Binomial, Poisson, and Multinomial Distributions

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.