Geometric Distribution (Optional) Quiz

PDF page 262: There are three main characteristics of a geometric experiment:

PDF page 262: 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.

PDF page 263: X = the number of independent trials until the first success.

PDF page 263: p = the probability of a success, q = 1 – p = the probability of a failure.

PDF page 263: 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

PDF page 263: and the probability of a failure q = 1 – p = 1 – 0.57 = 0.43. Both p and q remain

PDF page 263: 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

PDF page 263: 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

PDF page 263: 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 . 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 . 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.

PDF page 263: 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?

PDF page 264: 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?

PDF page 264: 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.

PDF page 265: 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 . We can use the formula to find . 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 . The instruction of TI83, 83+, 84, 84+ is given below.