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A dice game consists in rolling fair six-sided die: the roll is considered as success only when side four O side five (face N 4 or Face N 5) is on top, otherwise t...

Question

A dice game consists in rolling fair six-sided die: the roll is considered as success only when side four O side five (face N 4 or Face N 5) is on top, otherwise the die roll is considered as failure.1-Calculate (with justification) the following: Probability of a success Probability of a loss 2-Assuming that the die rolls are independent; what is the distribution (type and parameters) of the number of trials until getting the first success?3-What is the expected number of trials until the firs

A dice game consists in rolling fair six-sided die: the roll is considered as success only when side four O side five (face N 4 or Face N 5) is on top, otherwise the die roll is considered as failure. 1-Calculate (with justification) the following: Probability of a success Probability of a loss 2-Assuming that the die rolls are independent; what is the distribution (type and parameters) of the number of trials until getting the first success? 3-What is the expected number of trials until the first success? 4-What is the probability to win at most after 2 trials? 5-What is the probability to win at most after 3 trials? 6-Use_the_conditionalprobability rule to find the probability to win at most after 5 trials giving that no success is observed in the first three trials? 7-Compare the results of (6) and (4), comment on this



Answers

Find the probability of rolling a die five times and obtaining a 6 on the first two rolls, a 5 on the third roll, and a $1,2,3,$ or 4 on the last two rolls.

Now, here's ah, the answer for question. Um, your problem. Number 108 on in introductory statistics. Chapter for section For the question States that supposed that you are performing the probability experiment of rolling one fair, six sided die. Let f be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the 1st 4 or five as the outcome, and that defines P as the probability of success. So if you don't if occurs, Q is the probability of a failure. Eso going deaf does not occur. Letter it says, right? The description of the rain and variable X. That's that means, um, X is gonna equal the number of times that the dices rolled For us to achieve the 1st 4 or five part B, it says. What is the value that X can take on? So it's since X is just a number. Um, it's, you know, it's a number of times that it takes to roll a four or five so X can be any positive whole numbers. So, for example, 1234 five until you know you roll that four or five. Let her see it says, Find the values of P and Q. I remember from up above here it says that he is the probability of a success. So, um, the probably looking for the probability that we roll a four or five. Well, there's 14 and there's 15 So there's to total up, you know, um, successful outcomes. And there's six total. So two under six, which is 1/3 I accuse the probability of a failure. So therefore Q is gonna be not a four or five, which would be 123 or six. So that's four options out of six total so far six, which is 2/3 and then later de it says, find the probability that the first occurrence of you don't f rolling A 45 is on the second trial. So what we're looking for is the probability for success in the second trial. So you basically had to roll the dice twice, um, and their independent events, because the outcome of one does not affect the outcome of the other. So we're looking for the probability that we roll a Q. How failure and then the probability that we're old. Um, the success. Eso we got 4 to 6 for probably a few, and then we got two out of six for the probability of B. So that gives us eight out of 36 which is reduced to 2/9, which is 0.2, repeating as an answer.

And when a wait. We have a fair, six sided die, and we're interested in how many times we need to roll the die until we obtained the first four or five as an outcome. So a right. The description of the random variable x X is going to be the number of times the die is rolled. Yeah, until a four or five occurs. And B, what are the values that X could take on? How many times can you roll it until you get a four or five X? Could be one? It could be, too, Um, but essentially X could go all the way up to infinity. See, find the values of P and Q P is the probability of success, so P is going to be four or five. So there's two different times out of six possibilities, which is one third que is everything but a four or five. So there's four possibilities out of six, which is two thirds. And indeed we want to find the probability of the first occurrence. Yeah, is on the second trial, so that means the first trialing needs to be a failure, which is two thirds and the second trial needs to be a success, which is one third that gives us two nights, which is 20.22 repeating.

So in this question we are basically given the sample space of rolling a two sided um to dies basically. And for us to first identify certain events. So the event end is basically the event that the sum is at least nine. So we're going to look At all of the events in the entire sample space of Rolling two Dice. So for the sum to be at least nine, nothing in our first role, in our second role nothing as well. Third role, We have one event 3, six, 4th row, we have 4,5, 5th rule, we have 54 and six through. We have 63 So that constitutes all of the events where the sum is at least nine. Now T at least one of the dies is a truth at least one is it too? So we have 1 to at least one of the devices are too and then we have 21 two, two or three 4 to 5 26 We have 32 again and then we have 42 52 Nothing else over there. And then six two. And then we have the third event at For at least one of the dies is up five at least one is five. So we have 15 to fly 35 four or 5. And then we have my warning 525354. Fine fine. 56 and six. Bye. So we have our three events, N. T. And F. Now first we are asked to find the probability of and so the probability of end is basically 1234 over. There's 36 events Which is 1/9 and that's equal to point 11 The probability that the sum is at least part B. The probability of N. Even X. Is basically the probability according to conditional events. The quality of the intersect with death divided by the probability of. So we are given the event F. If it's basically this event, and if we have 123456789 10, 11. So we're just going to Use events. So we're going to take our denominators 11. And the probability of any given f. So we're going to see where the intersect. So the intersect at 45 and five, so that's true out of 11 and that's equal to point one heat. Now, the third we asked the probability of and given T. So we're given the event T. And we have 1234567 8, 9, 10, 11 Events in T. And the probability of any given T. So that let's see how many intersect here. So we have Let's see if 36 intersex. So it doesn't for 55 for 60, none of them, because at least one of the ties has to be true and that's never going to happen with this event. So We basically have zero. Our third Cancer here is zero. And were asked to determine from the previous answers whether or not the events N and F are independent and whether or not, and and t are independent. So we're going to compute here. So to see an N. F are independent. We're basically going to take the probability of 10 times the probability of f. So a probability of N. We have 1/9 probability of event F. So we have 123456789 10 11 11/36. So Times 11/36. So 1/9, times 11/36 is approximately point for three or now the probability of and intersect with that and intersect with F. We basically had two events out of 36. So that's two out of 36 Which is find 0.56. And so we say that they are not independent because the probabilities are not this is not equal to this. So then an F. Or not independent. Now R. N. And T. Independent. Let's take a look. So we have event. And here we have event. See here now the probability of event and intersecting with event T. was basically zero. So the probability of an intersect with teen was zero. Now the probability of end we know is 1/9. The probability of T. We know is 234567 You know it's 11 over 36. And we know that these are not equal. So again they are not independent. So and until they are not independent but they appear because their intersection is zero, they appear to be mutually exclusive. So we have our answer to part mm B. C. Andy

The scenario described in this question is basically one of rolling a die until we get a one and we want to know how many roles it's going to take. The number of rules that will take to get a one can be modelled as a geometric, random variable. And this is because each roll of the die can be broken down into either success, which is we get a one or failure. We get anything from 536 and each trial is independent of the other trials. The probability of getting a one on each trial does not change based on what you get on other trials, and the probability success for X trial is one out of six. What we say X is the number of rules. Until we get a one, we can say it is a geometric random variable with probability of success 1/6. And so for part, A were asked to describe how we would simulate rolling the die until we get a one so you could actually do this quite manually. Just with an actual die. You could keep rolling the die until you get a one and each time record the number of roles that was required until getting a one. So you might roll it once and get a two and then roll it a second time and get a one so you would record, too. Then you start over again. If you get a one on your four thrill the next time, then you recorded for and you just keep repeating that you could also define a random number generator in a computer application, with the probability of success of 1/6. So Step one is roll a die or use a random number generator To would be to repeat one until success. And here success is defined as rolling a one and then stop so three would be count the number of roles. So on your first simulation, you might get a one on the third, a role so then X is equal to three second simulation. You might get a one on the fifth role, so X is equal to five third one. You might get a one on the 10th. Roll X is equal to 10 and so on, and for B, it says to run at least 30 simulations or 30 trials, so that means we're going to do this. I am actually doing 50 from my example. So I'm gonna repeat this process up until 50. And I actually did this using our software, So I made a little program, and I'll just warn you that I'm not a expert programmer, so this may not be particularly well written, but I have verified that it works properly. So in case you want to take a stab at using our, um, this might give you a leg up on getting started and you can see that I've used 50 trials here and so I won't walk through the how the program works. That's beyond the scope of this exercise. But you can see I have 50 trials that have been executed here, and we can ignore the number, the numbers in brackets. So for 50 trials, these are the number of roles that it took each time before getting a one. So on the second trial, it actually took 17 rolls before getting one, which is pretty unlikely. But you can see normally it happens within six or seven rolls and pipe, Artzi says. Based on our simulation estimate, the probabilities that you might win on the first roll on the second roll, the third rule, etcetera. So basically, we're going to estimate the probability distribution for the geometric random variable with probability of success of 1/6 so we can make a table with X is equal to one to, and I'll make a catch all category for greater than or equal to eight. Since these probabilities air getting smaller and smaller as we get to higher numbers of roles, required it until success. And our strategy will be to use the relative frequency definition for a probability. And you may remember that the relative frequency would be equal in this case, equal to the number of simulations for which X equals I divided by the number of simulations. So, for example, if if X equals seven and you can see we had one 23 I think there are only three simulations with seven, so we would get seven divided by 50 which is the total number of simulations, and so we can repeat that step for all ex from one through 50. And if I do that with my simulation results, I get 0.18 and so that completes see our estimate of the probabilities for winning on one roll to roll three rolls, etcetera for her part dealer as to calculate the actual probabilities for the geometric a random variable. So we want to calculate probability of requiring X rolls before success. And that's given by this formula, where P is the probability success and Q is one minus p. So, for example, probably of success on the first roll is equal to P basically, which is equal to 1/6. And so we can tell you these up in the same table that we already created in part C. So just copy that here, and we'll call this rule the probability of X equaling X. And so if we repeat this calculation for all X, we get the following numbers. And so the second row is three. Calculations for the Probability model, which completes Part D and for Party, were asked to compare the distribution of outcomes in the simulation to the probability model. So we've already done a good job of this by placing them both side by side in this table so you can you can compare them on a one by one basis to see how close they are to each other, and you can see that it's actually quite close. We did 50 simulations and the valleys air quite close. This might be the furthest apart that the two ah strategies give. These two are even further apart, but it turns out that 50 simulations gave us a pretty good estimate of the probability distribution.


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