Garden Variety Example: Flip a coin $10$ times, record the number of "heads."
If you do this repeatedly, ($10$ flips, record number of "heads"), you'll notice some outcomes are more likely than others.
The random variable $X=\mbox{# of heads}$ has what is called a binomial distribution.
Bernoulli Trials
A Bernoulli Trial is a repeatable experiment that either results in "Success" or "Failure."
For each trial, the event "Success" has probability $p$, and the event "Failure" has probability __________.
Each trial is independent of the other.
Bernoulli Trial PDF
$$ \begin{array}{|c|c|} \hline k&P(X=k)\\\hline S&p\\\hline F&1-p\\\hline \end{array} $$
Example
Flipping a fair coin is an example of a Bernoulli trial. Defining "heads" as "success," we have $P(S)=0.5$ and $P(F)=0.5.$
$$ \begin{array}{|c|c|} \hline k&P(X=k)\\\hline S&0.5\\\hline F&0.5\\\hline \end{array} $$
Example
Rolling a "six" with a fair die is an example of a Bernoulli trial with $\displaystyle P(S)=\frac{1}{6}$ and $\displaystyle P(F)=\frac{5}{6}.$
$$ \begin{array}{|c|c|} \hline k&P(X=k)\\\hline S&\displaystyle \frac{1}{6}\\\hline F&\displaystyle \frac{5}{6}\\\hline \end{array} $$
Experiment
Flip a coin ten times. Record the number of "heads" (i.e. "successes").
Let $X$ be the random variable of the number of "successes" out of $10$ flips.
Repeat. Keep recording the number of successes $X$ each time.
Quick Question: What is the expected value of the distribution of $X$?
Binomial Distributions
The Binomial Distribution: A Fixed Number of Repeated Bernoulli Trials
Perform a series of $n$ Bernoulli trials with probability of $p$ of success and $q=1-p$ of failure.
The random variable $X$=number of successes out of $n$ trials has a Binomial Distribution and is said to be a Binomial Random Variable.
The probability of $k$ successes out of $n$ trials can be calculated as follows: $$P(X=k)=\left( \begin{array}{c} n \\ k \end{array}\right)p^kq^{n-k}.$$
Special Note #1: The symbol $\left( \begin{array}{c} n \\ k \end{array}\right)$ is used to denote the number of ways we can choose $k$ objects from $n$ objects and is given by the formula $$\left( \begin{array}{c} n \\ k \end{array}\right)=\frac{n!}{k!(n-k)!}.$$ For example, there are $\left( \begin{array}{c} 10 \\ 3 \end{array}\right)$, or $120$ ways to choose $3$ objects from a collection of $10$ objects. As long as you understand the meaning of this symbol, don't worry too much about how to calculate it. Leave that to software.
Special Note #2: Elsewhere, in particular on your calculator, you may see the notation $${}_{n}C_{k}$$ used in place of $\left( \begin{array}{c} n \\ k \end{array}\right).$
Any standard scientific calculator should be able to compute this.
Special Note #3: To calculate these probabilities, we generally use software.
Your TI-84 calculator will calculate binomial probabilities for you.
Examples
In our coin flip experiment, calculate the probability of $3$ successes ("heads") out of $10$ trials ("flips").
$$
\begin{array}{l}
P(X=3)=\left( \begin{array}{c} 10 \\ 3 \end{array}\right)0.5^3 0.5^7\\\\
=120 \cdot 0.5^{10}\\\\
\approx 0.1172
\end{array}
$$
On your TI-83/84 calculator, you may obtain the above by simply hitting the yellow 2nd button, then the VARS button
(for DISTR), scroll down to option 0: $\mbox{binompdf(}$ and hit ENTER. Then type in $\mbox{binompdf(10,0.5,3)}$ and hit ENTER. The result
should be the above decimal answer.
Calculate the probability that we observe $3$ successes OR LESS.
$$
\begin{array}{l}
P(X \leq 3)\\\\
=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\\\
=\left( \begin{array}{c} 10 \\ 0 \end{array}\right)0.5^{0} 0.5^{10}+\left( \begin{array}{c} 10 \\ 1 \end{array}\right)0.5^1 0.5^9+\left( \begin{array}{c} 10 \\ 2 \end{array}\right)0.5^2 0.5^8+\left( \begin{array}{c} 10 \\ 3 \end{array}\right)0.5^3 0.5^7\\\\
=1 \cdot 0.5^{10}+10 \cdot 0.5^{10}+45 \cdot 0.5^{10}+120 \cdot 0.5^{10}\\\\
\approx 0.1719
\end{array}
$$
On your TI-83/84 calculator, you may obtain the above by simply hitting the yellow 2nd button, then the VARS button
(for DISTR), scroll down to option A: $\mbox{binomcdf(}$ and hit ENTER. Then type in $\mbox{binomcdf(10,0.5,3)}$ and hit ENTER. The result
should be the above decimal answer.
Calculate the probability that we observe MORE THAN $3$ successes.
$$
\begin{array}{l}
P(X \gt 3)\\\\
=1-P(X \leq 3)\\\\
\approx 1 - 0.1719\\\\
= 0.8281
\end{array}
$$
Binomial Distribution: Notation, Mean $\mu$, and Standard Deviation $\sigma$
If $X$ is a binomial random variable over $n$ trials with probability $p$ of success and $q=1-p$ of failure, we write $X \sim B(n,p).$
The mean of $X$ is $\mu=np.$
The standard deviation of $X$ is $\sigma=\sqrt{npq}.$
Example
In our coin flip experiment, write the distribution of $X.$
Calculate the mean $\mu$ and standard deviation $\sigma.$
For our coin flip experiment, $n=10,$ $p=0.5,$ and $q=1-p=1-0.5=0.5.$
So, the mean is $$ \mu=np=10\cdot 0.5=5 $$ and the standard deviation is $$ \sigma=\sqrt{npq}=\sqrt{10\cdot 0.5\cdot 0.5}\approx 1.58113883 $$
So, the mean is $$ \mu=np=10\cdot 0.5=5 $$ and the standard deviation is $$ \sigma=\sqrt{npq}=\sqrt{10\cdot 0.5\cdot 0.5}\approx 1.58113883 $$
Example
A light bulb manufacturer knows that about $3\%$ of the light bulbs they manufacture are defective. Suppose you are a quality control inspector and you choose a random sample of $200$ bulbs.
Letting $X$ be the number of defective light bulbs out a a random sample of $200:$
(a) Calculate the mean $\mu$ and the standard deviation $\sigma$ of $X.$
Example (Continued)
(b) What is the probability that exactly $3$ bulbs in our sample are defective?
$$
\begin{array}{l}
P(X=3)=\left( \begin{array}{c} 200 \\ 3 \end{array}\right)0.03^{3} 0.97^{197}\\\\
\approx 0.0879
\end{array}
$$
On your TI-83/84 calculator, you may obtain the above by simply hitting the yellow 2nd button, then the VARS button
(for DISTR), scroll down to option 0: $\mbox{binompdf(}$ and hit ENTER. Then type in $\mbox{binompdf(200,0.03,3)}$ and hit ENTER. The result
should be the above decimal answer.
(c) What is the probability that $3$ bulbs OR LESS in our sample are defective?
$$
\begin{array}{l}
P(X \leq 3)\\\\
=P(X=0)+P(X=1)+P(X=2)+P(X=3)\\\\
=\left( \begin{array}{c} 200 \\ 0 \end{array}\right)0.03^{0} 0.97^{200}+\left( \begin{array}{c} 200 \\ 1 \end{array}\right)0.03^{1} 0.97^{199}+\left( \begin{array}{c} 200 \\ 2 \end{array}\right)0.03^{2} 0.97^{198}+\left( \begin{array}{c} 200 \\ 3 \end{array}\right)0.03^{3} 0.97^{197}\\\\
\approx 0.1472
\end{array}
$$
On your TI-83/84 calculator, you may obtain the above by simply hitting the yellow 2nd button, then the VARS button
(for DISTR), scroll down to option A: $\mbox{binomcdf(}$ and hit ENTER. Then type in $\mbox{binomcdf(200,0.03,3)}$ and hit ENTER. The result
should be the above decimal answer.
(d) What is the probability that MORE THAN $3$ bulbs in our sample are defective?
$$
\begin{array}{l}
P(X \gt 3)\\\\
=1-P(X \leq 3)\\\\
\approx 1 - 0.1472\\\\
= 0.8528
\end{array}
$$
Example (Continued)
Let's look at the simulated distribution $X \sim B(200,0.03)$