III. More Discrete Probability Distributions A. The Geometric Distribution Many actions in life are repeated until a success occurs. An accounting major will take the CPA exam as many times as it takes to pass it. You may have to call someone several times before they answer, but you keep trying until they do. Situations such as these can be represented by a geometric distribution. 1. A geometric distribution is a discrete probability distribution of a random variable x that satisfies the following conditions: a. A trial is repeated until a success occurs. b. The repeated trials are independent of each other. c. The probability of success p is constant for each trial. 2. The probability that the first success will occur on trial number x is found using the formula: 𝑃 𝑠 =𝑝( 𝑞) 𝑥−1 , where q = 1 – p 3. You can also find this probability with the calculator using the geometpdf function (2nd VARS E)

III. More Discrete Probability Distributions A. The Geometric Distribution Example 1 – Page 222 From experience, you know that the probability that you will make a sale on any given telephone call is 0.23. Find the probability that your first sale on any given day will occur on your fourth or fifth sales call. Solution: Since this is an “or” question, we simply add the probability of a sale on the fourth call to the probability of a sale on the fifth call. Using the formula, you have 𝑃 4 =.23 .77 3 =.10500 and 𝑃 5 =.23 .77 4 =.08085 Adding these, we get .105 + .08085 = .18585 Using the calculator, you access the geometpdf function at 2nd VARS E Input .23 for p and 4 for x. You get .105 for x = 4. Repeat for 5; 2nd VARS E, .23, 5 and get .08085. .105 + .08085 = .18585 These are the same answers you got doing it by hand.

III. More Discrete Probability Distributions A. The Geometric Distribution Try it Yourself 1 – Page 223 Find the probability that your first sale will occur before your fourth sales call. Since we want success before the 4th call, we want it to occur on the 1st, 2nd, or 3rd call. We can use either the formula or the calculator to find these probabilities and add them up. 𝑃 1 =.23 .77 0 =.23; 𝑃 2 =.23 .77 1 =.1771; 𝑃 3 =.23 .77 2 =.13637 .23 + .1771 + .13637 = .543467 We can also use the geometcdf function (2nd VARS F) on the calculator. Since we want the cumulative probability of up to 3 calls, enter 3 for the x- value. 2nd VARS F, .23 and 3 gives us the probability of .543467, which is exactly what we got doing it by adding 3 probabilities together.

III. More Discrete Probability Distributions B. Poisson Distribution In a binomial experiment you are interested in finding the probability of a specific number of successes in a given number of trials. Suppose instead that you want to know the probability that a specific number of occurrences takes place within a given unit of time or space. For instance, to determine the probability that an employee will take 15 sick days within a year, you can use the Poisson distribution. 1. The Poisson distribution is a discrete probability distribution of a random variable x that satisfies the following conditions: a. The experiment consists of counting the number of times, x, an event occurs in a given interval. The interval can be an interval of time, area, or volume. b. The probability of the event occurring is the same for each interval. c. The number of occurrences in one interval is independent of the number of occurrences in other intervals.

III. More Discrete Probability Distributions B. Poisson Distribution 2. The formula for finding the probability of exactly x occurrences in an interval is: 𝑃 𝑥 = 𝜇 𝑥 𝑒 −𝜇 𝑥! , where e is an irrational number approximately equal to 2.71828 and 𝜇 is the mean number of occurrences per interval unit. 3. The calculator will also help us with the Poisson distribution (2nd VARS C for pdf and D for cdf)

III. More Discrete Probability Distributions B. Poisson Distribution Example 2 – Page 223 The mean number of accidents per month at a certain intersection is three. What is the probability that in any given month four accidents will occur at this intersection? Solution Using x = 4 and 𝜇 = 3, the probability that 4 accidents will occur in any given month at the intersection is: 𝑃 4 = 3 4 𝑒 −3 4! =.16803; you can enter the e by pressing 2nd and LN On the calculator, 2nd VARS C gets you into the distribution. Enter 3 for the mean and 4 for the x-value. You get .16803, the same answer as above.

III. More Discrete Probability Distributions B. Poisson Distribution Try it Yourself – Page 224 What is the probability that more than four accidents will occur in any given month at the intersection? By hand, we need to find the probabilities for 1, 2, 3, and 4 accidents and add them up. We then subtract that from 1 to get the probability of more than 4 accidents occurring. P(0) = .04979; P(1) = .14936; P(2) = .22404; P(3) = .22404; P(4) = .16803 1 – (.04979 + .14936 + .22404 + .22404 + .16803) = .1847 Using the calculator, we use the Poissoncdf function to find the probability of up to 4 accidents occurring and then subtract that from 1. 2nd VARS D, 3 and 4 = .81526 1 - .81526 = .1847, the same answer we got doing it by hand.

There is a full page summary chart detailing the different discrete probability distributions and their formulas on page 225 in your book. I strongly recommend spending time looking that over and becoming familiar with the information it contains!!

Assignments: Classwork: Page 226; #1–8, 11–16 Homework: Pages 227; #17–24