Binomial Distribution

We begin with the binomial distribution. Here we consider an experiment that is repeated many times. There are two possible outcomes: $A$ and $B$. The probability for outcome $A$ is $p$ and the probability for outcome $B$ is $1-p$. We assume that each experiment has the same probability for each outcome and that there is no correlation between the outcome of one experiment and that of another. We may then ask the question, out of $N$ repetitions of the experiment, what is the probability that we get $A$ exactly $k$ times? For example, suppose $N=4$ and $k=3$. The answer is found by first asking for the probability for a particular sequence of outcomes $AABA$, for example. The probability is just the product of the probabilities for each event: $pp(1-p)p$. This statement makes use of the fact that there is no correlation between one experiment and another. Since our question doesn't ask for a particular order of outcomes, but just any order that yields $3$ $A$'s out of $4$ trials, we then ask how many different ways there are of getting $3$ $A$'s. We can enumerate them: $BAAA$, $ABAA$, $AABA$, and $AAAB$. Since the probability for each is the same, the probability for any of them is four times the probability for just one of them. So the probability is $4p^{3}(1-p)$ to get 3 $A$'s out of 4 trials. The general expression is called the binomial distribution. The probability for getting $k$ $A$'s (and $N-k$ $B$'s) out of $N$ trials is

$$P(k,N) = \frac{N!}{k!(N-k)!}p^{k}(1-p)^{N-k}.$$ (1)

Notice that the binomial probabilities generate the binomial series, which adds up to 1, as it should:

$$\sum_{k=0}^{N} P(k,N) = [p + (1-p)]^{N} = 1.$$ (2)