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Question 4: A dice is biased so that the probability mass function of the face value is as follows:face value probability mass function0.200.250.250.100.150.05Evalu...

Question

Question 4: A dice is biased so that the probability mass function of the face value is as follows:face value probability mass function0.200.250.250.100.150.05Evaluate the expected value and standard deviation of the face value of the dice, if we roll it once. Discuss the result 6) If the dice is rolled once what is the probability that the face value is less than 32 What is the probability that the face value is larger than O equal to 22 Ifa dice is rolled twice, what is the probability mass fu

Question 4: A dice is biased so that the probability mass function of the face value is as follows: face value probability mass function 0.20 0.25 0.25 0.10 0.15 0.05 Evaluate the expected value and standard deviation of the face value of the dice, if we roll it once. Discuss the result 6) If the dice is rolled once what is the probability that the face value is less than 32 What is the probability that the face value is larger than O equal to 22 Ifa dice is rolled twice, what is the probability mass function of the average value of the faces?



Answers

Two fair dice are rolled. Determine the probability that the sum of the faces is 12.

If 26 sided number cubes air rolled. What is the probability that the sum is 12? Let us start by trying to figure out how many possibilities there are. If we roll the first number. Cuba the first die. There are six possibilities once we roll the next number Q. There are six more. The fundamental coming principle tells us if we multiply these possibilities together. There are 36 total possibilities for what could happen in this experiment. So our final answer here will be something out of 36. How do we get us some of 12? One way we could do this is to write down the entire 36 member probability model or simple space. Or we could ask ourselves, How do we get 12 by rolling to die? What if we rolled a one first? Could we get in 11 on the next die off course Not so we can't start with one. So if you think about what would this probability matter look like? We'd start with one a two a three a four of five or six. But we just said you can't get a one and then 11 so we don't have to worry about starting with one. Could we get a two and then a 10 off course? Now you can't get a 10 on to die. If you start with a three, you would next need a night also impossible. If you started with a four, you would then need in eight Impossible. We also cannot get a five and then a seven. So we just said right now you can't start with 12345 or six. Well, what if we started with six Are possibilities for the next role or either a one a two a three a four a five or a six. The only possibility out of all 36 possibilities that makes sense is to start with a six and then get a six on the next roll. That is one possibility out of 36. So the probability that we get a summer 12 is one out of 36

Right, This is problem 20 of chapter five, Section two. And this problem. We want to look at the probability distribution of he some of to roll dice. So whenever we roll, two dice will just take the sum of their to face values. And we want to know the mean variance and standard deviation of this distribution. So right here, I've already laid out a table of all the possible dice outcomes. So, for instance, we roll a four and a three will have a sum of seven. You roller six and six will have us some of 12. And I've also broken this table down over here into the sums and the frequencies that sums show up in. So, for instance, if two dice have a sum of six, that's gonna happen about five times. Likewise, if they have some of 12 that's gonna happen once of six and six. So we would conduce is that with this is first noticed that, um, there are a total of 36 possible outcomes here, and we know the frequency of each of these sums here. So in order to get a probability distribution, we can go ahead and modify our frequency table and make it a probability table simply by taking these frequencies and inviting it by 36 the total number of outcomes. So, for instance, there's a one out of 36 probability that the sum of two dice is going to be, too. Um, we can do this for all of the rest them as well. So this is going to be won over 18. You're gonna have a 1/12. And they went over nine, five or 36 and a 1/6. And this basically just repeats in reverse orders of 5/36 and 1/9 over 12 over 18 1/36. This is our probability distribution. So what, we can do it? This is now we confined our mean variance and standard deviation. Let's go ahead and find our mean first. It's very good. Um, so are mean. You is just gonna be equal to two times 1/36 plus three times 1/18. Last four times, 1/12 from you hadn't skipped to the end here. Plus, uh, 12. There's 1/36 and this is going to give us a means of exactly seven. So the sum of two dies on average, is going to be seven. You grab our variance as well. That's going to be two squared times one over 36 plus three squared times, 1/18 plus four square times, 1/12. Once again, I'll skip to the end here, plus 12 square times one first 36 and don't get to subtract off seven squared. This will be a variance 5.8 33 repeating. And of course, we can grab our standard deviation just by taking the square root of 5 23 3 and we see we get a standard deviation equal to 2.415 And that is the probability distribution and the means variance and standard deviation of the data. Thank you.

Curious to calculate the probability that the some off faces of two fair dice equals three. We know that the only ways you could get three of your old one and two get it to go 1 36 4 year old to No. One and equal one of the 36. All you have to do is add up these probabilities. They get 1/18. Since we know this for some of the faces equals three, the probability of that is the sum of all the ways you can get three.

If 26 sided number cubes or fair die are rolled, what is the probability that the summer's three any time we're doing a probability? We want to know how many ways that can happen out of some total number of possibilities. If we're going to die, there are six possibilities. If we roll another one, there also six possibilities there. The number of the counting principle would show us their 36 total possibilities. So that's gonna be our denominator here. It's gonna be something out of 36. Well, if we want to ask ourselves, how do we get a sum of three? There's only there's very few ways it can happen. You could get a one on your first I and two on your second. Or you could get a two on your first die and the one on your second. There is no other way this can occur. We could write down all the possibilities. We could make a tree, but what you're gonna find here is let's see, we didn't make a tree. For example, if your first die was US three, you can add up to three because your next one could be a one of the least. You can't start with a four or five or six. All of these branches have to be discounted. You can do him if you start with one or you start with the two what I wrote above or the Onley possibilities that could work out. So since there are only two, our probability here would be two out of 36. You could simplify that as well, to call it one out of 18. But the probability some US three is two out of 36 which simplifies the one out of 18.


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