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