How to tell if a situation is binomial: 4 checks

By Jude Wallis · Published July 30, 2026

Check four conditions: the number of trials $n$ is fixed in advance, each trial ends in success or failure, the success probability $p$ is the same on every trial, and the trials are independent. If all four hold, the count of successes is binomial. If any one fails, it is not.

AP Statistics: Unit 2 (topics 2.10 The Binomial Distribution). Unit 2 topic 2.10 of the Fall 2026 AP Statistics course asks you to justify why a random variable is or is not a binomial random variable, defining it as a count of successes in repeated independent trials with two outcomes and a fixed success probability; the geometric distribution was removed from the exam in the same redesign but remains college content.

The four conditions in one place

A count is binomial when all four of these hold.

A fixed number of trials. You can write down $n$ (the number of trials) before the process starts. It is given to you, not something you wait to discover.
Two outcomes per trial. Every trial ends in success or failure. "Success" only labels the outcome you are counting, even when that outcome is a bad thing like a defect.
The same probability of success on every trial. The value $p$ (the success probability) does not drift as the trials go on.
Independent trials. The result of one trial does not change the probability on any other trial.

One more requirement sits underneath all four: the random variable has to be the count of successes, so $X$ takes whole-number values from 0 to $n$ . Many students remember the list as BINS, for Binary, Independent, Number of trials fixed, and Same probability.

A checklist you can run in one pass

Read the problem once and answer these five questions in order.

Can I write down $n$ right now from the wording? If not, the situation is not binomial.
Does each trial have exactly two outcomes I can label success and failure?
Is $p$ the same on every trial, or does something in the context push it up or down?
Does the result of one trial change the probability on another?
Am I counting successes, or am I counting or measuring something else?

If the answers line up, name the distribution explicitly as binomial with your values of $n$ and $p$ , then go to the formula in the binomial probability guide. Naming $n$ and $p$ is part of the justification, not an extra step.

Five scenarios, classified

Roll a fair six-sided die 20 times and count the 6s. Binomial. The number of trials is fixed at $n = 20$ , each roll is a 6 or not a 6, $p = 1/6$ on every roll, and the rolls do not affect one another.

Record how many minutes a student spends on homework on each of 5 nights. Not binomial. The two-outcomes condition fails, because minutes is a quantitative measurement rather than a success-or-failure result. Counting how many of the 5 nights exceed 60 minutes would repair it.

Flip a coin until the first head appears, and record how many flips that took. Not binomial. The fixed-number-of-trials condition fails, since the number of flips is the very thing being recorded. This is a geometric setting.

A player takes 15 free throws in a row, and her arm tires, so her make rate slips from about 80% early to about 65% late. Not binomial. The same-probability condition fails, because $p$ changes across the 15 shots.

Deal 10 cards from a standard 52-card deck and count the hearts. Not binomial. Independence fails: the chance the first card is a heart is $13/52 = 0.25$ , but it drops to $12/51 \approx 0.2353$ after a heart and rises to $13/51 \approx 0.2549$ after a non-heart.

The near miss: sampling without replacement

Scenario 5 is the near miss that catches most people. Whenever you sample people or objects without replacement, removing one changes the make-up of what is left, so the success probability shifts and strict independence fails. Every opinion poll is technically in this position, since nobody is surveyed twice.

Taken literally, that would rule out the binomial model for nearly every survey question ever asked, which is too strict to be useful. The fix is a size comparison, covered in the next section. For the underlying probability difference, see with replacement vs without replacement.

The 10% rule that rescues it

When you sample without replacement, the trials stay close enough to independent as long as your sample is a small slice of the population. The standard rule is

n \le 0.10N

where $n$ is the sample size and $N$ is the population size. In words, the population must be at least 10 times the sample.

Take a survey of 12 randomly chosen adults from a city of 400,000. Ten percent of that population is 40,000, and 12 sits far below it, so the shift in $p$ from one person to the next is much too small to matter. Treating the number of yes answers as binomial with $n = 12$ is reasonable.

The card deal in scenario 5 fails the rule. Dealing 10 cards would need a population of at least 100 cards, and a deck holds 52, so 10 cards is 19.2% of the deck. The AP course states the 10% condition formally in Unit 3 as a condition for the sampling distribution of a sample proportion, and the same reasoning is what lets you call a small sample from a large population binomial.

Binomial or geometric, and what the AP exam expects

The condition that most often separates these two is the fixed number of trials. A binomial setting fixes $n$ and counts successes. A geometric setting does not fix the number of trials and instead counts how many it takes to reach the first success, which is scenario 3 above. The side-by-side comparison lives in binomial vs geometric.

One honest note about the exam. The geometric distribution was removed from AP Statistics in the Fall 2026 redesign, with the first exam in May 2027, so you will not be asked for its formula. It remains standard content in college introductory statistics, and spotting a geometric setting is still the quickest way to rule out a binomial one. Course details are on the College Board course page.

How to write the justification on a free-response question

Topic 2.10 asks you to justify why a random variable is or is not binomial, so the wording is worth rehearsing. Tie each condition to the context instead of listing the letters B, I, N, and S on their own.

A model answer for a quality inspection reads like this.

There is a fixed number of trials, $n = 40$ chargers. Each charger is either defective or not defective. The probability that a charger is defective is 0.03 for every one of them, and the chargers were chosen at random from a shipment of 5,000, with 40 below 10% of 5,000, so the trials are approximately independent. Therefore $X$ , the number of defective chargers, is binomial with $n = 40$ and $p = 0.03$ .

If a condition fails, say which one and why, then stop. Do not compute a binomial probability for a setting you have just argued is not binomial.

Mistakes that cost points

Treating every yes-or-no situation as binomial. Two outcomes is one condition out of four.
Counting the wrong variable. If $X$ is the number of trials, or a total of measurements, it is not a binomial count even when the trials themselves behave.
Ignoring without-replacement sampling instead of checking the 10% rule and saying so.
Assuming $p$ holds steady when the context says otherwise, as with fatigue, practice effects, or changing weather.
Calling a situation binomial because the problem appeared in the binomial section of a textbook.

Once you have confirmed the four conditions, run the numbers with the binomial probability calculator and build speed on the binomial practice set.

Justify a binomial model, then use it

A shipment holds 5,000 phone chargers, and the manufacturer reports that 3% are defective. An inspector selects 40 chargers at random without replacement and counts the defective ones. Is this binomial, and if so, what are the mean, the standard deviation, and the probability of finding no defective chargers?

Fixed number of trials: yes, $n = 40$ chargers, stated in the problem.
Two outcomes: yes, each charger is defective or not defective. Label defective as the success, so $p = 0.03$ .
Same probability: yes, approximately. The 3% rate applies across the whole shipment.
Independent trials: the sampling is without replacement, so check the 10% rule. Ten percent of 5,000 is 500, and $40 \le 500$ , so the trials are close enough to independent.
All four conditions hold, so $X$ is binomial with $n = 40$ and $p = 0.03$ .
Mean: $\mu_X = np = 40(0.03) = 1.2$ defective chargers.
Standard deviation: $\sigma_X = \sqrt{np(1-p)} = \sqrt{40(0.03)(0.97)} = \sqrt{1.164} \approx 1.079$ chargers.
No defectives: $P(X = 0) = \binom{40}{0}(0.03)^0(0.97)^{40} = (0.97)^{40} \approx 0.2957$ .

Binomial with $n = 40$ and $p = 0.03$ . The mean is 1.2 defective chargers per sample of 40, the standard deviation is about 1.079 chargers, and $P(X = 0) \approx 0.2957$ , so roughly 29.6% of samples like this one contain no defective charger.

When the 10% rule fails, the binomial answer is wrong

A bag holds 10 marbles, 4 of them red. You draw 3 marbles without replacement and count the red ones. Show that this is not binomial, then compare the answer the binomial formula would give for exactly 1 red with the exact answer.

Two conditions do hold: the number of trials is fixed at $n = 3$ , and each draw is red or not red.
Test the constant-probability condition. The first draw has $P(\text{red}) = 4/10 = 0.40$ . If that marble is red, the second draw has $P(\text{red}) = 3/9 \approx 0.3333$ ; if it is not red, the second draw has $P(\text{red}) = 4/9 \approx 0.4444$ .
Because the probability moves with the earlier result, the trials are not independent and $p$ is not constant, so the count is not binomial.
Try the 10% rule as a rescue: 3 marbles out of 10 is 30% of the population, far above the 10% ceiling, so the binomial model is not even a good approximation here.
See what the binomial formula would give with $n = 3$ and $p = 0.40$ : $P(X = 1) = \binom{3}{1}(0.40)^1(0.60)^2 = 3(0.40)(0.36) = 0.432$ .
Get the exact value by counting sets: choose 1 of the 4 red marbles and 2 of the 6 non-red marbles, out of all sets of 3 chosen from 10. That is $\frac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}} = \frac{4 \times 15}{120} = \frac{60}{120} = 0.500$ .
Compare: $0.500 - 0.432 = 0.068$ , an error of about 13.6% of the true probability.

Not binomial. Drawing without replacement from only 10 marbles changes $p$ between draws, and 3 of 10 is 30% of the population, so the 10% rule does not rescue it. The binomial formula returns 0.432 for exactly 1 red while the exact probability is 0.500, an error of 0.068.

Frequently asked questions

Does a success have to be a good outcome?

No. In a binomial setting, success is just the label for whichever outcome you are counting. If you are counting defective parts, missed shots, or rainy days, that outcome is the success and $p$ is its probability.

What do I do when the sample is drawn without replacement?

Check the 10% rule. If the sample size is at most 10% of the population, $n \le 0.10N$ , the change in $p$ from trial to trial is small enough that a binomial model is reasonable, and you should say so in your justification. If the sample is a larger share of the population, the count is not binomial.

Is the geometric distribution still on the AP Statistics exam?

No. The geometric distribution was removed in the Fall 2026 redesign, with the first exam in May 2027. It is still standard content in college introductory statistics, and recognizing a geometric setting still helps you rule out a binomial one on the exam.

If a without-replacement count is not binomial, what is it?

College courses call the exact model the hypergeometric distribution, which counts successes when you draw a fixed number of items from a finite population without replacement. The AP course does not cover it. AP problems are written so the 10% rule holds, or they ask you to explain that the setting is not binomial.

Does a binomial model require a random sample?

The model itself requires independent trials with the same success probability. Random selection is usually what makes those two conditions believable, which is why a justification names the random selection and then checks the 10% rule. Without randomness you have no reason to claim independence.