In this problem, we're told that the number of visits that a person makes to an auto repair shop in a given your follows a personal distribution where the personal print parameter is based on the number of visits that they made in the previous year. So, for example, if they make if they made zero visits in the previous year, then the number of visits in the next year is a possible distribution with this parameter. And so for part, they were asked to construct the one step transition matrix for the number of repair shop visits by a randomly selected customer in the observed year. Now recall that postal distribution for X and parameter mu is given by this equation, and so the probability that except n plus one equals X Given except Ben was I visit. Is it possible with this probability distribution? So we could say that the one step probability of going from zero visits in the previous year two zero visits in the next year is equal to the probability of zero given this rate and so that rate amuse sub zero is this value and so that's equal to you, you to the miners 1.93 nine approximately times 1.939 to explain it. Zero over zero factorial, which comes out to approximately 0.144 And we continue with those count calculations for all single step transitions. Because there are five possible states, we know that the transition matrix is going to be five by five. So you would save a lot of time by calculating these probabilities using software. So that's what I did. And here's the one step transition matrix that resulted. And so this answer is part A. So for Part B were asked that if a customer makes two visits this year, what is the probability that customer makes two visits in the next year and two visits in the year after that? So it helps to index the is just Teoh. Make sure we make less mistakes. So we're looking for the one step probability that we go from two visits to two visits, and then we again go from two visits to two visits. So this represents two visits in the next year, and this represents two visits in the year after that, and they're both one step probabilities one step transitions because the first time we're just going from this year into the next year. So what? We know we went to two visits this year. What is the probability of two visits in the next year and then in the next year, Once we've gone to visits, it's just one more step to the next year. So it's going from two visits next year, 22 visits the year after. So that's one transition. So the one step transition probability from two visits to you to visit is 0.27 So this comes out to you. Zero point 27 squared and that equals 0.73 for Part C were asked if a customer makes no visits last year, what is the probability? They make a total of exactly two visits in the next two years? So we're looking for all of the ways to have a total of two visits over two transitions, assuming the current state is zero visit. So one way is to go ahead and make two visits in the next year, and then we would need to also make zero visits in the next year that is in the year after the next year. So next year we make two visits and the year after we make zero visits for a total of two visits and another way is to do one visit in each of the next years. So that's the problem. One step probability of going from 0 to 1 times the single step probability of going from 1 to 1. So here again we will have a total of two visits and the third way is to have zero visits in the next year, but have two visits in the year after that. And so this comes out to 0.27 times 0.148 plus zero point 279 time, 0.333 the 0.144 time, 0.27 and that comes out to about 0.172