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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. 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?
2. 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

Binomial experiment

Binomial probability distribution

Continuous

Continuous random variables

Discrete

Discrete random variables

Expected value

Geometric probability distribution

Linear Transformation Rule

Linear transformations

Poisson probability distribution

Probability distribution

Quantitative variables

Random variables

Re-centering

Rescale

Subtraction Rule

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