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# 4.6: The Geometric Probability Distribution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

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

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

The 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. The box below lists the main characteristics of the geometric random variable.

## Characteristics of the 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 do we wait until a success is observed? For example, in the world of business, the business owner wants to know the length of time a customer will wait for some type of service. Or, an employer, who is interviewing potential candidates for a vacant position, wants to know how many interviews he/she has to conduct until the perfect candidate for the job is found. Or, a police detective might want to know the probability of getting a lead in a crime case after 10\begin{align*}10\end{align*} people are questioned.

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

p(x)μσ2=(1p)x1p    x=1,2,3,=1p=1pp2\begin{align*}p(x) & = (1 - p)^{x - 1} p \ \ \ \ 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 probability distributions of the Geometric distributions. Note how the curve starts high and drops 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 juror being admitted, is 50%\begin{align*}50\%\end{align*}.

1. Find the mean and the standard deviation.
2. Find the probability that more than two prospective jurors must be examined before one is admitted to the jury.

Solution:

1. The mean and the standard deviation are,

μσ2=1p=10.5=2=1pp2=10.50.52=2\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\end{align*}

Thus

σ=2=1.41\begin{align*}\sigma = \sqrt{2} = 1.41\end{align*}

2. To find the probability that more than two prospective jurors will be examined before one is selected,

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

Obviously, this is an infinitely large sum so it is best to use the complementary rule:

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*}

Before we go any further, we need to find p(1)\begin{align*}p(1)\end{align*} and p(2)\begin{align*}p(2)\end{align*}. Substituting into the formula for p(x)\begin{align*}p(x)\end{align*}:

p(1)p(2)=(1p)11p=(.5)0(.5)=0.5=(1p)21p=(.5)1(.5)=0.25\begin{align*}p(1) & = (1 - p)^{1 - 1} p = (.5)^0 (.5) = 0.5 \\ p(2) & = (1 - p)^{2 - 1} p = (.5)^1 (.5) = 0.25\end{align*}

Then,

p(x>2)=1p(x2)=1(.5+.25)=0.25\begin{align*}p(x > 2) & = 1 - p(x \le 2) \\ & = 1 - (.5 + .25) = 0.25 \end{align*}

This result says that there is a 25%\begin{align*}25\%\end{align*} chance that more than two prospective jurors will be examined before one is admitted to the jury.

Technology Note

The TI-83/84 calculators have commands for the geometric distribution.

• Press 2nd and scroll down (or up) to geometpdf (press [ENTER] to place geometpdf on your home screen.) Type values of p\begin{align*}p\end{align*} and x\begin{align*}x\end{align*} separated by a comma and press [ENTER]
• Use geometcdf( for probability of at least x\begin{align*}x\end{align*} successes.

Note: it is not necessary to close the parentheses.

## Lesson Summary

1. Characteristics of the Geometric Probability Distribution

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

2. Probability distribution, mean, and variance of a Geometric Random Variable

p(x)μσ2=(1p)x1p   x=1,2,3,=1p=1pp2\begin{align*}p(x) & = (1 - p)^{x - 1} p \ \ \ 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 outcome

x=\begin{align*}x =\end{align*} The number of trials until the first S 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\begin{align*}1.5\end{align*}.
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. If the mean number of patients entering an emergency room at a hospital is 2.5\begin{align*}2.5\end{align*}. If the number of available beds today is 4\begin{align*}4\end{align*} 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 had determined that the probability of finding oil at a particular drilling operation is 20%\begin{align*}20\%\end{align*}. 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.)

1. 0.1255;\begin{align*}0.1255;\end{align*}
2. 0.3347\begin{align*}0.3347\end{align*}
1. 0.1088\begin{align*}0.1088\end{align*}
2. 8.19%\begin{align*}8.19\%\end{align*}

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