## Binomial, Poisson, and Multinomial Distributions

First, we will talk about binomial probabilities, how to compute their cumulatives, and the mean and standard deviation. Then, we will introduce the Poisson probability formula, define multinomial outcomes, and discuss how to compute probabilities by using the multinomial distribution.

The binomial distribution consists of the probabilities of each of the possible numbers of successes on $N$ trials for independent events that each have a probability of п (the Greek letter pi) of occurring. For the coin flip example, $N=2$ and $\Pi=0.5$. The formula for the binomial distribution is shown below:

\begin{align*}P(x)=\frac{N !}{x !(N-x) !} \pi^{x}(1-\pi)^{N-x}\end{align*}

where $P(x)$ is the probability of $x$ successes out of $N$ trials, $N$ is the number of trials, and $п$ is the probability of success on a given trial. Applying this to the coin flip example,

\begin{align*}\begin{aligned}&P(0)=\frac{2 !}{0 !(2-0) !}\left(.5^{0}\right)(1-.5)^{2-0}=\frac{2}{2}(1)(.25)=0.25 \\&P(1)=\frac{2 !}{1 !(2-1) !}\left(.5^{1}\right)(1-.5)^{2-1}=\frac{2}{1}(.5)(.5)=0.50 \\&P(2)=\frac{2 !}{2 !(2-2) !}\left(.5^{2}\right)(1-.5)^{2-2}=\frac{2}{2}(.25)(1)=0.25\end{aligned}\end{align*}

If you flip a coin twice, what is the probability of getting one or more heads? Since the probability of getting exactly one head is $0.50$ and the probability of getting exactly two heads is $0.25$, the probability of getting one or more heads is $0.50+0.25=0.75$.

Now suppose that the coin is biased. The probability of heads is only $0.4$. What is the probability of getting heads at least once in two tosses? Substituting into the general formula above, you should obtain the answer .64.