Binomial distributions

This is about probability of having exactly, at most, or at least $k$ successes in $n$ trials.

Updated 2026-05-13 on GDC (calculator) usage and removed substituting values into binomial distribution formula

When to Use
Definition
Calculator
Expected value
Variance and standard deviation
Mode and median
Tips
Practice

When to Use

If it exhibits the following symptoms, binomial distribution may be for you

for probability of having a particular number of “successes” out of $n$ independent tries, or
for the expected number of “successes” out of $n$ independent tries

For picking items from a large pool (in the ~thousands), selection without replacement can still be approximate by binomial distribution.

Definition

Random variable $X$ is binomially distributed is notated as

X \sim \text{B}(n, p)

with $n$ trials or attempts, and probability $p$ of success for each independent trial. It has the probability distribution

\text{P}(X = x) = \binom nx p^x(1-p)^{n-x}

where $\displaystyle \binom nx$ is the binomial coefficient or $^nC_x$ .

This formula is not on the formula booklet and not required to know But it is very similar to the one for binomial expansion.

The $\displaystyle \binom nx$ counts the different sequences of $x$ successes and $n - x$ failures.

Note: In $n$ tries, $x$ could take on $n + 1$ values from $0$ to $n$ .

Binomial distribution is about defining the success for a single trial, and treating different successes separately.

Calculator

Keep in mind very important that < vs ≤ are not interchangeable for binomial. For Nspire and Casio, write cumulative probability as P(a ≤ X ≤ b). For TI 84, this is P(X ≤ b) - P(X ≤ a - 1).

Arguments in $\color{grey}\text{grey}$ are optional.

	navigation	P(X = x)	P(a ≤ X ≤ b), or P(X ≤ x)	least x with P(X ≤ x) ≥ cdf	least n with P(X ≤ x) ≤ cdf
TI-84	2nd vars	A.binompdf(n, p, x)	B.binomcdf(n, p, x)	C.invBinom(cdf, n, p)
Nspire	menu 6.stats, then 5.distr	A.binomPdf(n, p, x)	B.binomCdf(n, p, $\color{grey}a$ , b)	C.invBinom(cdf, n, p)	D.invBinomN(cdf, p, x)
CG50 / 9860GIII	`I.DIST` / `2.STAT` or `1.Run/Mat` OPTN `stat`	Bpd(x, n, p)	Bcd(x, n, p)	InvB(cdf, n, p), `Run-Mat` and `Stat` only
CG50 only			Bcd( $\color{grey}a,$ b, n, p)

$a$ is by default 0, which reduces to the TI 84 / 9860GIII case.

Casio: Use I.Distribution app by default. Use 2. Stat if needing InvN. Only use 1.Run-Mat OPTN if one of n or p is unknown. The 9860GIII 2.Statistics and Bcd can only find P(X ≤ b) and not P(a ≤ X ≤ b), but the I.Distribution app works fine.

Example: A coin is manufactured to land on head 52% of the time and the rest on its tail.

a) 800 people are each flipping such a coin 10 times in a row. Find the probability that exactly one person gets all tails or all heads.

b) Find the number of people needed to have a 95% chance of at least two people each flipping at least eight tails or at least eight heads in ten flips.

a) Let $X$ be the number of heads in ten flips, such that ${X \sim \text{B}(10, 0.52)}$ .

\text{P}(X = 0) + \text{P}(X = 10) \approx 0.0020948

where we can store this in variable (Alpha letter) P, and when we find binompdf or bpd we can use our variable.

Then let $Y$ be number of times someone gets all tails or heads in ten flips, out of $800$ people. ${Y \sim \text{B}(800, 0.0020948)}$ .

\begin{align*} \text{P}(Y = 1) \approx 0.314 \qed \end{align*}

b) At least 8 tails or at least 8 heads means

\text{P}(X \leq 2) \text{P}(X \geq 8)

using the $X$ from previous part.

TI 84 can only do cumulative binomial distribution probability from 0 to $x$ , so this is

\begin{align*}&\,\text{P}(X \leq 2) + 1 - \text{P}(X \leq 7)\\ \approx&\,0.11218749 \end{align*}

where we can store this in another variable, eg Q.

Now we can define ${Z \sim \text{B}(n, 0.11218749)}$ for number of people getting at least 8 tails or at least 8 heads and solve

\text{P} (Z \geq 2) \geq 0.95

or equivalently

\text{P}(Z \leq 1) \lt 0.05

As $n$ is a discrete value, taking on only positive integers, this cannot be solved via solver, but rather through a table of values.

$n \quad$	$\text{P}(Z \leq 1)$
$39$	$0.0572$
$40$	$0.0519$
$41$	$0.0470$
$42$	$0.0426$

As $n = 41$ is the first value to drop below $0.0500$ , the answer is $n = 41 \qed$

Alternatively, you can also use round( or int( to first plot the function, replacing n with round(n).

Alternatively using TI Nspire, we can use invBinomN.

Expected value

The expected value is

\text{E}(X) = np

This is why

\text{expected count} = \text{\# trials}\times\text{probability}

Variance and standard deviation

\text{Var}(X) = np(1-p)

\sigma = \sqrt{np(1-p)}

Some ways that they could be used on exam is if you are given expected value and variance and you need to solve for $n$ and $p$ .

Mode and median

The mode and median are very close to $np$ . Precise formulas are beyond scope of the course. If asked on exam, use methods discussed in discrete random variables to investigate.

The median is not necessarily equal to the mode(s).

Tips

Does it say at most ( $\leq N$ ), at least ( $\geq N$ ), less than( $< N$ ), more than ( $> N$ ), or equal to ( $=N$ )?
See also probability formulas.
See also binomial expansion.
While for large $n$ , the binomial distribution can be approximated by a normal distributions, the normal approximation cannot reliably obtain correct answers to 3 sf.

Practice

Practice evaluating binomial probabilities using Bolt ⚡ Binomial.

Previous : Discrete random variables

Next : Normal distributions