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# 4.7: Geometric Probability Distribution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Know the definition of a geometric distribution.
• Identify the characteristics of a geometric distribution.
• Identify the type of statistical situation to which a geometric distribution can be applied.
• Use a geometric distribution to solve statistical problems.

## Introduction

In this lesson, you will learn both the definition and the characteristics of a geometric probability distribution. In addition, you will be presented with a real-world problem that you can solve by applying what you have learned.

### Geometric Probability Distributions

Like the Poisson and binomial distributions, a geometric probability distribution describes a discrete random variable. Recall that in the binomial experiments, we tossed a coin a fixed number of times and counted the number, X\begin{align*}X\end{align*}, of heads as successes.

A geometric distribution describes a situation in which we toss the coin until the first head (success) appears. We assume, as in the binomial experiments, that the tosses are independent of each other.

### Characteristics of a Geometric Probability Distribution

• The experiment consists of a sequence of independent trials.
• Each trial results in one of two outcomes: success, S\begin{align*}S\end{align*}, or failure, F\begin{align*}F\end{align*}.
• The geometric random variable X\begin{align*}X\end{align*} is defined as the number of trials until the first S\begin{align*}S\end{align*} is observed.
• The probability p(x)\begin{align*}p(x)\end{align*} is the same for each trial.

Why would we wait until a success is observed? One example is in the world of business. A business owner may want to know the length of time a customer will wait for some type of service. Another example would be an employer who is interviewing potential candidates for a vacant position and wants to know how many interviews he/she has to conduct until the perfect candidate for the job is found. Finally, a police detective might want to know the probability of getting a lead in a crime case after 10 people are questioned.

### Probability Distribution, Mean, and Variance of a Geometric Random Variable

The probability distribution, mean, and variance of a geometric random variable are given as follows:

p(x)μσ2=(1p)x1px=1,2,3,=1p=1pp2\begin{align*}p(x) &= (1-p)^{x-1} p \quad x=1, 2, 3, \ldots\\ \mu &= \frac{1}{p}\\ \sigma^2 &= \frac{1-p}{p^2}\end{align*}

where:

p=\begin{align*}p =\end{align*} probability of an S\begin{align*}S\end{align*} outcome

x=\begin{align*}x =\end{align*} the number of trials until the first S\begin{align*}S\end{align*} is observed

The figure below plots a few geometric probability distributions. Note how the probabilities start high and drop off, with lower p\begin{align*}p\end{align*} values producing a faster drop-off.

Example: A court is conducting a jury selection. Let X\begin{align*}X\end{align*} be the number of prospective jurors who will be examined until one is admitted as a juror for a trial. Suppose that X\begin{align*}X\end{align*} is a geometric random variable, and p\begin{align*}p\end{align*}, the probability of a juror being admitted, is 0.50.

a. Find the mean and the standard deviation of X\begin{align*}X\end{align*}.

b. Find the probability that more than two prospective jurors must be examined before one is admitted to the jury.

a. The mean and the standard deviation can be calculated as follows:

μσ2σ=1p=10.5=2=1pp2=10.50.52=2=2=1.41\begin{align*}\mu &= \frac{1}{p}=\frac{1}{0.5}=2\\ \sigma^2 &= \frac{1-p}{p^2}=\frac{1-0.5}{0.5^2}=2\\ \sigma&=\sqrt{2}= 1.41\end{align*}

To find the probability that more than two prospective jurors will be examined before one is selected, you could try to add the probabilities that the number of jurors to be examined before one is selected is 3, 4, 5, and so on, as follows:

p(x>2)=p(3)+p(4)+p(5)+\begin{align*}p(x > 2)=p(3)+p(4)+p(5)+ \ldots\end{align*}

However, this is an infinitely large sum, so it is best to use the Complement Rule as shown:

p(x>2)=1p(x2)=1[p(1)+p(2)]\begin{align*}p(x > 2) &= 1-p(x \le 2)\\ &= 1-[p(1)+p(2)]\end{align*}

In order to actually calculate the probability, we need to find p(1)\begin{align*}p(1)\end{align*} and p(2)\begin{align*}p(2)\end{align*}. This can be done by substituting the appropriate values into the formula:

p(1)p(2)=(10.5)11(0.5)=(0.5)0(0.5)=0.5=(10.5)21(0.5)=(0.5)1(0.5)=0.25\begin{align*}p(1) &= (1-0.5)^{1-1}(0.5)=(0.5)^0(0.5)=0.5\\ p(2) &= (1-0.5)^{2-1}(0.5) =(0.5)^1 (0.5)=0.25\end{align*}

Now we can go back to the Complement Rule and plug in the appropriate values for p(1)\begin{align*}p(1)\end{align*} and p(2)\begin{align*}p(2)\end{align*}:

p(x>2)=1p(x2)=1(0.5+0.25)=0.25\begin{align*}p(x > 2) &= 1-p(x \le 2)\\ &= 1 - (0.5 + 0.25)=0.25\end{align*}

This means that there is a 0.25 chance that more than two prospective jurors will be examined before one is admitted to the jury.

Technology Note: Calculating Geometric Probabilities on the TI-83/84 Calculator

Press [2ND][DISTR] and scroll down to 'geometpdf('. Press [ENTER] to place 'geometpdf(' on your home screen. Type in values of p\begin{align*}p\end{align*} and x\begin{align*}x\end{align*} separated by a comma, with p\begin{align*}p\end{align*} being the probability of success and x\begin{align*}x\end{align*} being the number of trials before you see your first success. Press [ENTER]. The calculator will return the probability of having the first success on trial number x\begin{align*}x\end{align*}.

Use 'geometcdf(' for the probability of at most x\begin{align*}x\end{align*} trials before your first success.

Note: It is not necessary to close the parentheses.

## Lesson Summary

Characteristics of a Geometric Probability Distribution:

• The experiment consists of a sequence of independent trials.
• Each trial results in one of two outcomes: success, S\begin{align*}S\end{align*}, or failure, F\begin{align*}F\end{align*}.
• The geometric random variable X\begin{align*}X\end{align*} is defined as the number of trials until the first S\begin{align*}S\end{align*} is observed.
• The probability p(x)\begin{align*}p(x)\end{align*} is the same for each trial.

Geometric random variable:

The probability distribution, mean, and variance of a geometric random variable are given as follows:

p(x)μσ2=(1p)x1px=1,2,3,=1p=1pp2\begin{align*}p(x) &= (1-p)^{x-1} p \quad x=1, 2, 3, \ldots\\ \mu &= \frac{1}{p}\\ \sigma^2 &= \frac{1-p}{p^2}\end{align*}

where:

p=\begin{align*}p =\end{align*} probability of an S\begin{align*}S\end{align*} outcome

x=\begin{align*}x =\end{align*} the number of trials until the first S\begin{align*}S\end{align*} is observed

## Review Questions

1. A prison reports that the number of escape attempts per month has a Poisson distribution with a mean value of 1.5.
1. Calculate the probability that exactly three escapes will be attempted during the next month.
2. Calculate the probability that exactly one escape will be attempted during the next month.
2. The mean number of patients entering an emergency room at a hospital is 2.5. If the number of available beds today is 4 beds for new patients, what is the probability that the hospital will not have enough beds to accommodate its new patients?
3. An oil company has determined that the probability of finding oil at a particular drilling operation is 0.20. What is the probability that it would drill four dry wells before finding oil at the fifth one? (Hint: This is an example of a geometric random variable.)

Keywords

If X\begin{align*}X\end{align*} and Y\begin{align*}Y\end{align*} are random variables, then:

μX+YμXY=μX+μY=μXμY\begin{align*}\mu_{X+Y} &= \mu_X + \mu_Y\\ \mu_{X-Y} &= \mu_X - \mu_Y\end{align*}

Binomial experiment
A binomial experiment is a probability experiment
Binomial probability distribution
The binomial probability distribution is: p(x)=(nx)px(1p)nx=(nx)pxqnx\begin{align*}p(x)=\binom{n}{x} p^x (1-p)^{n-x}=\binom{n}{x} p^x q^{n-x}\end{align*}.
Continuous
Random variables that can take on any of the countless number of values in an interval of real numbers are called continuous.
Continuous random variables
is one that can assume any value in some interval or intervals of real numbers
Discrete
Random variables that can assume only a countable number of values are called discrete.
Discrete random variables
is a random variables that assumes its values only at isolated points.
Expected value
The weighted mean of the possible values that a random variable can take on
Geometric probability distribution
A geometric probability distribution describes a discrete random variable.
Linear Transformation Rule
The linear transformation is an important concept in statistics, because many elementary statistical formulas involve linear transformations.
Linear transformations

μc+dXσc+dX=c+dμX=|d|σX\begin{align*}\mu_{c+dX} &= c+d \mu_{X}\\ \sigma_{c+dX} &= |d| \sigma_{X}\end{align*}

Poisson probability distribution
A Poisson probability distribution is useful for describing the number of events that will occur during a specific interval of time or in a specific distance, area, or volume.
Probability distribution
This is a complete description of all the possible values of the random variable
Quantitative variables
a quantitative variable that can be measured or observed,
Random variables
is a variable where value is determined by chance. random variables are denoted by capital letter.
Re-centering
If you add the same value to all the numbers of a data set, the shape and standard deviation of the data set remain the same, but the value is added to the mean. This is referred to as re-centering the data set.
Rescale
if you rescale the data, or multiply all the data values by the same nonzero number, the basic shape will not change, but the mean and the standard deviation will each be a multiple of this number.
Subtraction Rule
If X\begin{align*}X\end{align*} and Y\begin{align*}Y\end{align*} are random variables, then:

μX+YμXY=μX+μY=μXμY\begin{align*}\mu_{X+Y} &= \mu_X + \mu_Y\\ \mu_{X-Y} &= \mu_X - \mu_Y\end{align*}

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