Binomial Distribution Examples: Exact, At Least and At Most

Suppose you flip a coin five times and want exactly three heads. One particular sequence, HHTHT, is easy to calculate. But HHTHT is only one of several ways to get three heads. A binomial calculation does both jobs: it finds the chance of one sequence and counts all the sequences that meet your condition.

These binomial distribution examples show how to find an exact count, at least one success, and at most two successes. Before using the formula, check whether the situation actually fits it. For a gentler introduction to the language of chance, see understanding probability.

Check the four binomial conditions

The binomial model counts successes across repeated trials. OpenStax's binomial distribution lesson describes a model we can check in four parts:

1. A fixed number of trials: decide $n$ before observing the outcomes.
2. Two outcomes per trial: label each outcome success or failure for the question you are asking.
3. Independent trials: learning one result does not change the probability of another.
4. A constant success probability: the same $p$ applies on every trial.

If all four hold, let $X$ be the number of successes. Then $X$ can be any integer from zero through $n$. The label success does not mean the outcome is good. If you are counting defective parts, a defective part can be the mathematical success because it is the event you chose to count.

Consider five independent questions where each answer has probability $p=0.4$ of being correct. We fix five questions, call a correct answer a success, assume independence, and keep the chance at 0.4 each time. That is a binomial setup.

Build the probability formula

For exactly $k$ successes in $n$ trials, start with a single arrangement. If success has probability $p$, failure has probability $1-p$. Independence makes the probability of an arrangement with $k$ successes and $n-k$ failures equal to $p^k(1-p)^{n-k}$.

Now count the arrangements. There are $\binom{n}{k}$ ways to choose which $k$ trial positions contain a success. Each has the same probability, so:

$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k},\qquad \binom{n}{k}=\frac{n!}{k!(n-k)!}.$$

For example, three successes in five trials have $\binom{5}{3}=10$ arrangements. You do not need to list them all. If the counting term is unfamiliar, permutations and combinations explains why choosing positions is a combination rather than an ordering problem.

As a quick scale check, the mean success count is $np$, and the variance is $np(1-p)$ under this model, as given by OpenStax. These are summaries of the whole distribution, not substitutes for the probability of a particular count. For the difference between variance and standard deviation, see this guide.

Example 1: exactly three successes

Return to the five independent questions, each with $p=0.4$ for a correct answer. Find the probability of exactly three correct answers. The number of trials is $n=5$, and the requested count is $k=3$. Substitute those values before using a calculator:

$$\begin{aligned} P(X=3)&=\binom{5}{3}(0.4)^3(0.6)^2\\ &=10(0.064)(0.36)\\ &=0.2304. \end{aligned}$$

So the probability is $23.04\%$. The factor $10$ counts where the three correct answers can occur. Without it, $0.02304$ would be the probability of just one specific sequence, such as correct, correct, wrong, correct, wrong. That is the most common missing piece in an exact-count solution.

The exponents also check the setup: three successes plus two failures equals five trials. If your exponents do not add to $n$, stop before calculating. A useful calculation habit is to keep the exact decimal $0.2304$ through subsequent steps and round only when you state the answer.

Example 2: at least one success

Suppose four independent attempts each have a success probability of $0.3$. Find the probability of at least one success. This phrase includes one, two, three, and four successes. You could add four separate binomial terms, but their complement is shorter: zero successes.

$$\begin{aligned} P(X\geq1)&=1-P(X=0)\\ &=1-\binom{4}{0}(0.3)^0(0.7)^4\\ &=1-0.2401=0.7599. \end{aligned}$$

The answer is $75.99\%$. Read the complement in words before choosing it: “at least one” fails only when there are none. The complement of “at least two” would instead include zero and one. Subtracting only $P(X=0)$ in that case would give the wrong answer.

Notice that $P(X=0)=(1-p)^n$. The combination term is one because there is exactly one arrangement with no successes. This shortcut works because the trials share the same probability and are independent.

Example 3: at most two successes

Now imagine six independent trials, each with a success probability of $p=0.25$. “At most two” means zero, one, or two. Add precisely those three mutually exclusive cases:

$$\begin{aligned} P(X\leq2)&=P(X=0)+P(X=1)+P(X=2)\\ &=\binom{6}{0}(0.25)^0(0.75)^6 +\binom{6}{1}(0.25)^1(0.75)^5 +\binom{6}{2}(0.25)^2(0.75)^4\\ &=0.177978515625+0.35595703125+0.296630859375\\ &=0.83056640625\approx0.8306. \end{aligned}$$

That is about $83.06\%$. Here direct addition uses three terms. The complement, $1-P(X\geq3)$, would require four terms, so it is less convenient. For a large threshold, compare the number of terms on both sides and calculate the shorter side. A calculator's cumulative-binomial function can help, but you still need to set the inequality correctly. A function that returns $P(X\leq2)$ includes two; a function that returns $P(X<2)$ does not.

If intermediate decimals are rounded too early, their sum can move by a few units in the final decimal place. Keep full precision during the addition, then round the final probability to the requested accuracy. The answer should remain between zero and one.

A situation that is not binomial

You draw three cards from a standard deck without replacement and count aces. There is a fixed number of draws, and each draw can be labeled ace or not ace. But after the first draw, the deck has changed. If you drew an ace, the next ace probability is $3/51$; if you did not, it is $4/51$. The success probability is no longer fixed, and the draws are dependent. The binomial formula with $n=3$ and $p=4/52$ is therefore not an exact model.

Putting each card back and reshuffling before the next draw would restore the constant probability and independence assumption for this simplified example. This distinction matters more than memorizing a formula: first decide what is being sampled and whether a trial changes the next one.

Try three questions, then check the solutions

1. Exactly two. A game has five independent rounds and a $0.2$ chance of winning each round. What is the probability of exactly two wins?

Solution: $n=5$, $k=2$, and $p=0.2$. There are $\binom{5}{2}=10$ arrangements of two wins. Thus

$$P(X=2)=10(0.2)^2(0.8)^3=10(0.04)(0.512)=0.2048.$$

The chance is $20.48\%$. Counting just one arrangement would have produced $0.02048$, one tenth of the correct answer.

2. At least two. Four independent attempts each have a success probability of $0.5$. What is the probability of at least two successes?

Solution: “At least two” excludes zero and one. With $p=0.5$, the complement is easy:

$$\begin{aligned} P(X\geq2)&=1-P(X=0)-P(X=1)\\ &=1-\binom{4}{0}(0.5)^4-\binom{4}{1}(0.5)^4\\ &=1-\frac{1}{16}-\frac{4}{16}=\frac{11}{16}=0.6875. \end{aligned}$$

The answer is $68.75\%$. Subtracting only zero successes would instead answer “at least one.”

3. At most one. Three independent checks each have a positive-result probability of $0.1$. What is the probability of at most one positive result?

Solution: Include zero and one, then add:

$$\begin{aligned} P(X\leq1)&=\binom{3}{0}(0.1)^0(0.9)^3 +\binom{3}{1}(0.1)^1(0.9)^2\\ &=0.729+0.243=0.972. \end{aligned}$$

The answer is $97.2\%$.

For more work on probability fundamentals, use Math Zen's probability practice. The most valuable first step on every problem is still the four-condition check.