Binomial Distribution

Intuition

Imagine flipping a coin $n$ times and counting heads. More generally, any time you repeat a yes/no experiment a fixed number of times with the same success probability on each trial, the total number of successes follows a binomial distribution. It’s one of those distributions you reach for almost reflexively once you recognize the setup: fixed trials, two outcomes, independence, constant $p$.

Definition

A random variable $X$ follows a binomial distribution $X\sim B(n,p)$ if it represents the number of successes in $n$ independent Bernoulli trials, each with success probability $p$ and failure probability $q=1-p$.

$X$ can take integer values $x=0,1,2,\dots ,n$.

Four conditions must hold for a binomial model to apply:

1. The number of trials $n$ is fixed in advance.
2. Each trial has exactly two outcomes (success or failure).
3. Trials are mutually independent.
4. The probability of success $p$ is constant across all trials.

Key Formulas

Probability Mass Function (PMF):

$b(x;n,p)=\left(\genfrac{}{}{0ex}{}{n}{x}\right)p^x{q}^{n-x},x=0,1,\dots ,n$

where $\left(\genfrac{}{}{0ex}{}{n}{x}\right)=\frac{n!}{x!(n-x)!}$ is the binomial coefficient.

Mean:

$\mu =np$

Variance:

${\sigma }^2=npq$

Standard Deviation:

$\sigma =\sqrt{npq}$

The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{x}\right)$ counts the number of ways to choose which $x$ of the $n$ trials are successes.

Example

Four identical electronic components are subjected to a shock test. Each has a 75% chance of surviving ($p=0.75$). What is the probability that exactly 2 survive?

$P(X=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)(0.75{)}^2(0.25{)}^2=6\cdot 0.5625\cdot 0.0625=0.2109$

So there is roughly a 21% chance that exactly 2 of the 4 components survive.

The expected number of survivors is $\mu =4\cdot 0.75=3$, with variance ${\sigma }^2=4\cdot 0.75\cdot 0.25=0.75$.

We can also compute the probability that at least 3 survive:

$P(X\ge 3)=P(X=3)+P(X=4)=\left(\genfrac{}{}{0ex}{}{4}{3}\right)(0.75{)}^3(0.25{)}^1+\left(\genfrac{}{}{0ex}{}{4}{4}\right)(0.75{)}^4(0.25{)}^0$

$=4\cdot 0.4219\cdot 0.25+1\cdot 0.3164\cdot 1=0.4219+0.3164=0.7383$

So there is about a 74% chance that 3 or more components survive the test.

Why It Matters in CS

Anytime you’re counting “how many out of $n$” in a system, you’re probably looking at a binomial. Packet losses across $n$ transmissions, bit errors in a block of encoded data, defective chips on a wafer - all binomial if the trials are independent with constant $p$.

The place you’ll encounter it most directly is A/B testing. When 5,000 users visit a page and 312 convert, that conversion count is $\text{Bin}(5000,p)$. The entire statistical significance calculation rests on this model. Understanding the binomial also tells you why small-sample A/B tests are so unreliable: the variance $npq$ is large relative to the mean when $n$ is small, so the observed conversion rate swings wildly between runs.

Tip

When $n$ is large and $p$ is moderate, computing $\left(\genfrac{}{}{0ex}{}{n}{x}\right)$ directly overflows most integer types. In practice you’d use the normal approximation ($np\ge 5$ and $nq\ge 5$) or work in log-space.

Related Notes

• Probability Distributions - overview that introduces the binomial alongside other distributions
• Geometric Distribution - models trials until the first success rather than counting successes in $n$ trials
• Poisson Distribution - approximates the binomial when $n$ is large and $p$ is small

Note: Probability Distributions covers the binomial at an overview level. This note provides a deeper treatment with worked examples and CS applications.