**The Binomial Distributions**Worksheet

**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 $X \sim B(200,0.03)$