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Q 2 _ Let (N(t))tzo he Poisson process with rate A > 0_ Find the probability that there are three arrivals in (1,2] given that there is at least one arrival in [...

Question

Q 2 _ Let (N(t))tzo he Poisson process with rate A > 0_ Find the probability that there are three arrivals in (1,2] given that there is at least one arrival in [0, 2].

Q 2 _ Let (N(t))tzo he Poisson process with rate A > 0_ Find the probability that there are three arrivals in (1,2] given that there is at least one arrival in [0, 2].



Answers

Refer to Exercise $3.122 .$ Assume that arrivals occur according to a Poisson process with an average of seven per hour. What is the probability that exactly two customers arrive in the twohour period of time between a. 2: 00 P.M. and 4: 00 P.M. (one continuous two-hour period)? b. 1: 00 P.M. and 2: 00 p.M. or between 3: 00 p.M. and 4: 00 P.M. (two separate one-hour periods that total two hours)?

To solve this problem, we have to rely on problem number 1 32. The previous problem. And in that particular problem, we were talking about the fact that there was an average of one car entering a tunnel in a two minute period. And you had to determine the probability that we had greater than three cars entering the tunnel In that one, earn that two minute time period. So what we would have to do is we'd have to let the variable why represent the number of cars entering that tunnel In a two minute period. So if you looked at the probability distribution for that, you would have been applying the formula. The probability of a poussin event occurring would be lambda raised to the Y power times E. To the negative lambda power all over. Why? Factorial. And we have already defined that λ was one and our Why value? We could have no cars entering the tunnel in a two minute period or one or two or three or four or five. And it keeps going infinitely with small probability, but it does keep going infinitely up to infinity. So if we were to apply this formula With why being zero, you would get approximately a probability of .368. And then if you use that same formula and changed why to be one, we would get .368. And if we let y equal to, we'd get about .184 and if we let y equal three, we'd get approximately .061. Now our interest lies with all the values greater than three. So we're looking for the sum of all of these values. Well, because that list infinitely goes on and on and on, we can't just apply that formula and infinite number of times. So what we need to recall is that if we add up the probabilities in a probability distribution, we will always get a sum of one. So if we're interested in Here, we would do one take away the probability that why was less than or equal to three, which means we would have to sum up these values and they're going to give you approximately 8.981. So if we do 1 -8.981, we are going to get the probability that the some that are that the number of cars going through the tunnel being greater than three would be approximately .01 nine. So therefore the answer from problem 32 or 132 was .019. Now we get into our problem, we want to solve problem number 1 33. And in problem number 1 33 were changing from a binomial or sorry, from a Poisson distribution and a pleasant probability now into a binomial probability. Yeah. And we're changing into a probability that is by no meal because we are going to look at 10 independent observations. So now we have that N is 10. And when you're doing a binomial probability, you have to define four different variables. You have the number of trials, we have our probability of success, we have Q. Which is our probability of failure and then we have our why value or are X value? We can call it any variable we want, which is when the successes occurred. So that that would be the number of successes. So in this particular problem, we want to find the probability that more than three cars went through the tunnel at least one of those 102 minute intervals. So we're gonna say X represents The number of two minute intervals With more than three cars. And we are trying to determine the probability of X. So of those 10 trials, we might have had none of the trials with more than three cars passing through the tunnel or one or two or three All the way up to 10. And our probabilities of interest are we want the probability that X is at least one, so that would be greater than or equal to one. So our areas of interest would be all of these. All right. So that means we could apply the binomial probability problem once for each of those values. Or again, we could use the fact that in a discrete probability distribution the sum of P of X always has to equal one. So if we were to apply our formula and find the probability of zero, then for us to get the some from 1 to 10, we could say one minus the probability that X was equal to zero. So therefore we need to apply that probability formula. And the probability formula for a binomial probability is the probability the event will occur will be n C. X times P. To the X power times Q. To the n minus X power. So we are going to do one and we're going to apply that formula. So our end was 10. See to the zero Power Because our X0 our probability of success relies back onto the previous problem. Our probability of success, Which we are defining as more than three cars through the tunnel was .019. So therefore we're going to have .019 raised to the X power which again is zero. And then our probability of failure. Our probability of failure. If success plus failure has to be equal to one will be the same as one minus our probability of success. So we could do one minus 10.19 And then when we do n minus X. N was 10 X zero. So we're gonna have to the 10th power. If you were to utilize your calculator you would find that the combination of 10 items taken zero at a time would be one. And we know based on our algebra properties that anything raised the zero power is also one. So in essence we are just Going to solve the quantity of 1 -119 to the 10th power. And When we do that, we are going to get a probability of approximately .174 five. So again, in order to solve this problem, we did have to rely back on the answer that you got in problem number 132, that the probability of success, which we are defining as having more than three cars go through the tunnel was .019. And the probability when we randomly Do 10 different observations of getting More than three going through that tunnel on at least one of those 102 minute intervals would end up being .1745.

In this question were given a Poisson process with a rate of three per hour. And in the first part of the question, what is the probability of having 10 events occur in the interval from 0 to 5 hours? Given that we've had four events in the time interval from 0 to 2 hours, So the numbers of events that are occurring in these processes are Poisson random variables, and we know that because it's been described as a Poisson process. Remember that for a Poisson random variable, the probability of so many events occurring within a given time interval is equal to the following. So for part A, we're looking for the probability that the number of events occurring within five hours is equal to 10, given that the number of events that occur in the first two hours is for so the time intervals that we're talking about our from 0 to 2 hours and from 0 to 5 hours. And as you can see these air overlapping time intervals the first time interval is a subset of the second one. So from our probability laws, we can say that this is equal to the probability of the number of events occurring between zero and five hours, equaling 10 an intersection, with a number of events occurring between zero and two hours equaling four, divided by the probability that the number of events occurring in the first two hours is equal to four now. Another way to look at this is if we are given that four events have occurred between zero and two hours, then for 10 events to have occurred by five hours, we need an additional six events to occur between the times of two hours and five hours. So we can write this as the probability of the number of events occurring between five and two hours is equal to six. So now we're dealing with two intervals. One is from 0 to 2 and the second is from two 25 And these air non overlapping intervals and for a personal process, the number of events that occur and non overlapping intervals eyes independent, which means that we can write the intersection of the two events in the numerator as the product of two probabilities. So we could say that the probability of the number of events occurring in three hours being equal to six times the probability that the number of events that occurs in two hours is equal to four and then divided by the probability at the number of events that occurs in two hours is equal to four. And obviously this gets a lot nicer now that we can factor these out. So now we can just solve for the probability of having six events in three hours, given a process rate of three per hour. So this is e to the negative Lambda T, which comes out to negative nine. So Lambda is three and t history times lambda t to the exponents six over six factorial. And this comes out to 0.91 and then next for B were asked to consider a more general case where we have to fix times. So we have from zero until s hours and also from zero until T hours and two entered Your number of arrivals where we have m is smaller than M. And so were asked to find the probability that we receive that there are any events by Time t given that we have seen em events by time s, we can write this as the probability that by time t there have been any events given that by time s there have been m events as with part a weaken split this into two non overlapping time intervals. So let's say this is time s. And by time s we have seen m events. So in order to have any events by time T we must between time s and time t see and minus m events this way at time as we have had em events and by time t we will have a total of any events and so we can write this as follows. So the probability of having em events by time s intersected with the probability intersected with having and minus M events in the time interval from S t. T. So, just in case it wasn't 100% obvious what I just did here I replaced this event with the probability of having and minus am events in the interval from s to t. So that is T minus s hours. Number of events is and minus M so these two probabilities are equivalent. If we know if it's given that we had em events in s hours. The probability of having any events in T Ours is the same as the probability of having n minus M events in the next T minus s hours. And now, because the time from zero to us and the time interval from STT are non overlapping time intervals, we can rewrite the probability as follows and as you can see weaken, cross these out and then this probability is simply e to the Lambda T or each of the negative Lambda t. So lambda is three and he is t minus s. That's the duration times lambda t to the exponents and minus m over n minus m factorial.

For this question were given a possum process with a rate of Lambda. And we're also told that on the interval from zero to end, we had any events. So from right so say this is time n when you in this time that we have had and events that is given to us and so were asked between zero and one What is the probability that we have X events so x of the and events. So since it is a porcelain process, the number of events that occur in the given time is denoted as x of T. So what we're looking for is the probability that the number of events that occur between zero and one so in one unit of time, is equal to X. Given that given that we had given that the number of events that occurred between zero and N is equal Thio in. So this is our conditional probability given that between zero and n we had any events taking place. What is the probability that between zero and one x of those events take place in that time period? And so, by our probability laws, this is the probability that number of events taking place by one time unit secretly X intersected with the event that any events take place within the time from zero to, um divided by probability of any events taking place between zero and end time. So another way to look at that is to if we have actual events taking place between zero and one, we also then need to have had and minus X events taking place between Time one and M. So we can rewrite this probability in the numerator as probability that by time one we have X events times the probability in the duration and minus one. So that's from time. One until time and we have n minus X events. And we can simply multiply these two probabilities together since non overlapping time intervals are independent for opposing process. And we have to keep our denominator, which is the probability that we have any events by time N. And so we can work this out t e to the negative Lambda Times one. So that's just even negative. Lambda Times landed times one to the exponents X over X factorial times E to the negative lambda and now the time duration is and minus one, so we can have landed times and minus one. All of this to the exponents and minus X divided by end minus X factorial. And the denominator is still probability that by time and we have had any events And so let's right this out in more detail. So that's e to the negative round to end times lambda in to the exponents 10 divided by in factorial. So this is equal to the following. So this should be in brackets here like that. So all I've really done in this part is collected like terms from from this more cumbersome equation here. And so this gets a bit simpler. We can cancel these out. Lambda two exponents n is seen right here, and this comes out to you and minus one to the exponents and minus x times and factorial divided by X factorial times n minus X factorial times end to the exponents and so this is the answer. But you might also want to note that it is also equal to and shoes X times one over m to the exponents X times and minus one over em to the exponents and minus X, which is the binomial distribution for n Bernoulli trials with probability of success, one over n and probability of failure and minus one over in. Which makes sense because if you think of the duration, say from zero to end and some sub interval, which is 10 to 1, we're told that we have had any events between Time zero and N and for a plus on process. If we know that we've had any events, those events are equal, equally likely to have occurred anywhere along this time interval, so they're uniformly distributed. And for a uniformed distribution, the likelihood of any individual event falling between zero and one is equal to one divided by the total range of time. So that's n so basically, the probability of falling in this interval here is one divided by the total range of time, which is n. So that's our probability of success. And so Q is one minus that which is equal to and minus one over in, which also makes sense because this duration here is and minus one, and so the probability of an individual event falling in that time duration is and minus one divided by the total time, which is m, which is what we have here for the probability of failure. So another way to have answered this question would be to make this rationalization and then just presented as a binomial distribution. So now, for Part B, we're once again told that in the time interval from zero to end, we have any events that have occurred. So we're asked, what is the limiting conditional distribution of the number of events and the time interval from 0 to 1 as an approaches infinity. So remember, the probability X by one is equal to X is equal to what we had found in part a. So this part right here So we can rewrite this a little bit differently, though we can rewrite this as and minus one over em to the exponents and minus X times one over n to the exponents X times in factorial over Expect Auriol times in minus X factorial So all I have done here is I've just taken this end to the exponents m and rewritten it as and if exponents and minus x times into the exponents X So these air in the denominator over here and over here. And so the only mistake I think I've made so faras I should have expressed this as the conditional probability. So this would be probability number of events by time one equals X, given that the number of events by time and is equal to, um and so this is the conditional probability and we want the limiting conditional probability as n approaches infinity. So if we take the limit as an approaches infinity so we can just rearrange this a little bit. So we have and minus one over in the exponents and minus X times in factorial over and it exponents x times and minus X factorial times one over x factorial. So that's just some minor rearrangement of the previous expression. And so now if we take the limit when an approaches infinity, this component comes out to one and this component comes out to one and this component comes out to one over x factorial. So it's not too hard to see for this one. I mean, if we have, if Anna's approaching infinity, we basically have infinity over infinity so that that fraction is one for this one. As an approaches infinity in factorial is like ended exponents n and to the exploding exes like end of exploring X and end minus X factorial is like end to the exponents and minus X. So this product would give you and the exponents end, which is what we have in the numerator as well. So that comes out to one. And so the final answer comes out to one over x factorial.

Yeah. This problem you're dealing with us on probability distribution. You're told that lambda 03 Now for Poisson probability distribution, the probability of why is given my landed at the Y. Either than any of lambda overweight factorial assistance. Land is three, this is three to the Y. Even the negative third over wife act or so. Let's find what we were asked for you. We want to first find the probability that X is equal to two. Mhm. So the probability that I'll use why since I wrote why here? But probably why is too would be three to the second. Either the negative third over two factor we just plug into for why? It's the three of the second even their third divided by two factorial gives us 0.2240 That's probability wise too naturally want the probability that it is less than or equal three. Mhm. This is equal to the probability of zero plus the probability of one most. The probability of two. Also the probability of three. And so we just need to put all of these into our distribution function. And so the probability of zero is 3 to 0. Either the negative third over zero fat tour probability of one is three to the one either the negative third over one factory. And then just putting all these into our probability distribution function. You have three of the second, even the negative third over two factorial. And the probability of three is three to the third. Either the negative sir, over three Fattal. So here we're just trying to type all of these in straight like they are on our calculator to evaluate this. And when I do okay. This should give us approximately 0.6472 Well that's probability. And what is the probability that acts is greater than zero. I use in our complement rule this is the same as the one minus the probability that act is less than or equal to zero. And since X only takes on values of zero or larger, this is just one minus the probability that act zero. And so this is one minus 3 to 0. Either the negative third over zero factorial. Mhm. That's why we evaluate this. And when we do this would you have us 0.9502 That would be we want the mean and the standard deviation. The expected value of a person is always lambda as much as three. Standard deviation is the square root of the variance. For a pleasant distribution the variance is always land this, this is route three. And so that's our standard deviation standard deviation is route three.


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