Let $X$ be a discrete random variable. Then $X \sim \mathrm{Poisson}(\lambda)$ means $X$ takes on a Poisson distribution with parameter $\lambda$.
The pmf of the Poisson distribution is \[f(x) = \mathrm{P}[X=x] = \dfrac{\lambda^x e ^{-\lambda}}{x!}\] where $ e$ is the base of the natural logarithm.
If the question gives you an average rate, rather than an exact probability, we cannot use the binomial distribution so we must use the Poisson distribution. In these types of questions, $\lambda$ is this average rate. For example, if a lecturer makes three mistakes per lecture, then $\lambda$ would equal $3$.
If $X$ is a discrete random variable, the binomial distribution may not always be the most efficient way of reaching an answer. For example, if $n$ is $100000$, calculators may struggle to determine the binomial coefficient needed to perform the binomial distribution. At times like this we rely on the Poisson distribution. The Poisson distribution is preferable to the binomial distribution when $n$ is large and $p$ is small. However the binomial distribution will give an exact result whereas the Poisson distribution will only give an approximation.
Given a discrete random variable $X$, the mean $\mu$ is equal to the expected value of $X$, $\mathrm{E} [X]$. In this case \[\mu = \mathrm{E} [X] = \lambda\] and the variance $\mathrm{Var}[X] = \mathrm{E} [X^2] - \mu $ is \[\mathrm{Var}[X] = \lambda .\]
Let $X$ and $Y$ be two independent Poisson random variables with $X \sim \mathrm{Poisson}(\lambda_1)$ and $ Y \sim \mathrm{Poisson}(\lambda_2)$.
i.e it is true that $\mathrm{P}[X=x$ and $ Y=y] = \mathrm{P}[X=x]\cdot \mathrm{P}[Y=y]$ for all $x, y \in \mathbb{N}$.
Then the random variable $Z= X+Y$ is Poisson distributed with parameter $\lambda_1 + \lambda_2$. This means $Z \sim \mathrm{Poisson}(\lambda_1 + \lambda_2)$. So \[\mathrm{P}[Z=z] = \dfrac{(\lambda_1 + \lambda_2)^z e ^{\lambda_1 + \lambda_2}}{z!}.\]
It is found that $1$% of the lie detector tests on a popular daytime TV show return false results. Calculate the probability that in a sample of $200$:
a) exactly $2$ lie detector tests return the wrong result.
b) $2$ or more lie detector tests return the wrong result.
a) $n=200$ is large and $p=0.01$ is small so we choose to approximate using the Poisson distribution.
Let $X$ be the discrete random variable where $X =$ “the number of lie detector results which are wrong.” $\lambda = np = 200 \times 0.01 = 2$. Then $X \sim \mathrm{Poisson}(2)$. We want the probability that $2$ results are wrong so we find $\mathrm{P}[X=2]$ by substituting into the pmf as follows: \[\mathrm{P}[X=2] \approx f(2) = \dfrac{2^2 e ^{-2} }{2!} = 0.2706.\]
So we have a probability of $0.27$ that exactly $2$ lie detector tests returned the wrong result.
How does this compare to the binomial distribution? Lets try $X \sim \mathrm{Bin}(200,0.01)$ \[\mathrm{P}[X=2] = f(2) = \displaystyle \binom {200}{2} 0.01^2 \times 0.99^{198} = 0.2720.\]
So it seems the Poisson distribution is a very good estimator for the binomial distribution when $n$ is large and $p$ is small.
b) Again $X \sim \mathrm{Poisson}(2)$ but this time we are looking for $\mathrm{P}[X \geq 2]$. Note that \[\mathrm{P}[X \geq 2] = 1-\mathrm{P}[X <2] = 1- (\mathrm{P}[X =0] + \mathrm{P}[X =1] ) .\] Using the pmf of the Poisson distribution we can work out \[\mathrm{P}[X =0] \approx \dfrac{2^0 e ^{-2} }{0!} = 0.1353\] and \[\mathrm{P}[X =1] \approx \dfrac{2^1 e ^{-2} }{1!} = 0.2707 .\]
So \begin{align} \mathrm{P}[X \geq 2] &= 1-\mathrm{P}[X <2] \\ &\approx 1- (0.1353 + 0.2707) \\ &= 0.5940. \end{align}
In this video, Daniel Organisciak computes a probability using the Poisson Distribution.
In this video, Dr. Malcolm Farrow uses the Poisson distribution to estimate the binomial distribution.
In this video, Dr Lee Fawcett solves a problem using the Poisson distribution.
This workbook produced by HELM is a good revision aid, containing key points for revision and many worked examples.