-card deck.
The changing conditional probabilities show why “there are two outcomes per draw” is not enough. A heart on one draw changes what is left for the next draw. Notice, too, that the 10% condition is a guideline for considering an approximation, not a switch that turns dependent draws into independent ones.
Worked Example: Stopping at the First Success
Worked Example: Stopping at the First Success
A volunteer calls people from a list until someone agrees to test a new community garden app. Each person either agrees or declines. Let \(T\) be the number of people called, including the person who first agrees. Is \(T\) a binomial count?
State. One trial is calling one person. Define success as that person agreeing to test the app. The variable \(T\) counts the calls made up to and including the first agreement.
Plan. Check BINS, paying special attention to whether the number of calls is fixed before the calling starts.
Do. Binary: Each call results in agreement or decline. Independent: The description does not establish independence; responses could be related, for example if people have discussed the app. Even if independence were assumed, another condition fails. Number fixed: The number of calls is not fixed in advance. It depends on when the first agreement occurs. If the first person agrees, \(T=1\); if the first four decline and the fifth agrees, \(T=5\). Same probability: The situation does not provide enough information to establish a common probability of agreement. The list or the people’s awareness of earlier calls could also affect the chances.
Conclude. \(T\) is not binomial as described because the number of trials is not fixed. The binary outcomes do not remedy this failure. Independence and a common probability would also need support before they could be assumed.
A process that ends at the first success fixes the number of successes at one, but it does not fix the number of trials. This distinction is central: a binomial count counts successes in a predetermined number of trials. It does not count how many trials are needed to reach a target.
Worked Example: Independent Trials with Different Success Chances
Worked Example: Independent Trials with Different Success Chances
A testing lab checks 10 sample containers. Assume the results are independent. Because the testing conditions change partway through the session, the chance that a container passes is 0.92 for each of the first five containers and 0.97 for each of the last five. Let \(P\) be the number of containers that pass. Is \(P\) binomial?
State. One trial is checking one container. Define success as the container passing. The variable \(P\) counts passes among the 10 containers.
Plan. Apply BINS separately. The stated independence assumption is evidence for the Independent condition, but check whether one common probability applies to all 10 trials.
Do. Binary: Each container passes or does not pass. Independent: The problem states that the results are independent. Number fixed: Exactly 10 containers are checked. Same probability: This condition fails. The success probability is 0.92 for the first five trials and 0.97 for the last five, so no single probability applies to every trial.
Conclude. \(P\) is not binomial as described because the probability of success is not the same on every trial. Independence, two outcomes, and a fixed number of trials do not make up for that failure.
This example shows why “independent” and “same probability” must be checked separately. Independence means one outcome does not affect the others. It does not mean that trials conducted under different conditions have the same success probability.
Worked Example: Replacing Each Card
Worked Example: Replacing Each Card
A student draws a card from a standard deck, records whether it is a heart, returns it to the deck, and thoroughly shuffles before the next draw. The student repeats this process for eight draws. Let \(R\) be the number of hearts. Is an exact binomial model appropriate, assuming the shuffling and draws work as described?
State. One trial is one draw. Define success as drawing a heart. The variable \(R\) counts hearts in the eight draws.
Plan. Apply BINS, comparing this process with drawing without replacement.
Do. Binary: Each draw results in a heart or not a heart. Independent: Replacing the card and shuffling before the next draw means one result does not change the cards available for the next draw. Under the stated assumptions, the draws are independent. Number fixed: There are exactly eight draws. Same probability: Each draw has the same probability of a heart, \(13/52=0.25\), because the full deck is restored each time.
Conclude. All four BINS conditions are met under the stated assumptions, so an exact binomial model is appropriate for \(R\). Replacing and reshuffling preserves the same chance of success and prevents one draw from changing the next draw’s available cards.
The word “cards” alone does not determine whether a binomial model applies. The sampling method matters: drawing without replacement changes the available deck, while replacing and shuffling restores it for the next trial.
Common Mistakes and AP Exam Tip
A strong explanation identifies the failed condition and gives the relevant detail from the situation. “Not binomial” by itself is incomplete; name what fails and why. When a prompt supports an assumption, state it. When a condition is unknown, do not present it as established.
- Stopping after identifying two outcomes. Binary outcomes meet only one BINS condition. A full-credit answer also checks independence, a fixed number of trials, and the same probability of success.
- Calling a stopping rule a fixed number of trials. “Continue until someone agrees” does not fix the number of calls. Explain that the trial count depends on the results.
- Saying that draws without replacement are independent. Removing an item changes the remaining pool. Explain how that can affect a later outcome, then consider whether the 10% condition supports an approximation.
- Treating the 10% condition as proof of independence. It is a guideline for whether an approximation may be reasonable. It does not make dependent draws exactly independent.
- Combining independence with the same probability. The lab trials are independent by assumption, but their success chances differ. Name the Same-probability condition as the one that fails.
- Giving a verdict without evidence. “The probability changes” is clearer when paired with the actual reason, such as cards being removed or test conditions changing.
A concise full-credit response for the eight-card example would say: “Each draw is a heart or not a heart, and eight draws are planned, but drawing without replacement makes the draws dependent and changes the probability of a heart. Because \(8/52\approx0.1538>0.10\), the 10% condition does not support a binomial approximation here.” For a stopping-rule example, explicitly state that the number of trials depends on when success occurs.
Key Takeaway
Look for changes in the available population, changes in trial conditions, and rules that stop the process after an outcome. Each can reveal a failed BINS condition. A good explanation names the failure rather than relying on the vague claim that the situation “doesn’t look binomial.”
Check Your Understanding
For each situation, identify whether it is binomial as described. Name any failed BINS condition and explain your reasoning.
- A student draws five marbles from a jar without replacement. Each marble is red or not red. The jar contains 20 marbles. Which conditions fail, and does the 10% condition support a binomial approximation?
- A survey worker asks people to sign up until three people agree. Let \(X\) be the number of people asked. Which BINS condition is clearly not met?
- A machine independently checks 12 items. The probability of a defect is 0.03 for the first six items and 0.05 for the last six. Explain why independence alone does not make the defect count binomial.
- A player draws a card, records whether it is a spade, replaces it, and shuffles before each of 10 draws. Assuming those steps are carried out properly, which features make a binomial model appropriate?
- A student says, “Selecting without replacement always makes a binomial approximation impossible.” Explain why this statement is too strong, using the 10% condition in your answer.