Binomial distribution for n = 3

We first derive the probability function for one specific binomial distribution,

\[ X \;\; \sim \;\; \BinomDistn(n=3, \pi) \]

The eight possible sequences of three successes and failures are:

{SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}

Since the successive Bernoulli trials are independent, we can multiply the probabilities of successes and failures to get the probability for each such sequence. For example,

\[ P(\text{SSF}) \;=\; \pi \times \pi \times (1-\pi) \;=\; \pi^2(1-\pi)^1\]

The following table shows the probabilities for all possible sequences.

Sequence P(Sequence) Number of
successes, X
SSS \(\pi^3(1-\pi)^0\) 3
SSF \(\pi^2(1-\pi)^1\) 2
SFS \(\pi^2(1-\pi)^1\) 2
FSS \(\pi^2(1-\pi)^1\) 2
SFF \(\pi^1(1-\pi)^2\) 1
FSF \(\pi^1(1-\pi)^2\) 1
FFS \(\pi^1(1-\pi)^2\) 1
FFF \(\pi^0(1-\pi)^3\) 0

Adding the probabilities for each possible value of \(X\), the number of successes, gives the following probability function:

x p(x)
3 \(1 \times \pi^3(1-\pi)^0\)
2 \(3 \times \pi^2(1-\pi)^1\)
1 \(3 \times \pi^1(1-\pi)^2\)
0 \(1 \times \pi^0(1-\pi)^3\)

Note that the coefficient for \(x=2\) is the number of different sequences of successes and failures resulting in 2 successes. This is the number of different ways to choose 2 out of the 3 Bernoulli trials to be successes,

\[ {3 \choose 2} = \frac {3!} {2!(3-2)!}\]

A similar formula holds for the other possible values, \(x\), so the whole probability function can be expressed with a single mathematical function,

\[ p(x)= {3 \choose x} \pi^x(1-\pi)^{1-x} \quad \quad \text{for } x=0, 1, \dots, 3 \]

General formula for binomial probability function

We now generalise with a formula that holds for any \(n\).

Binomial probability function

If \(X\) has a \(\BinomDistn(n, \pi)\) distribution, its probability function is

\[ p(x)= {n \choose x} \pi^x(1-\pi)^{n-x} \qquad \text{for } x=0, 1, \dots, n \]

Since each Bernoulli trial is independent, any sequence of \(n\) successes and failures containing \(x\) successes and \((n-x)\) failures has probability

\[ \pi^x(1-\pi)^{n-x} \]

Since there are \(\displaystyle{n \choose x}\) different sequences with \(x\) successes and \((n-x)\) failures, the probability of \(x\) successes is

\[ p(x)= {n \choose x} \pi^x(1-\pi)^{n-x} \qquad \text{for } x=0, 1, \dots, n \]