4.8 The binomial distribution

Introduction

Many questions about chance count successes in a fixed number of attempts: how many of 20 seeds germinate, how many of 150 passengers turn up for a flight, how many children of two carrier parents inherit a genetic condition. When every attempt has the same chance of success and the attempts do not influence each other, the number of successes follows the binomial distribution, one of the most widely used discrete probability distributions.

The binomial distribution is a special case of the discrete random variables of subtopic 4.7, Discrete random variables, with one great advantage: its whole distribution is fixed by just two numbers, the number of trials $n$ and the probability of success $p$. From those two numbers come every probability, the mean $np$ and the variance $np(1 - p)$, so a single line of notation such as $X \sim \mathrm{B}(150, 0.92)$ describes a situation completely.

When is a situation binomial?

Trials with two outcomes

The binomial distribution is built from repeated trials, each with exactly two possible results. A tossed coin lands heads or tails; a manufactured bulb works or is faulty; a passenger turns up or does not; a free throw scores or misses. Such a trial is named after Jacob Bernoulli, who studied repeated trials of this kind.

The two outcomes are labelled "success" and "failure", but the labels say nothing about which result is good. When counting faulty bulbs, a faulty bulb is the "success", because it is the outcome being counted. The probability of failure is the complement of success, $1 - p$, sometimes written $q$.

A trial with more than two results can still be treated as a Bernoulli trial by grouping them. Rolling a die has six outcomes, but "rolling a six" against "not rolling a six" is a Bernoulli trial with $p = \frac{1}{6}$.

Independent trials

The second key idea is that the trials do not affect each other. Whether the first bulb in a box is faulty should tell nothing about whether the second is faulty. This is the idea of independent events from subtopic 4.6, Combined events, applied to a whole sequence of trials.

Independence is what allows probabilities along a sequence of trials to be multiplied. For three independent trials with success probability 0.3, the probability of success, then failure, then success is $0.3 \times 0.7 \times 0.3 = 0.063$, with no adjustment for what happened earlier.

The four conditions

A random variable $X$ that counts successes follows a binomial distribution when all four of these conditions hold.

1. There is a fixed number of trials, $n$, decided before the trials begin.

2. Each trial has two outcomes, success or failure.

3. The probability of success, $p$, is the same for every trial.

4. The trials are independent.

Then $X$, the number of successes in the $n$ trials, can take the values $0, 1, 2, \ldots, n$. A short way to remember the conditions is "fixed $n$, two outcomes, constant $p$, independent".

Real situations rarely meet the four conditions perfectly. A binomial distribution is then a mathematical model whose assumptions should be stated and judged. Two passengers booked together are likely to arrive together, so their trials are not independent; a free-throw shooter may tire, so $p$ is not quite constant. The model can still be useful when the departures from the conditions are small, and stating the assumptions makes clear what the conclusions depend on.

The binomial distribution

Building the distribution with a tree diagram

The probabilities of a binomial distribution come from a tree diagram. With three trials, each branch is a success (S) with probability $p$ or a failure (F) with probability $q = 1 - p$, and because the trials are independent, the probability of each path is the product of its branches.

Every path with exactly one success contains one $p$ and two $q$s, so each has probability $pq^2$; only the order differs. There are three such paths (SFF, FSF, FFS), so the probability of exactly one success is $3pq^2$. Grouping all eight paths by the number of successes gives the whole distribution for $n = 3$.

The general pattern follows the same logic. For exactly $r$ successes in $n$ trials, every path has $r$ successes and $n - r$ failures, so it has probability $p^r q^{n-r}$. The probability of $r$ successes is that path probability multiplied by the number of paths that have exactly $r$ successes.

Counting the paths: binomial coefficients

The number of paths with exactly $r$ successes in $n$ trials is the number of ways to choose which $r$ of the $n$ trials are the successes. For $n = 3$ and $r = 1$, the single success can be in position 1, 2 or 3: three ways.

With binomial coefficients, the binomial probability formula is

In practice binomial probabilities are found with the GDC, but the formula shows why the GDC gives the values it does. The GDC also calculates $\binom{n}{r}$ directly (often labelled nCr), so no hand method for the coefficients is needed.

Pascal's triangle

The binomial coefficients form a famous pattern. Writing the numbers of paths for $n = 0, 1, 2, \ldots$ in rows gives a triangle in which every number is the sum of the two numbers diagonally above it.

The addition rule has a simple reason. A sequence of $n$ trials with $r$ successes either ends in a success, in which case the first $n - 1$ trials hold $r - 1$ successes, or ends in a failure, in which case they hold $r$ successes. So the number of paths for $(n, r)$ is the sum of the numbers for $(n - 1, r - 1)$ and $(n - 1, r)$, the two entries above it. Each row is also symmetric, because choosing which $r$ trials are successes is the same as choosing which $n - r$ are failures.

Source: Wikipedia, "Pascal's triangle" and "Yang Hui", accessed 2026.

The binomial distribution and its notation

Putting the pieces together gives the distribution.

The statement $X \sim \mathrm{B}(n, p)$ is read "$X$ follows a binomial distribution with parameters $n$ and $p$". The two parameters fix everything: $n$ is a positive integer, $p$ is a probability between 0 and 1, and $X$ can take each whole number from 0 to $n$. In a context, the random variable must be defined in words: "Let $X$ be the number of faulty bulbs in a box of 25. Then $X \sim \mathrm{B}(25, 0.04)$."

Calculating binomial probabilities with the GDC

Exact and cumulative probabilities

The GDC has two binomial functions, and choosing the right one is the main skill.

• The binomial probability function (often called "pdf") gives $\mathrm{P}(X = r)$, the probability of exactly $r$ successes.

• The binomial cumulative function (often called "cdf") gives $\mathrm{P}(X \le r)$, the probability of at most $r$ successes: the sum $\mathrm{P}(X = 0) + \mathrm{P}(X = 1) + \ldots + \mathrm{P}(X = r)$.

On a TI-84 Plus CE, both are in the distributions menu (2nd VARS), as binompdf and binomcdf, with inputs in the order: number of trials, probability of success, value. On a TI-Nspire they are under Menu → Statistics → Distributions (Binomial Pdf and Binomial Cdf). On a Casio fx-CG50 they are in MENU → Statistics → DIST → BINOMIAL. Some GDCs ask for a lower bound and an upper bound for the cumulative function instead of a single value, which allows $\mathrm{P}(a \le X \le b)$ to be entered directly.

Turning every question into "at most"

A GDC that gives only $\mathrm{P}(X \le r)$ can answer any inequality, once the inequality is rewritten. Because $X$ takes only whole numbers, "fewer than 7" is the same as "at most 6", and the complement of "at least 7" is "at most 6".

The rule behind the table: subtract $\mathrm{P}(X \le k)$ where $k$ is the largest value not wanted. For $X \ge 6$, the unwanted values are 0 to 5, so the answer is $1 - \mathrm{P}(X \le 5)$. For $3 \le X \le 6$, the unwanted values below the range are 0 to 2, so subtract $\mathrm{P}(X \le 2)$ from $\mathrm{P}(X \le 6)$.

What to write down

Because the GDC does the calculation, the written working has to show the model and the probability being found. Three things are written: the random variable in words (if the question has not defined it), the distribution with its parameters, and the probability statement in mathematical notation, followed by the value.

Calculator commands such as binomcdf(25, 0.04, 3) are not mathematical notation; the statement $\mathrm{P}(X \le 3)$ says the same thing in a form any reader can follow.

Counting failures instead of successes

Any question about successes can be turned into one about failures: $r$ successes in $n$ trials is the same event as $n - r$ failures. If $X \sim \mathrm{B}(n, p)$ counts successes, then $Y = n - X$ counts failures and $Y \sim \mathrm{B}(n, 1 - p)$. Either variable gives the same answer, as long as the inequality is converted correctly.

Finding the number of trials

Sometimes $n$ is the unknown: "how many items must be tested to be 95% sure of finding at least one fault?" The probability of at least one success comes from the complement of no successes, which in $n$ trials is $(1 - p)^n$.

The resulting inequality is solved with the GDC, either by solving the equation and rounding in the right direction or by scrolling a table of values of $1 - (1 - p)^n$ for $n = 1, 2, 3, \ldots$.

The shape of the binomial distribution

The shape of $\mathrm{B}(n, p)$ depends on $p$. The bar charts for $n = 10$ and three values of $p$ show the pattern.

• When $p = 0.5$, success and failure are equally likely, so the distribution is symmetric about $\frac{n}{2}$.

• When $p < 0.5$, small numbers of successes are most likely, and the distribution is skewed to the right (a long tail towards large values).

• When $p > 0.5$, the distribution is skewed to the left.

• $\mathrm{B}(n, p)$ and $\mathrm{B}(n, 1 - p)$ are mirror images, because counting successes in one is counting failures in the other.

In every case the peak is at or next to $np$, the mean. As $n$ increases, the skew becomes less noticeable and the bars take on a smooth, symmetric bell shape, even when $p$ is not 0.5. That bell shape is the normal distribution of subtopic 4.9.

The Galton board

In the 1870s Francis Galton built a device, later described in his book Natural Inheritance (1889), in which a binomial distribution forms by itself. Balls fall from a funnel through rows of evenly spaced pins. At each pin a ball bounces left or right with equal probability, independently of its other bounces, and at the bottom it lands in a slot determined by how many times it bounced right.

With $n$ rows of pins, the number of right bounces is $\mathrm{B}(n, 0.5)$, and the number of routes to each slot is a number from row $n$ of Pascal's triangle. The middle slots can be reached by the most routes, so the balls pile up highest there, and the pile forms the bell shape that Galton used to illustrate how many small random effects add up.

Mean and variance of the binomial distribution

The mean: np

A fair coin tossed 100 times is expected to land heads about 50 times; a die rolled 60 times is expected to show about 10 sixes. In general, $n$ trials with success probability $p$ are expected to produce $np$ successes. This is the expected number of occurrences from subtopic 4.5, and for the binomial distribution it is the expected value.

An informal justification: each single trial produces 1 success with probability $p$ and 0 successes with probability $1 - p$, so it contributes $1 \times p + 0 \times (1 - p) = p$ successes on average. Adding the contributions of $n$ trials gives $np$. Applying $\mathrm{E}(X) = \sum x\,\mathrm{P}(X = x)$ to the table for $\mathrm{B}(4, 0.3)$ confirms it:

The variance: np(1 − p)

The mean says where the distribution is centred; the variance says how spread out it is. For a random variable, the variance measures how far the values typically lie from the mean over many repetitions, in the same way that the variance of a data set (subtopic 4.3) measures the spread of the data. Its square root is the standard deviation, in the same units as $X$.

The formula makes sense at the extremes. If $p = 0$ or $p = 1$, every trial has a certain result, $X$ always takes the same value, and $np(1 - p) = 0$: no spread. The factor $p(1 - p)$ is largest when $p = 0.5$, when each trial is most uncertain, and the variance grows in proportion to $n$, because more trials allow more room for variation.

Finding n and p from the mean and variance

Because the mean and variance are both formulas in $n$ and $p$, knowing their values gives two equations. Dividing the variance by the mean cancels $n$:

Since $1 - p$ is between 0 and 1, the variance of a binomial distribution is always smaller than its mean. If a count in real data has a variance larger than its mean, the trials are probably not independent or $p$ is not constant, which is evidence that a binomial model does not fit.

Applications

Overbooking a flight

Airlines know that some passengers who buy tickets do not turn up, so they often sell more tickets than there are seats. The binomial distribution measures the risk of having too many passengers. The figures that follow are hypothetical.

Overbooking is a typical use of the binomial distribution as a decision tool. The airline balances two costs: empty seats, which lose revenue, and too many passengers, who must be compensated. The binomial model turns each choice of $n$ into a probability that can be weighed against those costs, combining the expected values of subtopic 4.7 with binomial probabilities.

Case study: inheriting cystic fibrosis

Cystic fibrosis (CF) is a genetic condition inherited in an autosomal recessive way: a person has CF only if both copies of the relevant gene are faulty. A person with one faulty copy is a carrier and is usually healthy. When both parents are carriers, each parent passes on a faulty copy with probability $\frac{1}{2}$, independently, so each child has probability $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of inheriting CF. The same probability applies to every pregnancy, whatever happened in earlier ones.

For a carrier couple with 4 children, the number of children with CF, $X$, meets the four conditions: a fixed number of children, CF or not, the same probability 0.25 each time, and independence between pregnancies. So $X \sim \mathrm{B}(4, 0.25)$.

The probability that none of the four children has CF is $0.75^4 = 0.316$, so the probability that at least one does is $1 - 0.316 = 0.684$. The expected number with CF is $4 \times 0.25 = 1$, and the most likely single outcome is also exactly 1 child (probability 0.422). A common misunderstanding is that "one in four" guarantees exactly one affected child in a family of four; the distribution shows that this happens less than half the time.

The binomial distribution also models carriers in a population. Among people of European descent, about 1 in 25 is a carrier, and about 1 in 3,000 newborns of that ancestry has CF.

Source: Wikipedia, "Cystic fibrosis", accessed 2026.

In a randomly chosen group of 2000 such people, the number of carriers is modelled by $C \sim \mathrm{B}(2000, 0.04)$.

So such a group would usually contain about 80 carriers, give or take about 9; the probability of 100 or more is $1 - \mathrm{P}(C \le 99) = 0.0153$. If two people with no family history have a child, and each is a carrier with probability $\frac{1}{25}$ independently, the probability that the child has CF is

The model's 1 in 2,500 is close to the observed 1 in 3,000, but not equal. Carrier rates differ between groups (estimates range from about 1 in 20 to 1 in 30), people do not choose partners at random with respect to ancestry, and newborn figures are averages over regions with different carrier rates. The comparison shows how a simple model built from independent probabilities can explain the size of a real rate while leaving room for the assumptions to be questioned.

Exam focus

Binomial questions appear on both Paper 1 (2–7 marks) and Paper 2, often as the last part of a longer question in which $p$ is found first from another part, for example a normal-distribution probability (subtopic 4.9) or a probability from a tree diagram. The command terms are find, calculate, write down (the mean $np$), state (an assumption or the distribution), determine and justify (whether a binomial model is suitable). Recurring shapes are:

• write down or find the expected number of successes, $np$;

• find $\mathrm{P}(X = r)$, then a cumulative probability with a strict or reversed inequality ("fewer than", "more than", "at least");

• find the smallest $n$ for a required probability, or $n$ from a required mean;

• state an assumption needed for the binomial model, in context;

• combine a binomial probability with expected value or cost (Paper 2).

The GDC does the calculation; the working that earns marks is $X \sim \mathrm{B}(n, p)$ with numbers and the probability statement with the correct bound.

Part (a) is answered in context: "people attend independently of each other" earns credit, but "the trials are independent" may not, because the question asks about these people. Assumptions are about the real situation, not about the formula.

When $p$ has been found in an earlier part, use its unrounded value from the GDC, not the 3-significant-figure version. A binomial probability with large $n$ can change in its third significant figure when $p$ is rounded, and the final accuracy mark may be lost.

Practice questions

1. For each situation, state whether $X$ can be modelled by a binomial distribution, giving a reason. (a) $X$ is the number of girls among the next 30 babies born in a large hospital. (b) $X$ is the number of aces when 4 cards are dealt from a standard pack. (c) $X$ is the number of rolls of a die needed to get a six. [3]

2. $X \sim \mathrm{B}(12, 0.35)$. Find (a) $\mathrm{P}(X = 4)$, (b) $\mathrm{P}(X \le 4)$, (c) $\mathrm{P}(X \ge 6)$, (d) $\mathrm{P}(3 < X \le 7)$. [6]

3. A footballer scores each penalty kick with probability 0.78, independently. He takes 15 penalties in a training session. (a) Write down the expected number she scores. [1] (b) Find the probability that she scores at least 12. [2] (c) Find the probability that she scores fewer than 10. [2]

4. $X \sim \mathrm{B}(50, 0.12)$. Calculate the mean, the variance and the standard deviation of $X$. [3]

5. A binomial random variable has mean 7.2 and variance 5.04. Find $n$ and $p$. [4]

6. A test has 20 multiple-choice questions, each with 5 options. A student guesses every answer. (a) Find the expected number of correct answers. [1] (b) The pass mark is 8. Find the probability that the student passes. [2]

7. A fair die is rolled $n$ times. Find the smallest value of $n$ for which the probability of at least one six is greater than 0.9. [3]

8. Seeds of a plant germinate with probability 0.9, independently. A packet holds 50 seeds and is guaranteed to have at least 42 that germinate. (a) Find the probability that a packet fails the guarantee. [2] (b) A garden centre buys 10 packets. Find the probability that at most one packet fails the guarantee. [3]

9. A class has 20 students, 3 of whom are left-handed. Explain why the number of left-handed students among 4 chosen at random from the class does not follow a binomial distribution, and why the number of left-handed people among 4 chosen at random from a large city can be modelled by one. [2]

Summary

• A Bernoulli trial has two outcomes, success (probability $p$) and failure (probability $1 - p$).

• $X \sim \mathrm{B}(n, p)$ counts successes when there is a fixed number $n$ of trials, two outcomes, a constant $p$ and independent trials; $X$ takes the values $0, 1, \ldots, n$.

• $\mathrm{P}(X = r) = \binom{n}{r}p^r(1 - p)^{n - r}$

• Binomial probabilities are found with the GDC: the probability function for $\mathrm{P}(X = r)$ and the cumulative function for $\mathrm{P}(X \le r)$.

• Convert inequalities using whole numbers: $\mathrm{P}(X < k) = \mathrm{P}(X \le k - 1)$, $\mathrm{P}(X \ge k) = 1 - \mathrm{P}(X \le k - 1)$

• Write $X \sim \mathrm{B}(n, p)$ and the probability statement, never calculator commands.

• $\mathrm{E}(X) = np$ and $\mathrm{Var}(X) = np(1 - p)$; the standard deviation is $\sqrt{np(1 - p)}$

• The distribution is symmetric when $p = 0.5$, skewed right for small $p$ and left for large $p$, and becomes bell-shaped as $n$ grows.

• A binomial model's conclusions depend on its assumptions, which should be stated and judged in context.

Self-check

• To do:

  I can state the four conditions for a binomial distribution and check them in a context.

• To do:

  I can explain why sampling without replacement from a small group is not binomial.

• To do:

  I can decide which outcome is the "success" and write $X \sim \mathrm{B}(n, p)$ with the correct numbers.

• To do:

  I can explain, with a tree diagram, why $\mathrm{P}(X = 1) = 3p(1 - p)^2$ when $n = 3$.

• To do:

  I can read binomial coefficients from Pascal's triangle and say what they count.

• To do:

  I can find $\mathrm{P}(X = r)$ and $\mathrm{P}(X \le r)$ on my GDC.

• To do:

  I can rewrite "fewer than", "more than", "at least" and "between" as cumulative probabilities with the correct bound.

• To do:

  I can answer a question by counting failures instead of successes.

• To do:

  I can find the smallest number of trials needed for a probability of at least one success.

• To do:

  I can calculate the mean, variance and standard deviation of a binomial distribution and interpret them.

• To do:

  I can find $n$ and $p$ from a given mean and variance.

• To do:

  I can describe the shape of $\mathrm{B}(n, p)$ for different values of $p$.

• To do:

  I can state an assumption of a binomial model in context and comment on whether it is realistic.

• To do:

  I can write binomial working in notation, without calculator syntax.

[2]2 marks

(b) No. Taking three together is the same as taking them one at a time without replacement. After a green counter is taken, the probability of green falls from $\frac{6}{10}$ to $\frac{5}{9}$, so $p$ is not constant.

(c) No. The number of trials is not fixed in advance: it is the quantity being measured.