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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.

Now, let's determine the actual probability function for this distribution.

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*}

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*}

The following interactive cell illustrates the range of choices for a Binomial setup with fixed N and fixed p:

Example 7.2.3. Flipping a coin a fixed number of times.

Let's consider a simple example for flipping coins. Indeed, suppose you flip a coin exactly 20 times and need to determine the probability of getting exactly 10 heads.

This 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*}

Note that the mean for this distribution is also 10 so one might expect 10 heads in general

If you rather would prefer to determine the probability of getting 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.

Checkpoint 7.2.4. WebWork - Binomial.
Example 7.2.5. Testing critical components.

Often one will test a critical system components for failure and toward that end collect a sample of 100 of these components from the manufacturer. Suppose the component is listed as having a p = 0.01 probability of breaking and you want to know the likelihood that at most 1 of the tested components actually fails when tested. You find it reasonable to presume that different components succeed or fail independently of each other. So, you can model this situation with a binomial distribution.

If \(X\) measures the number of components that fail when tested, the specific probability function is given by

\begin{equation*} f(x) = \binom{100}{x} 0.01^x 0.99^{100-x}. \end{equation*}

The probability that at most one component fails is then given by

\begin{equation*} F(1) = f(0) + f(1) = \binom{100}{0} 0.01^0 0.99^{100} + \binom{100}{1} 0.01^1 0.99^99 \\ = 0.99^{100} + 100 \cdot 0.01 \cdot 0.99^{99} = 0.99^{99}(0.99 + 100 \cdot 0.01 \\ = 0.99^{99} \cdot 1.99 \approx 0.73576. \end{equation*}

Utilize the interactive cell below to compute \(f(x)\) and \(F(x)\) for the Binomial distribution

For the mean 1,

\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 2,

\begin{gather*} \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{gather*}

Using the change of variables \(k=x-2\) and \(m = n-2\) yields a binomial series

\begin{gather*} = 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{gather*}

The skewness 3 and kurtosis 4 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 symbolically confirm the general formulas for the Binomial distribution.

Once again, you can see that as one of the parameters is allowed to increase without bound this distribution must become symmetrical and bell-shaped.

For skewness, take the limit of the skewness result above

\begin{equation*} \lim_{n \rightarrow \infty} \frac{n(2p-1)(p-1)p}{(-n(p1)p)^{3/2}} = \lim_{n \rightarrow \infty} C \frac{n}{n^{3/2}} = 0. \end{equation*}

Similarly for kurtosis

\begin{equation*} \lim_{n \rightarrow \infty} -\frac{6p^2 - 6p + 1}{n (p - 1) p} +3 = 0 + 3 = 3. \end{equation*}
Checkpoint 7.2.8. WebWork - Binomial.

You can of course get specific values and graph the Binomial Distribution using R as well...