Theorem 7.2.1 Derivation of Binomial Probability Function
For R = {0, 1, ..., n},
\begin{equation*}
f(x) = \binom{n}{x}p^x(1-p)^{n-x}
\end{equation*}
Section 7.2 Binomial Distribution
Consider a sequence of n independent Bernoulli trials with the likelihood of a success p on each individual trial stays constant from trial to trial with \(0 \lt p \lt 1 \text{.}\) If we let the variable \(X\) measure the number of successes obtained when doing a fixed number of trials n with \(R = \{ 0, 1, ..., n \}\text{,}\) then the resulting distribution of probabilities is called a Binomial Distribution.
You can of course get specific values and graph the Binomial Distribution using R as well...
Proof
Since successive trials are independent, then the probability of X successes occurring within n trials is given by
\begin{equation*}
P(X=x) = \binom{n}{x}P(SS...SFF...F) =
\binom{n}{x}p^x(1-p)^{n-x}
\end{equation*}
Theorem 7.2.2 Verification of Binomial Distribution Formula
\begin{equation*}
\sum_{x \in R} f(x) = \sum_{x=0}^n \binom{n}{x}p^x(1-p)^{n-x} = 1.
\end{equation*}
Proof
Using the Binomial Theorem with a = p and b = 1-p yields
\begin{equation*}
\sum_{x=0}^n \binom{n}{x}p^x(1-p)^{n-x} = (p + (1-p))^n = 1
\end{equation*}
Utilize the interactive cell below to compute f(x) and F(x) for the Binomial distribution
Theorem 7.2.3 Binomial Distribution Statistics
For the Binomial Distribution
\begin{equation*}
\mu = np
\end{equation*}
\begin{equation*}
\sigma^2 = np(1-p)
\end{equation*}
\begin{equation*}
\gamma_1 = \frac{1-2p}{\sqrt{np(1-p)}}
\end{equation*}
\begin{equation*}
\gamma_2 = \frac{1-6p(1-p)}{np(1-p)} + 3
\end{equation*}
Proof
For the mean,
\begin{align*}
\mu & = E[X] \\
& = \sum_{x=0}^{n} {x \binom{n}{x} p^x (1-p)^{n-x}}\\
& = \sum_{x=1}^{n} {x \frac{n(n-1)!}{x(x-1)!(n-x)!} p^x (1-p)^{n-x}}\\
& = np \sum_{x=1}^{n} {\frac{(n-1)!}{(x-1)!((n-1)-(x-1))!} p^{x-1} (1-p)^{(n-1)-(x-1)}}
\end{align*}
Using the change of variables \(k=x-1\) and \(m = n-1\) yields a binomial series
\begin{align*}
& = np \sum_{k=0}^{m} {\frac{m!}{k!(m-k)!} p^k (1-p)^{m-k}}\\
& = np (p + (1-p))^m = np
\end{align*}
For the variance,
\begin{align*}
\sigma^2 & = E[X(X-1)] + \mu - \mu^2 \\
& = \sum_{x=0}^{n} {x(x-1) \binom{n}{x} p^x (1-p)^{n-x}} + np - n^2p^2\\
& = \sum_{x=2}^{n} {x(x-1) \frac{n(n-1)(n-2)!}{x(x-1)(x-2)!(n-x)!} p^x (1-p)^{n-x}} + np - n^2p^2\\
& = n(n-1)p^2 \sum_{x=2}^{n} {\frac{(n-2)!}{(x-2)!((n-2)-(x-2))!} p^{x-2} (1-p)^{(n-2)-(x-2)}} + np - n^2p^2
\end{align*}
Using the change of variables \(k=x-2\) and \(m = n-2\) yields a binomial series
\begin{align*}
& = n(n-1)p^2 \sum_{k=0}^{m} {\frac{m!}{k!(m-k)!} p^k (1-p)^{m-k}} + np - n^2p^2\\
& = n(n-1)p^2 + np - n^2p^2 = np - np^2 = np(1-p)
\end{align*}
The skewness and kurtosis can be found similarly using formulas involving E[X(X-1)(X-2)] and E[X(X-1)(X-2)(X-3)]. The complete determination is performed using Sage below.
The following uses Sage to determine the general formulas for the Binomial distribution.
Flipping Coins
Suppose you flip a coin exactly 20 times. Determine the probability of getting exactly 10 heads and then determine the probability of getting 10 or fewer heads.
SolutionThis is binomial with n = 20, p = 1/2 and you are looking for f(10). With these values
\begin{equation*}
f(10) = \binom{20}{10} \cdot \left ( \frac{1}{2} \right )^{10} \cdot \left ( \frac{1}{2} \right )^{20-10} = \frac{46189}{262144} \approx 0.176
\end{equation*}
Notice, the mean for this distribution is also 10 so one might expect 10 heads in general. Next, to determine the probability for 10 or fewer heads requires F(10) = f(0) + f(1) + ... + f(10). There is no "nice" formula for F but this calculation can be performed using a graphing calculator, such as the TI-84 with F(x) = binomcdf(n,p,x). In this case, F(10) = binomcdf(20,1/2,10) = 0.588.