Conditions for a Binomial Distribution

What Are the Conditions for a Binomial Distribution?

A count follows a binomial distribution when four things are true: the number of trials is fixed in advance, each trial has only two outcomes, the probability of success is the same on every trial, and the trials are independent. Statisticians call a situation that meets all four a binomial experiment or a binomial setting.

If you remember only one line, remember the letters FBIT: Fixed number of trials, Binary outcomes, Independent trials, and the same probability on each Trial. When all four hold, the number of successes X has a binomial distribution with parameters n (the number of trials) and p (the probability of success). Its mean is np and its variance is np(1 − p).

The checker above walks through the four conditions for any scenario you describe and tells you which one fails, if any. This page explains each condition in plain terms, shows three worked examples (one that passes, one that passes only approximately, and one that fails), and ends with a practice quiz. Once a situation passes, the binomial distribution calculator does the arithmetic.

How Do You Use This Binomial Checker?

Describe your scenario by answering four questions, one for each condition.

1. Fixed trials. Choose Yes if you know the number of trials before you start, such as 10 flips or 25 inspected items. Choose No if you keep going until something happens.
2. Two outcomes. Choose Yes if each trial ends in success or failure. If there are more outcomes but you only care about one of them, that still counts as Yes, because everything else is “failure”.
3. Same probability. Choose Yes if the chance of success is identical on every trial.
4. Independence. Choose Yes if one trial tells you nothing about the next. If you are sampling without replacement, pick that option instead and enter the sample size n and the population size N so the 10% condition can be checked.

Optionally enter n and p. When the scenario is binomial, the tool then reports the mean, variance and standard deviation. The quick buttons load ready-made scenarios, including the ones used in the examples below. Press Clear to go back to the starting example.

Condition 1: Is the Number of Trials Fixed?

The number of trials, n, must be decided before the experiment starts. Ten coin flips, 25 items inspected on a production line and 200 voters surveyed all have a fixed n. “Flip until you get heads” does not, because the number of flips is itself random.

This condition is the one that separates the binomial from the geometric distribution. Both start from the same kind of success-or-failure trial. The binomial fixes the number of trials and counts the successes. The geometric fixes the number of successes (exactly one) and counts the trials. Rolling a die 12 times and counting sixes is binomial. Rolling until the first six appears is geometric, with a mean of 6 rolls, and the geometric distribution calculator handles it.

A quick test: can you write the possible values of X as 0, 1, 2, …, n for a known n? If the largest value depends on how the experiment turns out, or has no upper limit, the number of trials is not fixed.

Condition 2: Does Every Trial Have Exactly Two Outcomes?

Each trial must end in one of two outcomes, called success and failure. The words are labels, not judgments. Counting defective parts makes “defective” the success, even though nobody wants one.

Real trials often have more than two raw outcomes. A die has six faces, a survey answer might be one of five grades, and a patient might recover, stay the same or get worse. You can still use a binomial model when you only care about one outcome and lump the rest together. Rolling a die and asking “is it a six?” has two outcomes: six (probability 1/6) and not six (5/6). Grading a student and asking “is it an A?” works the same way.

What you cannot do is track several outcomes at once and still call it binomial. If you want the number of ones, twos and threes in the same ten rolls, you need the multinomial distribution. A binomial count follows one chosen outcome only.

Condition 3: Is the Probability of Success the Same Every Time?

The probability of success, p, must stay constant from the first trial to the last. A fair coin has p = 0.5 on every flip. A factory line with a steady defect rate of 5% has p = 0.05 for every item.

Constant probability fails when something about the process changes between trials. A golfer who learns the green gets more accurate with each putt, so p rises. A machine that wears out produces more defects as the day goes on, so p rises too. A hot streak or a tired player changes p in the same way. If p drifts, the number of successes does not follow a binomial distribution, even if every other condition holds.

This condition is also easy to confuse with the next one. Constant probability is about whether p is the same for every trial, while independence is about whether one trial affects another. Drawing cards without replacement breaks both at once, since removing a heart both changes the chance of a heart next time and ties the draws together.

Condition 4: Are the Trials Independent?

The outcome of one trial must not change the probability of another. Coin flips are independent: a run of five heads does not make a tail more likely. Draws from a deck are not, unless each card goes back and the deck is reshuffled.

Two situations commonly break independence. The first is sampling without replacement from a small population. Take a bag of 10 marbles with 4 red. The chance that the first marble is red is 4/10, so the chance that two in a row are red with replacement is 0.4 × 0.4 = 0.16. Without replacement it is 4/10 × 3/9 = 0.1333. The second draw depends on the first.

The second is real-world linkage. People in the same household share germs, students in one class share a teacher, and plants on one plot share soil. If you test six members of one household for a cold, an infection in one raises the chance of an infection in the others. Those six trials are not independent, however convenient the arithmetic would be.

What Is the 10% Condition for Sampling Without Replacement?

When you sample without replacement from a large population, the trials are not strictly independent, but they are close enough. The usual rule: if the sample size n is no more than 10% of the population size N, treat the trials as independent and use the binomial model.

The reason is simple. Removing one person from a population of 400,000 changes the proportions in the remainder by an amount far too small to notice. Removing one card from a deck of 52 changes them a lot. For a sample of 300 voters from 400,000, the sample is 0.075% of the population and the rule is passed by a wide margin. For 5 cards from 52 it is borderline, since 5 ≤ 5.2. For 4 marbles from a bag of 12 it fails badly, because 4 is 33% of the bag.

The exact distribution for sampling without replacement is the hypergeometric distribution. Its standard deviation is smaller than the binomial one by the factor √((N − n)/(N − 1)), the finite population correction. For 300 voters out of 400,000 that factor is 0.9996, so the correction is invisible. For 5 cards out of 52 it is 0.96, and for 5 items out of 20 it is 0.89. The hypergeometric distribution calculator gives the exact answer when the 10% condition fails.

Example 1: Do Ten Coin Flips Form a Binomial Experiment?

Flip a fair coin 10 times and let X be the number of heads. Go through the four checks.

All four pass, so X follows a binomial distribution with n = 10 and p = 0.5. The mean is np = 5 heads. The variance is 10 × 0.5 × 0.5 = 2.5, so the standard deviation is 1.5811. The probability of exactly five heads is C(10, 5) × 0.5¹⁰ = 252/1024 = 0.2461. The binomial probability calculator returns the same number.

Example 2: Are Hearts in a Five-Card Hand Binomial?

Deal 5 cards from a standard 52-card deck without replacement and let X be the number of hearts. There are two outcomes (heart or not), a fixed number of trials (5) and, at the first draw, p = 13/52 = 0.25. The catch is the fourth condition. The cards are drawn without replacement, so the trials are not independent.

Apply the 10% condition. The sample is 5 cards and 10% of the deck is 5.2, so 5 ≤ 5.2 and the rule just passes. A binomial model with n = 5 and p = 0.25 is acceptable, and it is only an approximation. Here is how close it gets.

The largest gap between the two columns is about 0.016, in the “0 hearts” row. That is good enough for a rough answer but not for a precise one, and for a textbook problem that says “assume binomial” you should follow the instruction. If you need the true probabilities, use the exact column. The mean is 1.25 hearts in both models, but the exact standard deviation is 0.8603 against 0.9682 for the binomial, a drop of 11%.

Example 3: Why Do Defective Bulbs From a Small Box Fail the Test?

A box holds 20 light bulbs and 5 of them are defective. You pick 5 bulbs at random without putting any back, and X is the number of defective bulbs in your pick. The first bulb is defective with probability 5/20 = 0.25. Whether the second is defective depends on what you drew first, so the draws are dependent.

This time the 10% condition fails: 10% of 20 is 2, and the sample is 5. The binomial model with p = 0.25 is a poor fit.

The binomial model overstates the chance of drawing no defective bulb by 0.0436, a relative error of more than 20%. The real standard deviation is 0.8603 against the binomial 0.9682. When a sample takes a quarter of the population, the exact hypergeometric distribution is the right tool.

Which Distribution Do You Use When a Condition Fails?

A failed condition does not leave you stuck. It points to a different distribution.

Customers arriving at a shop in an hour make a good example of the first row. There is no fixed number of trials, since customers could keep coming, so the count is modelled with a Poisson distribution instead. Poisson is also the usual limit of a binomial when n is large and p is small.

How Do You Answer “Does This Procedure Result in a Binomial Distribution?”

Many textbooks phrase this question as “determine whether the given procedure results in a binomial distribution” or “assume that a procedure yields a binomial distribution.” Answer it in four short steps, and always write the reason for a No.

1. Name the trial. What single repeated action is being counted? A flip, a draw, a shot, one person surveyed.
2. Name the success. What event are you counting? Write it as a yes-or-no statement, such as “the card is a heart”.
3. State n and p. n is the number of trials and p is the probability of a success on one trial.
4. Test the four conditions. Give a one-line reason for each, and if you sample without replacement, state the 10% check.

Suppose the question says: “A survey of 500 randomly chosen adults from a city of 2 million asks whether they own a car; X is the number who do.” The trial is one adult, the success is “owns a car”, n = 500 and p is the proportion of car owners. The sample is without replacement, but 500 is far below 10% of 2 million, so the trials are close enough to independent. The answer is yes, approximately binomial.

Test Yourself: Which of These Is Binomial?

Twelve scenarios follow. Pick an answer for each one, then press Check answers to see your score and a short reason for every question.

What Mistakes Do People Make When Checking the Conditions?

The first is treating “random” as the same as “independent”. A random sample drawn without replacement from a small population is random, but the draws are still dependent.

Second, missing a hidden trial. In the question about two coins tossed together ten times, the trial is one pair of tosses, and a success is “both heads” with p = 0.25. Counting each coin as a trial gives the wrong n and p.

Third, forgetting that you can group outcomes. A die does not have two outcomes, but “six or not six” does. If the question counts only one outcome, the second condition is satisfied.

Fourth, mixing up binomial and geometric. If the question asks for “the number of rolls until…”, the number of trials is the random quantity, and the model is not binomial. If it asks for “the number of sixes in 12 rolls”, the number of trials is fixed and the model is.

Fifth, believing a distribution is binomial because the numbers look neat. A binomial model is justified by how the data are produced, not by the pattern of the counts. Always return to the four questions.

Sixth, using the 10% condition as a blanket excuse. It only repairs independence. It does nothing for a probability that changes, a number of trials that is not fixed, or outcomes that cannot be grouped into two.

Once It Is Binomial, How Do You Find Probabilities, the Mean and the Standard Deviation?

With a confirmed binomial setting, the probability of exactly k successes in n trials is:

A basketball player makes 80% of her free throws and takes 10 shots in a game. Shots are fixed in number, each is a make or a miss, p = 0.8 is assumed constant, and the shots are taken to be independent. The mean is 8 makes, the standard deviation is 1.2649, and the probability of making all 10 is 0.8¹⁰ = 0.1074. The probability of making at least 8 is 0.6778. A student guessing every answer on a 20-question test with four options each has n = 20 and p = 0.25, a mean of 5 and a standard deviation of 1.9365.

Excel / Sheets   =BINOM.DIST(8, 10, 0.8, FALSE)         exactly 8
                 =1 - BINOM.DIST(7, 10, 0.8, TRUE)       at least 8
R                dbinom(8, 10, 0.8);  1 - pbinom(7, 10, 0.8)
Python (SciPy)   binom.pmf(8, 10, 0.8);  binom.sf(7, 10, 0.8)

For ranges of values the binomial CDF calculator adds up the terms for you, the binomial distribution table lists whole rows for small n, and the variance calculator and standard deviation calculator cover spread for data you have measured directly.

Key Points to Remember

• A binomial setting needs four things: a fixed number of trials, two outcomes, a constant probability of success and independent trials (FBIT).
• Binary does not mean the raw trial has only two outcomes. You can group outcomes into “the one I count” and “everything else”.
• Fixed n separates binomial from geometric, negative binomial and Poisson situations.
• Sampling without replacement breaks independence. If n ≤ 10% of N, the binomial model is a fair approximation. If not, use the hypergeometric distribution.
• Five cards from a 52-card deck pass the 10% test by a hair, and the binomial error is still up to 0.016 in a single probability.
• Once all four conditions hold, X has mean np, variance np(1 − p) and P(X = k) = C(n, k) p^k (1 − p)^{n − k}.

Frequently Asked Questions

Sources: Penn State STAT 500, section on binomial random variables; NIST/SEMATECH e-Handbook of Statistical Methods, section 1.3.6.6.18 (binomial distribution); SciPy documentation for scipy.stats.binom and scipy.stats.hypergeom; NumPy documentation.

Related Calculators

Passed the checks, or need a different model? These do the next step, or browse the full list of free statistics calculators:

Binomial Distribution CalculatorProbabilities, mean and spread once the conditions hold. Binomial Probability CalculatorExactly, at least or at most k successes. Hypergeometric Distribution CalculatorThe exact model when sampling without replacement. Geometric Distribution CalculatorTrials until the first success.