Contents

Geometric Distribution (Optional)

There are three main characteristics of a geometric experiment:

Repeating independent Bernoulli trials until a success is obtained. Recall that a Bernoulli trial is a binomial experiment with number of trials n = 1. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bull's-eye until you hit the bull's-eye. The first time you hit the bull's-eye is a success so you stop throwing the dart. It might take six tries until you hit the bull's-eye. You can think of the trials as failure, failure, failure, failure, failure, success, stop.
In theory, the number of trials could go on forever. There must be at least one trial.
The probability, p, of a success and the probability, q, of a failure do not change from trial to trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is $\frac{1}{6} \frac{1}{6}$ . This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = $\frac{5}{6} \frac{5}{6}$ , the probability of a failure. The probability of getting a three on the fifth roll is $(\frac{5}{6}) (\frac{5}{6}) (\frac{5}{6}) (\frac{5}{6}) (\frac{1}{6}) (\frac{5}{6}) (\frac{5}{6}) (\frac{5}{6}) (\frac{5}{6}) (\frac{1}{6})$ = .0804.

X = the number of independent trials until the first success.

p = the probability of a success, q = 1 – p = the probability of a failure.

There are shortcut formulas for calculating mean μ, variance σ², and standard deviation σ of a geometric probability distribution. The formulas are given as below. The deriving of these formulas will not be discussed in this book.

μ = \frac{1}{p}, σ^{2} = (\frac{1}{p}) (\frac{1}{p} - 1), σ = \sqrt{(\frac{1}{p}) (\frac{1}{p} - 1)} μ = \frac{1}{p}, σ^{2} = (\frac{1}{p}) (\frac{1}{p} - 1), σ = \sqrt{(\frac{1}{p}) (\frac{1}{p} - 1)}

Example 4.16

Suppose a game has two outcomes, win or lose. You repeatedly play that game until you lose. The probability of losing is p = 0.57.

If we let X = the number of games you play until you lose (includes the losing game), then X is a geometric random variable. All three characteristics are met. Each game you play is a Bernoulli trial, either win or lose. You would need to play at least one game before you stop. X takes on the values 1, 2, 3, . . . (could go on indefinitely). Since we are measuring the number of games you play until you lose, we define a success as losing a game and a failure as winning a game. The probability of a success $p = .57 p = .57$ and the probability of a failure q = 1 – p = 1 – 0.57 = 0.43. Both p and q remain the same from game to game.

If we want to find the probability that it takes five games until you lose, then the probability could be written as P(x = 5). We will explain how to find a geometric probability later in this section.

Try It 4.16

You throw darts at a board until you hit the center area. Your probability of hitting the center area is p = 0.17. You want to find the probability that it takes eight throws until you hit the center. What values does X take on?

Example 4.17

A safety engineer feels that 35 percent of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions.

If we let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions, then X is a geometric random variable. All three characteristics are met. Each accident report she reads is a Bernoulli trial: the accident was either caused by failure of employees to follow instructions or not. She would need to read at least one accident report before she stops. X takes on the values 1, 2, 3, . . . (could go on indefinitely). Since we are measuring the number of reports she needs to read until one that shows an accident caused by failure of employees to follow instructions, we define a success as an accident caused by failure of employees to follow instructions. If an accident was caused by another reason, the report is defined as a failure. The probability of a success p = .35 and the probability of a failure $q = 1 - p = 1 - .35 = .65 q = 1 - p = 1 - .35 = .65$ . Both p and q remain the same from report to report.

If we want to find the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions, then the probability could be written as $p = .35 p = .35$ . If we want to find how many reports, on average, the safety engineer would expect to look at until she finds a report showing an accident caused by employee failure to follow instructions, we need to find the expected value E(x). We will explain how to solve these questions later in this section.

Try It 4.17

An instructor feels that 15 percent of students get below a C on their final exam. She decides to look at final exams (selected randomly and replaced in the pile after reading) until she finds one that shows a grade below a C. We want to know the probability that the instructor will have to examine at least 10 exams until she finds one with a grade below a C. What is the probability question stated mathematically?

Example 4.18

Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55 percent of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?

This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).

Problem

a. Let X = the number of ________ you must ask ________ one says yes.

Show solution

Solution

a. Let X = the number of students you must ask until one says yes.

Problem

b. What values does X take on?

Show solution

Solution

b. 1, 2, 3, . . ., (total number of students)

Problem

c. What are p and q?

Show solution

Solution

c. p = .55; q = .45

Problem

d. The probability question is P(_______).

Show solution

Solution

d. P(x = 4)

Try It 4.18

You need to find a store that carries a special printer ink. You know that of the stores that carry printer ink, 10 percent of them carry the special ink. You randomly call each store until one has the ink you need. What are p and q?

Notation for the Geometric: G = Geometric Probability Distribution Function

X ~ G(p)

Read this as X is a random variable with a geometric distribution. The parameter is p; p = the probability of a success for each trial.

Example 4.19

Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?

Let X = the number of computer components tested until the first defect is found.

X takes on the values 1, 2, 3, . . . where p = .02. X ~ G(.02)

Find P(x = 7). There is a formula to define the probability of a geometric distribution $P (x) P (x)$ . We can use the formula to find $P (x = 7) P (x = 7)$ . But since the calculation is tedious and time consuming, people usually use a graphing calculator or software to get the answer. Using a graphing calculator, you can get $P (x = 7) = .0177 P (x = 7) = .0177$ . The instruction of TI83, 83+, 84, 84+ is given below.

Using the TI-83, 83+, 84, 84+ Calculator

Go into 2nd DISTR. The syntax for the instructions are as follows:

To calculate the probability of a value P(x = value), use geometpdf(p, number). Here geometpdf represents geometric probability density function. It is used to find the probability that a geometric random variable is equal to an exact value. p is the probability of a success and number is the value.

To calculate the cumulative probability P(x ≤ value), use geometcdf(p, number). Here geometcdf represents geometric cumulative distribution function. It is used to determine the probability of “at most” type of problem, the probability that a geometric random variable is less than or equal to a value. p is the probability of a success and number is the value.

To find $P (x = 7) P (x = 7)$ , enter 2nd DISTR, arrow down to geometpdf(. Press ENTER. Enter .02,7). The result is $P (x = 7) = .0177 P (x = 7) = .0177$ .

If we need to find $P (x \leq 7) P (x \leq 7)$ enter 2nd DISTR, arrow down to geometcdf(. Press ENTER. Enter .02,7). The result is $(x \leq = 7) = .1319 (x \leq = 7) = .1319$ .

The graph of X ~ G(.02) is

This graph shows a geometric probability distribution. It consists of bars that peak at the left and slope downwards with each successive bar to the right. The values on the x-axis count the number of computer components tested until the defect is found. The y-axis is scaled from 0 to 0.02 in increments of 0.005.

Figure 4.2

The previous probability distribution histogram gives all the probabilities of X. The x-axis of each bar is the value of X = the number of computer components tested until the first defect is found, and the height of that bar is the probability of that value occurring. For example, the x value of the first bar is 1 and the height of the first bar is 0.02. That means the probability that the first computer components tested is defective is .02.

The expected value or mean of X is $E (X) = μ = \frac{1}{p} = \frac{1}{.02} = 50 E (X) = μ = \frac{1}{p} = \frac{1}{.02} = 50$ .

The variance of X is $σ^{2} = (\frac{1}{p}) (\frac{1}{p} - 1) = (\frac{1}{.02}) (\frac{1}{.02} - 1) = (50) (49) = 2,450 σ^{2} = (\frac{1}{p}) (\frac{1}{p} - 1) = (\frac{1}{.02}) (\frac{1}{.02} - 1) = (50) (49) = 2,450$

The standard deviation of X is $σ = \sqrt{σ_{}^{2}} = \sqrt{2,450} = 49.5 σ = \sqrt{σ_{2}} = \sqrt{2,450} = 49.5$

Here is how we interpret the mean and standard deviation. The number of components that you would expect to test until you find the first defective one is 50 (which is the mean). And you expect that to vary by about 50 computer components (which is the standard deviation) on average.

Try It 4.19

The probability of a defective steel rod is .01. Steel rods are selected at random. Find the probability that the first defect occurs on the ninth steel rod. Use the TI-83+ or TI-84 calculator to find the answer.

Example 4.20

Problem

The lifetime risk of developing pancreatic cancer is about one in 78 (1.28 percent). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G $(\frac{1}{78}) (\frac{1}{78})$ or X ~ G(.0128).

What is the probability that you ask 10 people before one says he or she has pancreatic cancer?
What is the probability that you must ask 20 people?
Find the (i) mean and (ii) standard deviation of X.

Show solution

Solution

P(x = 10) = geometpdf(.0128, 10) = .0114
P(x = 20) = geometpdf(.0128, 20) = .01
1. Mean = μ = $\frac{1}{p} \frac{1}{p}$ = $\frac{1}{.0128} \frac{1}{.0128}$ = 78
2. $σ = \sqrt{σ^{2}} = \sqrt{(\frac{1}{p}) (\frac{1}{p} - 1)} = \sqrt{(\frac{1}{.0128}) (\frac{1}{.0128} - 1)} = \sqrt{(78) (78 - 1)} = \sqrt{6,006} = 77.4984 \approx 77 σ = \sqrt{σ^{2}} = \sqrt{(\frac{1}{p}) (\frac{1}{p} - 1)} = \sqrt{(\frac{1}{.0128}) (\frac{1}{.0128} - 1)} = \sqrt{(78) (78 - 1)} = \sqrt{6,006} = 77.4984 \approx 77$
  The number of people whom you would expect to ask until one says he or she has pancreatic cancer is 78. And you expect that to vary by about 77 people on average.

Try It 4.20

The literacy rate for a nation measures the proportion of people age 15 and over who can read and write. The literacy rate for women in Afghanistan is 12 percent. Let X = the number of Afghani women you ask until one says that she is literate.

What is the probability distribution of X?
What is the probability that you ask five women before one says she is literate?
What is the probability that you must ask 10 women?
Find the (i) mean and (ii) standard deviation of X.