Geometric Distribution Quiz

PDF page 217: The geometric probability density function builds upon what we have learned from the binomial distribution. In this case the experiment continues until either a success or a failure occurs rather than for a set number of trials. There are three main characteristics of a geometric experiment.

PDF page 217: repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.

PDF page 217: 1 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 = 0.0804 ⎝6⎠⎝6⎠⎝6⎠⎝6⎠⎝6⎠

PDF page 217: 1 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 = 0.0804 ⎝6⎠⎝6⎠⎝6⎠⎝6⎠⎝6⎠

PDF page 218: of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P(x = 5).

PDF page 218: A safety engineer feels that 35% 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. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is 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? 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. X takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find P(x ≥ 3). ("At least" translates to a "greater than or equal to" symbol).

PDF page 219: c. p = 0.55; q = 0.45

PDF page 218: a. Let X = the number of students you must ask until one says yes.

PDF page 219: for x = 1, 2, 3, .... The expected value of X, the mean of this distribution, is 1/p. This tells us how many trials we have to expect until we get the first success including in the count the trial that results in success. The above form of the Geometric distribution is used for modeling the number of trials until the first success. The number of trials includes the one that is a success: x = all trials including the one that is a success. This can be seen in the form of the formula. If X = number of trials including the success, then we must multiply the probability of failure, (1-p), times the number of failures, that is X-1. By contrast, the following form of the geometric distribution is used for modeling number of failures until the first success:

PDF page 219: for x = 0, 1, 2, 3, .... In this case the trial that is a success is not counted as a trial in the formula: x = number of failures. The expected value,

PDF page 219: 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 = 0.02. X ~ G(0.02)

PDF page 219: Find P(x = 7). Answer: P(x = 7) = (1 - 0.02) × 0.02 = 0.0177.

PDF page 219: The probability that the seventh component is the first defect is 0.0177. The graph of X ~ G(0.02) is: