What is a Bernoulli trial? One trial, two outcomes

A Bernoulli trial in one sentence

A Bernoulli trial is a single random action that can come out only two ways. Three things have to be true.

1. Exactly two outcomes. You label one of them success and the other failure. The labels are bookkeeping, not judgment: "the part is defective" can be the success if that is what you are counting.
2. A fixed success probability $p$. Here $p$ is the probability of success on this trial. The failure probability is $1 - p$, often written $q$.
3. One trial. A Bernoulli trial is the atom. Repeating it is what builds the distributions you actually use.

One free throw, one coin flip, one surveyed voter answering yes or no, one manufactured part that passes or fails inspection: each is a Bernoulli trial. A question with three or more outcomes is not one, though you can force it into shape by collapsing the categories, for example treating "red" as success and "any other color" as failure.

No topic in the Fall 2026 AP Statistics course is titled after the Bernoulli trial. The setting arrives inside Unit 2 topic 2.10, The Binomial Distribution, so learn the three conditions rather than the label. The name is worth knowing anyway, because college courses and textbooks use it constantly.

The mean and standard deviation of one trial

Write the trial as a random variable $X$ that takes the value 1 for success and 0 for failure. That 0 or 1 coding is what makes the algebra work, because counting successes then becomes plain addition.

The mean of a discrete random variable is $\mu_X = \sum x_i P(x_i)$, so

The mean of a single Bernoulli trial is the success probability itself. The variance follows from $\sigma_X^2 = \sum (x_i - \mu_X)^2 P(x_i)$:

so the standard deviation is $\sigma_X = \sqrt{p(1-p)}$. With $p = 0.7$ the mean is 0.7, the variance is $0.7 \times 0.3 = 0.21$, and the standard deviation is $\sqrt{0.21} \approx 0.4583$.

Notice where $p(1-p)$ peaks. It is largest at $p = 0.5$, where it equals 0.25, and it shrinks toward 0 as $p$ approaches either end. A trial you can barely predict is the most variable one, and that same $p(1-p)$ shows up later inside every standard error for a proportion.

Stacking trials: how the binomial is built

Run $n$ independent Bernoulli trials that all share the same $p$, and let $X_i$ be the 0 or 1 result of trial $i$. Define

Each $X_i$ contributes 1 when that trial succeeds and 0 when it fails, so the sum is literally the number of successes. That count is a binomial random variable, written $X \sim B(n, p)$. The binomial distribution is not a separate idea from the Bernoulli trial; it is the answer to "how many of these trials succeeded".

The formula falls out of the same picture. Any one specific sequence with $k$ successes and $n-k$ failures has probability $p^k(1-p)^{n-k}$, because independent trials multiply. Then you count how many sequences have $k$ successes, which is the binomial coefficient $\binom{n}{k}$:

Make that concrete with $n = 5$ and $p = 0.7$. There are $2^5 = 32$ possible success-failure sequences. Exactly 10 of them contain 3 successes, and each of those 10 has probability $0.7^3 \times 0.3^2 = 0.343 \times 0.09 = 0.03087$. So $P(X = 3) = 10 \times 0.03087 = 0.3087$, which is what $\binom{5}{3}p^3(1-p)^2$ returns and what the binomial probability calculator returns.

Where np and the binomial standard deviation come from

Because a binomial count is a sum of $n$ Bernoulli trials, its mean and standard deviation are the single-trial values scaled up.

Means always add, whether or not the pieces are independent. Each trial contributes a mean of $p$, so

Variances add too, but only when the pieces are independent, which is exactly why independence is one of the binomial conditions. Each trial contributes a variance of $p(1-p)$, so the total variance is $np(1-p)$ and

Check it on the $n = 5$, $p = 0.7$ example. The mean is $5 \times 0.7 = 3.5$ successes, and adding the five single-trial variances gives $5 \times 0.21 = 1.05$, so $\sigma_X = \sqrt{1.05} \approx 1.0247$. Computing the mean and standard deviation straight from the six probabilities of the distribution returns the same 3.5 and 1.0247.

The standard deviation formula takes a square root, so standard deviations do not add. Five trials do not have five times the standard deviation of one; they have $\sqrt{5}$ times it. That square-root growth is the same behavior you see in sampling distributions.

What breaks a Bernoulli trial

Three failures come up repeatedly, and each one has a fix or a workaround.

More than two outcomes. Rolling a die has six. Collapse the categories to rescue it: "rolls a 6" versus "does not roll a 6" is a Bernoulli trial with $p = \frac{1}{6}$.

A changing $p$. A basketball player who tires over 40 attempts does not have one fixed $p$. Neither does a survey that switches question wording partway through. If $p$ moves, the trials are not identical and the binomial formula does not apply.

Dependence. Sampling without replacement is the classic case. Drawing 4 cards from a standard deck and counting hearts is not binomial, because after a heart leaves the deck the next draw has a different probability. The exact chance of exactly 1 heart in 4 cards is 0.4388, while treating it as binomial with $p = 0.25$ gives 0.4219, an error of about 0.017. Here the sample is $\frac{4}{52} \approx 7.7\%$ of the deck, which clears the usual 10% guideline.

That 10% condition is the standard rescue: when you sample less than 10% of the population without replacement, the drift in $p$ from draw to draw is small enough to work with. Close enough is not exact, though, and the cards show the size of what you are accepting. Clearing the guideline at 7.7% of a 52-card deck still leaves the binomial answer 0.017 away from the truth, because the condition bounds the drift rather than removing it. The full checklist is in how to tell if a situation is binomial.

Bernoulli, binomial, and geometric

Three distributions grow out of the same trial, and they differ only in what you count.

The distinguishing question is what is fixed. In a binomial setting the number of trials is fixed and the number of successes varies. In a geometric setting the number of successes is fixed at one and the number of trials varies. With $p = 0.7$, the geometric probability that the first success arrives on trial 3 is $0.3 \times 0.3 \times 0.7 = 0.063$: two failures, then a success.

One scope note. College Board removed the geometric distribution from the AP exam with the Fall 2026 course, along with several other topics, as listed on the College Board AP Statistics future revisions page. The binomial distribution stayed, as topic 2.10. The geometric distribution is still standard content in a college introductory course, and binomial vs geometric covers the comparison.

One more descendant matters for inference: when $n$ is large enough, a binomial count is close to a normal curve, which is the whole content of the normal approximation to the binomial.