Bernoulli Trial vs Trial

A Bernoulli trial is a single trial of a chance process with exactly two outcomes, one labeled success and the other failure, and a success probability $p$ that is fixed for that trial. Code it as a random variable taking 1 for success and 0 for failure and it becomes the smallest random variable worth studying, with $\mu_X = p$ and $\sigma_X = \sqrt{p(1-p)}$.

One free throw from a 78 percent shooter is a Bernoulli trial with $p = 0.78$. Its mean is 0.78 makes, which is a long-run share rather than anything a single shot can produce, and its standard deviation is $\sqrt{0.78(0.22)} = \sqrt{0.1716} \approx 0.414$.

"Every card I draw is red or black, so each draw is a Bernoulli trial with $p = 0.5$" is where this goes wrong. Two outcomes is not the whole condition. Deal one card and keep it, and the next draw has $p = 25/51 \approx 0.4902$ or $26/51 \approx 0.5098$ depending on what came first, so the draws no longer share a single $p$ and stacking them does not build a binomial count. Independence and a constant $p$ are what make trials interchangeable, and only the first draw is safe on its own.

There is no $n$ inside a Bernoulli trial. The moment a question asks how many successes in 20 attempts you have moved to the binomial, and the scaling is not the one people guess. Means add, so $\mu = 20(0.78) = 15.6$ makes. Standard deviations do not, so $\sigma = \sqrt{20(0.78)(0.22)} \approx 1.853$ rather than $20(0.414) = 8.28$. Variances are what add, and the square root then grows with $\sqrt{n}$.

No topic in the Fall 2026 AP Statistics course is named after the Bernoulli trial. The setting is stated inside topic 2.10 of Unit 2, The Binomial Distribution, so the conditions are what get tested and the name is worth carrying for later courses.

In statistics a trial is not a court case and not a whole clinical study. It is one repetition of a chance process, and it ends with exactly one outcome from the sample space. Flip a coin once and you have run one trial, landing heads or tails. The problem defines what one trial is, and writing that definition down is the first move, because every count that follows is measured in those units.

Flip a fair coin 20 times and you have run 20 trials. If 12 come up heads, the relative frequency is $12/20 = 0.60$ while the probability of heads is still 0.5 and the expected count was $20(0.5) = 10$. Trials are called independent when the result of one leaves the probabilities for the next unchanged, which fair coin flips satisfy.

Here is the error that quietly ruins simulations. To estimate the chance that a family with three children has at least one girl, one trial is three births, not one birth. "I flipped the coin 100 times, so I ran 100 trials" is wrong twice over: 100 flips supply 33 complete trials with one flip left dangling, and dividing by 100 understates every proportion. A trial runs until the question has a single yes or no answer, so a trial that takes three flips consumes three flips. The exact answer here is $1 - (0.5)^3 = 0.875$, and listing the 8 equally likely birth orders shows 7 of them contain a girl.

The binomial setting demands a fixed number of trials that are independent and share one probability of success. Sampling without replacement breaks the last two: pull chips from a bag and each trial changes what the next one faces, which is why the ten percent condition exists.

Trials are the counting unit of simulation, topic 2.3, Estimating Probabilities Using Simulation.