The Binomial Distribution

Starting point: what you already know

• Probability basics: $P(\text{not }A)=1-P(A)$, and for independent events $P(A\text{ and }B)=P(A)\times P(B)$.
• Combinations (Combinations chapter): $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$ counts the ways to choose $r$ things from $n$.
• Discrete random variables (Mathematics Advanced): a table of values and probabilities that add to 1, with mean $E(X)=\sum xP(X=x)$ and variance $\mathrm{Var}(X)=E(X^2)-[E(X){]}^2$.
• The binomial theorem: $(q+p{)}^n=\sum (\genfrac{}{}{0ex}{}{n}{r})p^r{q}^{n-r}$.

Bernoulli trials

A Bernoulli trial has exactly two outcomes, called success and failure. The probability of success is $p$, and of failure is $1-p$, sometimes written $q$.

Let $X=1$ for a success and $X=0$ for a failure. Then:

• $E(X)=0\times (1-p)+1\times p=p$
• $\mathrm{Var}(X)=E(X^2)-p^2=p-p^2=p(1-p)$

The binomial distribution

Now repeat a Bernoulli trial $n$ times and count the successes, $X$. This is a binomial distribution, written $X\sim \mathrm{Bin}(n,p)$, when:

• there are a fixed number $n$ of trials,
• each has only two outcomes,
• the trials are independent,
• $p$ is the same every time.

Why it works: one particular order of $r$ successes and $n-r$ failures has probability $p^r(1-p{)}^{n-r}$. There are $\left(\genfrac{}{}{0ex}{}{n}{r}\right)$ different orders.

Mean and variance

This makes sense: $X$ is the sum of $n$ separate Bernoulli trials, each with mean $p$ and variance $p(1-p)$.

For the basketball player: $E(X)=3$ shots, $\mathrm{Var}(X)=2.1$, $\sigma \approx 1.45$ shots.

Quick summary

• A Bernoulli trial has two outcomes, with mean $p$ and variance $p(1-p)$.
• Binomial needs: fixed $n$, two outcomes, independent trials, same $p$.
• $P(X=r)=\left(\genfrac{}{}{0ex}{}{n}{r}\right)p^r(1-p{)}^{n-r}$.
• $E(X)=np$ and $\mathrm{Var}(X)=np(1-p)$.
• "At least one" is $1-P(X=0)$.

Quiz