Binomial Distribution — Formula, Conditions & Example

A binomial distribution gives the probability of exactly $k$ successes in $n$ trials — but only when the situation passes four conditions. The single most important judgment is deciding whether those conditions hold, because the arithmetic can look correct while the model itself is wrong.

The four conditions, at a glance

Condition                When it fits                 When it fails
Fixed number of trials   n known in advance           trials continue until some event
Two outcomes per trial   success / failure defined    three or more tracked outcomes
Independent trials       sampling with replacement    sampling without replacement (small set)
Constant probability p   p same every trial           p changes from trial to trial

If every row lands in the left column, a binomial model applies. If any lands on the right, use a different model.

What the model means and what each part does

Repeat the same kind of trial $n$ times. On each trial, label one outcome a success and the other a failure. If the success probability is $p$ on every trial, the count of successes $X$ can follow a binomial distribution, written $X \sim \text{Bin}(n,p)$, where $n$ is the number of trials, $p$ is the per-trial success probability, and $X$ counts how many successes occur. This is a counting model: it asks how many successes happened, not which trial succeeded.

In the formula $P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$, each part has a job: $\binom{n}{k}$ counts the ways to arrange $k$ successes among $n$ trials, $p^k$ gives the probability of those successes, and $(1-p)^{n-k}$ gives the probability of the remaining failures. It works for $k = 0, 1, 2, \dots, n$.

A note on the conditions: a die roll still fits "two outcomes" if you define success as something like "rolling a $6$," and "independent trials" fits sampling with replacement but usually not sampling without replacement from a small group.

Worked example: exactly 3 heads in 5 flips

A biased coin lands heads with probability $0.6$, and you flip it $5$ times. Let heads be success, so $n=5$, $p=0.6$, $k=3$:

Compute each part:

So

The probability of exactly $3$ heads is $0.3456$, or $34.56\%$. The model is valid here because the experiment has a fixed $n$, two outcomes per flip, independent trials, and the same $p=0.6$ each time.

For "at least one success," the complement is often faster than summing many terms. With $X \sim \text{Bin}(5,0.6)$:

because "at least one success" is the complement of "zero successes."

Common confusions and how to keep them straight

• Ignoring the conditions. Using the formula when trials are not independent — for example, drawing without replacement from a small set while pretending $p$ never changes — gives a wrong model.
• Misreading "success." Success does not have to mean something good; it is simply the outcome you chose to count.
• Mixing up "exactly," "at least," and "at most." "Exactly $3$" is one term, "at least $3$" is several terms, and "at most $3$" is a different sum.

When the binomial distribution is used

It applies when you count repeated yes-or-no outcomes: defective vs. not defective, pass vs. fail, click vs. no click, heads vs. tails. It is useful in quality control, survey sampling under the right assumptions, reliability questions, and basic probability models.

To test the conditions and the formula together, take $8$ flips of a coin with $p=0.4$. Find $P(X=2)$, then $P(X \ge 1)$ using the complement, and finally note what would change if the trials were no longer independent.