Geometric Distribution (Optional) Quiz

PDF page 279: There are three main characteristics of a geometric experiment:

PDF page 279: 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 280: 1 − p. For example, the probability of rolling a three when you throw one fair die is . This is true no matter how many 6 times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls 5 one through four, you do not get a face with a three. The probability for each of the rolls is q = , the probability of a 6 ⎛5⎞⎛5⎞⎛5⎞⎛5⎞⎛1⎞ failure. The probability of getting a three on the fifth roll is = .0804. ⎝6⎠⎝6⎠⎝6⎠⎝6⎠⎝6⎠ X = the number of independent trials until the first success.

PDF page 280: p = the probability of a success, q = 1 – p = the probability of a failure.

PDF page 280: 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

PDF page 280: failure as winning a game. The probability of a success p = .57 and the probability of a failure q = 1 – p = 1 –

PDF page 280: 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.

PDF page 280: 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

PDF page 280: q = 1 − p = 1 − .35 = .65 . Both p and q remain the same from report to report.

PDF page 281: 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).

PDF page 281: a. Let X = the number of students you must ask until one says yes.

PDF page 282: 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 282: 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) , enter 2nd DISTR, arrow down to geometpdf(. Press ENTER. Enter .02,7). The result is