Let $X$ be the discrete random variable which counts the number of successes obtained from $n$ Bernoulli trials. Then we denote $X \sim \mathrm{Bin}(n,p)$ to mean $X$ follows the *binomial distribution* with $n$ trials, each of which have a probability $p$ of success.

The probability mass function of the binomial distribution is \[f(x) = \mathrm{P}[X=x] = \displaystyle\binom nx p^x (1-p)^{n-x}\text{.}\]

If $X$ is a binomially distributed discrete random variable then the mean of $X$ \[\mu = \mathrm{E} [X] = np\] and the variance of $X$ is \[\mathrm{Var}[X]= np(1-p)\text{.}\]

A student rolls three dice. What is the probability that he gets two sixes?

Let $X$ be the discrete random variable denoting the number of sixes obtained. We want the probability of obtaining two sixes so we are concerned with $\mathrm{P}[X=2]$. Because we have $n=3$ trials and a probability of success $p=\frac{1}{6}$, $\displaystyle X \sim \mathrm{Bin}(n,p)$ or, more specifically, $X \sim \mathrm{Bin}(3, \frac{1}{6})$. Now, using the probability mass function of the binomial distribution, \[\mathrm{P}[X=2] = f(2) = \binom 32 \times \left( \frac{1}{6} \right)^2 \times \left( 1-\frac{1}{6}\right) ^{3-2} = 3 \times \left( \frac{1}{6} \right)^2 \times \left( \frac{5}{6} \right) ^1 = \frac{5}{72} \text{.}\]

Compare this to how we worked it out earlier. It is already more efficient. Now what if we rolled the die 40 times and we wanted to find the probability of obtaining 19 sixes? With the binomial distribution it is simply a case of substituting into the probability mass function and will be just as quick as the above calculation.

The probability that any player on the England football team scores a penalty is $0.4$.

From $5$ penalties;

- a) What is the probability that England score fewer than $2$ penalties?
- b) What is the probability that England score $2$ or more penalties?

Let $X$ be a discrete random variable where $X =$ “the number of penalties England score”.

a) We know that \[\mathrm{P}[X<2] = \mathrm{P}[X=0]+\mathrm{P}[X=1].\] Now \begin{align} \mathrm{P}[X=0] &= f(0) =\binom 50 \times 0.4^0 \times 0.6^{5} = 1 \times 1 \times \ 0.6^5 = 0.0778\text{,}\\\\ \mathrm{P}[X=1] &= f(1) = \binom 51 \times 0.4^1 \times 0.6^{5-1} = 5 \times 0.4 \times 0.6^4 = 0.2592. \end{align} Therefore \[\mathrm{P}[X<2] = \mathrm{P}[X=0]+\mathrm{P}[X=1] = 0.0778+0.2592 = 0.3370\text{.}\] b) First notice that \begin{align}\mathrm{P}[X\geq 2] &= 1 - \mathrm{P}[X<2] \\ &= 1 - 0.3370\text{.} \end{align}

Therefore \[\mathrm{P}[X\geq 2] = 0.6630 .\]

In this video, Dr Lee Fawcett solves a problem using the binomial distribution.

In this video, Daniel Organisciak finds probabilities using the binomial distribution.

This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.

- Binomial distribution at Pennsylvania State University