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4EEnd) OZt42019/9/24 772241njuc m 0924ruiy Pana gungiatadEmail instucotYou have 14 attempts remaining 33"0 Your overall recorded scte timcs allempled this problem nate noxSubmit AnswersPieview My AnswetsNote: You cari eamn panial Credit on this problempant"? both Georgc and Jackic getting (wiopudi the probability (If the roles ere iilled (3) What1883 pan? George Arid Jackie gets exuclly onc (2) How many diiterent WEYS can incsp Toles ne filled26080 (1) How many different ways can these

4 EEnd) OZt 42019/9/24 77224 1njuc m 0924ruiy Pana gungiatad Email instucot You have 14 attempts remaining 33"0 Your overall recorded scte timcs allempled this problem nate nox Submit Answers Pieview My Answets Note: You cari eamn panial Credit on this problem pant"? both Georgc and Jackic getting (wiopudi the probability (If the roles ere iilled (3) What 1883 pan? George Arid Jackie gets exuclly onc (2) How many diiterent WEYS can incsp Toles ne filled 26080 (1) How many different ways can these roles be filled from trese auditioners" ole availablc; and 2 child rales available Jemdle availat:le Aaainb chilren audition The casling director has PuB akppr Guraq (emales audition,ono oi them '30joag) buioq Wayl trales audition one alrisse 'wbiqoud SI41 JO ] Keid involvirg aud tioning review exeicisos problem 35 from Ue Chaptcr Rework (luiod H} Waiqold KON Problem List Previous Problem Homework31: Problem 17 homework31 la19-bl-matn-mi18-8566 webwork 'oerjol anned In MatheMAtIcAL association Of AMerica MAA Tvgso



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Circular Permutations - Problems In Figure $9-4 \mathrm{c},$ the letters $A, B, C,$ and $D$ are arranged in a circle. Though these may seem to be different permutations, they are considered the same permutation because the letters have the same position with respect to each other. That is, each of the four letters has the same letter to its left and the same letter to its right. An easy way to calculate the number of different circular permutations of $n$ elements is to fix the position of one element and then arrange the other $(n-1)$ elements with respect to it (Figure $9-4 \mathrm{d}) .$ So, for the letters $A, B, C,$ and $D,$ the number of circular permutations is $n=1 \cdot \underline{3} \cdot \underline{2} \cdot \underline{1}=\underline{6}$. (FIGURE CAN'T COPY) In how many different ways could King Arthur's 12 knights be seated around the Round Table?

So if we have nine pennies And we want to know how many different ways can we have five pads And four tails come up and we have a total of nine pennies. If they were all unique and different, but you're not going to be able to tell the difference between those. So we'll have the five heads divided by those four tails. The number of ways that you can end up having repeated patterns when we do that. That ends up being 126. Uh that actually is the same thing as a combination of nine shoes five, but we'll get to that later.

No we're doing circular permutations and we have the letters A b c d E that would be formed in a circle and we have five letters. So we're going to take and -1 factorial. So we're going to have four factorial or 24 ways to put those letters in a circle. Next we have Q l m x t M And that is a total of six letters. But to put him in circular permutations, its end -1 factorial. So five factorial, which is 120 different ways. And then we have log, a rhythm log are is um And when we count how many letters we have nine letters there, but again there's N -1 factorial or eight factorial ways to put those into a circular permutation. And that is 40,000 320.

In this exercise, we simulate the coupon collector problem. The scenario here is cereal boxes come with one coupon each, and there are 10 unique coupons that may exist in each cereal box for each cereal box. Each of the coupons is equally likely to be in it, but each box only comes with one coupon. And so we're interested in the number of cereal boxes that we must purchase in order to collect all 10 coupons. Yeah, now I've produced some code in our which looks like this. Now I've produced some code in our which looks like this, and the way the code works is I represent the 10 unique coupons with the vector of the numbers one through 10, and then I start off each replication by sampling 10 from the number of coupon. So that represents buying the 1st 10 boxes because we know we need at least 10 boxes in order to collect all 10 coupons. And then I check to see if there exists any coupon that is not in the boxes that we already have. So that is the statement here. It's The statement is true if any of the coupons does not exist in our collection of boxes. So while this is true, we keep sampling one more cereal box where one more coupon and depend that to our list of cereal boxes that we have or the list of coupons that we've accumulated from the cereal boxes. And we simulate this 10,000 times So once. Once this condition is not satisfied, that's when we break the wild group, and that means that we have all 10 coupons. So we run the simulation 10,000 times, and for each replication, we upend the number of cereal boxes that we had to buy in order to collect all 10 coupons for part A were asked the probability of collecting all 10 coupons in the 1st 10 cereal boxes. So if X is the random variable, that is the number of cereal boxes that we had to collect in order to collect all 10 coupons, we're looking for the probability that X is equal to 10, which will be approximately equal to or hopefully approximately equal to our estimate. Where this probability, based on our simulation, the code weaken type in our is as follows. This will give us the probability that we get the all time coupons in the 1st 10 cereal boxes, and the number I got with the simulation was zero point 00 03 That was running it with 10,000 replications. Next for Part B were as to use counting techniques to find the exact probability for part A. So that is the exact probability that we get all 10 coupons in the 1st 10 cereal boxes. Now the probability of an event A is the number of ways to have that event divided by the total number of outcomes or the number of outcomes in the sample space. So for Event A, which is having all 10 coupons in the 1st 10 cereal boxes, there is 10 ways to select the first coupon. The second coupon has to be different. So there is nine ways to select that one eight ways to select the third coupon and so on. So the number of ways to have all different coupons is 10 factorial. That's the number of ways to have all different coupons in 10 cereal boxes and then for the denominated of the total number of ways to have coupons intense cereal boxes. There's 10 coupons that air possibility for the first one 10 that there is a possibility for the second one 10 that are a possibility for the third one and so on. So that number is 10 to the exponents 10 and this comes out to zero point 000 36 three. And so comparing these numbers are simulation did fairly well, but not great. You could increase the accuracy by increasing the number of replications. In this simulation, we're part C. We're asked for the probability that we require more than 20 boxes to collect all 10 coupons. The code that we can use in our is this. And when I run this command, I get a probability of zero point 7857 And lastly for Part D. We're told that on average it takes 29.3 cereal boxes to collect all 10 coupons and were asked to calculate the probability of collecting 0, 10 coupons in less than 29.3 cereal boxes. Since X is an integer value, this is also equal to the probability that it is at most 29. The command in our is this. And for this simulation, this came out to 0.5943

I This again is one experiment that you have to kill out. Specially interesting experiment, uh, which we drive from Mom under Carlos Metal finding ability. And, um, in this case, you are going to use a simple definition of probability which says the probability off in event is it close to the number of successful outcomes off the studio Sambo space. So that's what German Jews. And you can use your calculator your t i a n four plus, and then you you sent it to the correct amount. And in this guess you want to be waking on. Ah, two days that are being award and we are going to repeat the experiment 50 times. Uh then after 50 times, you have to go to their doubts. And, um, when there's out that you have noted, the first part of the person is that you're supposed to find the probability that this some is nine or more. So you just count how many name or a number which is greater than nine is a sound? What do you find? And in the second case, you also supposed to find the probability that the sum is less than these are in session. So which is you? Some off true Islam or three Summer four is on the show. Life's on six, okay, and seven is not included there. But in the offense tickets where you wanted to some off 99 is included. Or more so the sum is always up to talk toward his eyes, so it will be 9 10 11 and 12 that you'll be checking off. You come all the substance Slocum's. Then you divide by by 50 petitions that you give that and you actually see that there is also always fluctuating for the first person. They will be fluctuating on, uh, between 5 18 and in the second case around faithful 12.


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