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Cnat[azot erantte Hemawrk Probability bag contains 12 blue chips , 6 yellow chips, and (yelow) red chips - Find cach probability. wlo rtplukemu P(blue) (gicon)P(not...

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Cnat[azot erantte Hemawrk Probability bag contains 12 blue chips , 6 yellow chips, and (yelow) red chips - Find cach probability. wlo rtplukemu P(blue) (gicon)P(not red)P(not Uue)P(orange)P(red and lhen yelow)P(blue and thun red)P(yellow. FndaInen yello")P(yellow blue}P(blue or red)P(blue or yellow or rcd)Find the odds for each problcm: Odds for drawing yellow chip:Odds for drawing blue chip:Odds against drating red chip:Odds aga Ist drawing - bluechipUso the chan below answer tho jollawin

Cnat [azot erantte Hemawrk Probability bag contains 12 blue chips , 6 yellow chips, and (yelow) red chips - Find cach probability. wlo rtplukemu P(blue) (gicon) P(not red) P(not Uue) P(orange) P(red and lhen yelow) P(blue and thun red) P(yellow. FndaInen yello") P(yellow blue} P(blue or red) P(blue or yellow or rcd) Find the odds for each problcm: Odds for drawing yellow chip: Odds for drawing blue chip: Odds against drating red chip: Odds aga Ist drawing - blue chip Uso the chan below answer tho jollawina qucstions Internot Users Monc interneruser BEE Vnat I5 Ine probabilily of selecting Vacor2 Geson Ie aqe 1 2-40 range 61-807 41-E0 61-80 Over EJ 200 Huubar ul usett If one inlutnel user If ore internet user Is celecid #alasIni selecieo wat Is Ine probabilty gele cunad probabllty of selecting dceon the a9e range 41- person who nokin Ihe a90 range 61-80? If ore intemet User is sclected, what is Ihe probability of selecting pemon the age range 21-40? Mon internot user Is seleciedu proeatbatvorat isete selecting Derson under 21 Ovut 807



Answers

Exercise 39$($ Sect. 2.3$)$ describes the game Plinko from The Price is Right. Each contestant drops between one and 5 chips down the Plinko board, depending on how well s/he prices several small items. Suppose the random variable $C=$ number of chips earned by a contestant has the following distribution:
$$\begin{array}{c|cccc}{c} & {1} & {2} & {3} & {4} & {5} \\ \hline p(c) & {.03} & {.15} & {.35} & {.34} & {.13}\end{array}$$
The winnings from each chip follow the distribution presented in Exercise $39 .$ Write a program to simulate Plinko; you will need to consider both the number of chips a contestant earns and how much money is won on each of those chips. Use your simulation estimate the answers to the following questions:
(a) What is the probability a contestant wins more than $\$ 11,000 ?$
(b) What is a contestant's expected winnings?
(c) What is the corresponding standard deviation?
(d) In fact, a player gets one Plinko chip for free and can earn the other four by guessing the prices of small items (waffle irons, alarm clocks, etc.). Assume the player has a $50-50$ - chance of getting each price correct, so we may write $C=1+R,$ where $R \sim \operatorname{Bin}(4,5) .$ Use this revised model for $C$ to estimate the answers to $(a)-(c)$

Okay, so a box gardens turn red chips, 20 blue chips, 30 green. Uh if five drawn from a box find the probability of the indicated situations, the first one is that the probability That four of them are agreeing? So the probability of four green is equal to. Um So we want we are selecting oh from 30 green and we want four of them to be green. So what that means is from the remaining uh blue plus red, which is state. Again we will then get the last one. Uh then divided by the sample space of selecting uh five chips from 16. Use our calculator. Uh This would be Approximately 0.15. And the second chaos is where we can get at least two red, which means Greater or equal to two. Let's So there are many uh it's longer. If we we say we take two or 3 or four reds, it's a little bit longer. So the best is to take a compliment Which is one that. They are less than to rand's. So less than two reds means that I am going to uh to have either zero reds. If it's zero rates, what does it mean? It means the I've paid picked from 3rd green and 20 blue which makes it 50. Uh and I select all the fall. Okay. And uh No, no I select all the five. And if I do that it means uh then I will have to say this over the sample space. So this in this case we've got zero red and alternatively we can pick one red. And the probability of picking one. Read simply means From the 59 red. We'll pick four and then from the red And How Many Leads? 10? We select one and all this is over the sample space. So this would be equal to Settle .1 Night. And lastly the probability that at most two blue at most simply means Less or equal to two. So what it means is that we can have zero blue. Zero blue means we take all the others which are 10 blocks for uh which are 10 plus 30 which has 40. Uh And then we are selecting all the five 90 blue balls. Then this is over the total sample space, which is uh Or we can pick um 1. 9 Blue Ball. And How Many Blues Do We Have? 20? And from the blue. No, we can pull pick for 90 blue and one below over the sample space, which is 6 to c. five. Then the last is when we also peak three non blue. And from the blue we pick two and offer the sample space. And if you put this in the calculator, You get 0.87

For this problem, we are told that a box contains 10 red chips, 20 blue chips and 30 green chips. Uh If five chips are drawn from the box, we are asked to find the probability of drawing the indicated chips for part A were asked the probability of drawing exactly for green. So to draw exactly for green. So I'm going to call that just probability of that event. That will be the probability of drawing well for green Times The Probability of Drawing one, Not Green. Yeah, so the probability of drawing four green would be the probability of drawing for individual grains. So the probability of drawing one would be 30/60 because 10 plus 20 plus 30 60 we have 30 green ships, Then times are second ship 29/59, Then times 28/58, then times 27/57 where the order here doesn't matter. Hence, you know, it doesn't really matter which position we put each multiplication. And then lastly we have the probability of choosing something that isn't great. So that will be then we would have, What would that be. Once we exclude any of the green ships, we have 30 different possibilities and we can write that down as 30/56 is being the last thing in the product there, which comes out to 135 over 4484 Which is approximately three for part B. We're asked to find the probability of um drawing at least to read. So we can write the probability of that event As being equal to 1- the probability of pulling one. Red minus minus the probability of pulling zero red. So there will be one minus the probability or actually it will be one minus now for calculating out the probability of choosing one red, we have 10/60. Then we want to figure out probability of choosing anything other than red for the next four. So that will be 50 times 49 times 48 Times 46. Or sorry, 47, Divided by 59 at Times 58 times 56. That's 66 times 56 times 50. uh oops. Again I missed the number so 57 and then 56 it should be again, order doesn't actually matter. But anyways then we also want that probability of drawing no reds. So that'll be 50 times 49 times 48 Times 47 times 46, Divided by uh 60 Times 59 times 58 times 50 times 57 Times 56 in the Denominator. So calculating that out it comes to, I'm just going to give the percentage chance or the percentage of value there will be 52.7% approximately. Then lastly for part C were asked to find the probability of drawing at most two blue, so that will be the probability, So the probability of that event will be the probability of drawing zero blue. Okay. Plus Probability of Drawing one Blue. Yeah. Plus probability of drawing to blue. Oh, So for zero blue, how many blue chips are there? There are 20. So that would be um uh that would be 40 Times 39 times 38 times. 37 times 36. Divided by 60 times 59 times 58 Times 57, 57 times 56. Then we would have plus the probability of drawing one blue. So that's going to be Let's see here 20 blue chips. So that would be 20/60 times Um 40/59 times 39/58 times 38 over 57 times, 37 over 56. And then we have plus 20/60 Times 19/59 times. I'll write it down below 40/58 Times 39/57 times 38 over 56. And I'm going to pause for one moment here. All right, so the result there is that the probability will be about 22.1 8%

For asked answer questions about the game Plan CO. From the television game show. The price is right. So in this game, intestines have the opportunity to get chips to be dropped down onto a pegboard into slots labeled with cash amounts. We're told that every contestant is given one ship automatically and can earn up to four more chips by correctly guessing the prices of certain small items. Now, let P denote the probability a contestant correctly guesses the price of a prize. Then it follows that the number of chips they contestant earns they're invariable X can be modeled as X equals one plus n, where N is a random variable, which is by normally distributed with parameters for and N. In part, a were asked, determine the expected value of X and the variance of X. The number of chips a contestant earns. We have the expected value of X. Well, this is the same as the expected value of one plus end, which by linearity is the same as one, plus the expected value of N. And because an is by normally distributed, this is the same as one plus en, which is four times p Likewise, we have that the variance of X is the variance of one plus n. And so using property is a variance of a linear combination. This is one squared or one terms the variance of end because and is by noon really distributed. This is N, which is four times p times one minus p. Next, in part B, the ship amounts were given the amount of money one on the clink of board and were asked to determine the mean and variance of the winnings from a single chip. So we'll let w denote the winnings from one ship. We're going to use the probability mass function for W. So we have that the expected value of W This is the outcomes w could be. So we have the w could be anything from zero or 100 or 500 or 1000 or 10,000 and then time teach probability. So we have zero looking at the table times 0.39 plus looking at the table the outcome 100 times It's probability, which is three and so on up to the outcome 10,000 times its outcome 0.23 And after calculating we get. This is equal to $2598. Likewise, we have that the variance W well, this could be computed as the expected value of X squared minus the expected value of X. Sorry, W I mean, did I have w squared? How to find expected value of X w squared. We have the possible outcomes that w squared. So we have zero again times 00.39 plus 100 squared times point of three and so on. Up to 10,000 squared times Probability 10,000, which is 0.23 minus our calculation from before 2500 and 98 And this comes out to be 16 million 518,000 196 Next in part C were given a random variable y denoting the total winnings of a randomly selected contestant were given that the conditional mean and variance of the variable y Given that a player gets X chips arm you x and Sigma squared x respectively, where mu and sigma squared come from part B. We were asked to find expressions for the unconditional, mean and standard deviation of why Well, we have that by the law of total expectation unconditional, mean, expected value of why is equal to the expected value of the expected value of why given X, this is the expected value of well, we're given that the expected value of why given X is Mu X and so using linearity. This is the same as mu times the expected value of X which using calculations from before we have that new is 2000 598 and expected that you have x from part A was one plus four p. Likewise, we have that by the law of total variance, we have that the variance of why is the variance of the expected value of why given X plus expected value of few variants of why given X and from party we have that this is the variance of oh, sorry from this part were given that expected value of like given x again is mute X, and we're given that the variants of why given X is Sigma Squared X and by linearity for expectation by homogeneity for variance. This is mu squared variance of X plus sigma squared times, expected value of X substituting from previous problems you squared is 2598 squared times the variance of X, which is four p times one minus p plus Yeah, the Sigma squared is 2598. Or sorry, that's the wrong one. Sigma squared is 16,518,000 196 and the expected value of X was one plus four p. No simplifying and taking the square root. You get that. These standard deviation of why is thes square root of 165 million. We're sorry wants 16,518,000 196 plus 93 million 71,000 200 p minus 26 million 998,000 416 p squared. Finally, in Part D were asked to evaluate the answers from part C for P, equaling 0.5 and one well, using the formulas from Part C, we have that when P is equal to zero, the expected value of why is equal to 2598 times one plus four times zero, which is 2598 dollars, and we have that's thestreet nerd. Deviation of why, when P is equal to zero, this is going to be these square roots of 165 or sorry, 16,518,196. So this is simply 4000 and $64. So what is the interpretation of when P equals zero? This means that the contestant guesses incorrectly, always. So. If a contestant always guesses wrong, they we'll still get exactly one chip and the answers from part B apply. Now, let's consider when P is equal 2.5. Then we had the expected value of why, according to our formula, is 7794 dollars in the standard deviation of why is 7000 $504 Now? Let's consider when P is equal toe one. In this case, the expected value of why is 12,990 dollars, and the standard deviation of why is 9000 and $88 interpreting these last two results. In fact, all three results we see that as test instability to get chips improves. This is when P is increasing, then so does the contestants expected payout. We see that it increases from 2598 to 7794 up to 12,990. We see that the variability around that expectation also increases in a way we can reason about. This is because as the contestant has greater ability to get chips, they also have more opportunities. So mawr choices to make and therefore more vulnerability. Sorry, variability. But we have the standard deviation does not quite increase linearly with peace. So we see that it increases from 4000 to 7000 and then to 9000. So an increase of about 3000 500 between the first two and then only an increase of about 1000 500 for the second to

We are given that a box can paint them red chips, 20 chips and 50 green ships. And we need to draw fire chips from that box first. The provocative that all chips are blue at least one is green at most. One is ready. So let's draw right for sample space or sample space will be drawing five chips out of 30-plus 2015, less than 16. So we have to draw or select five chairs out of 60. So this is our sample space now for a number of a number of a all blue chips. So how many blue chips? We have 20. And from this 20 we have to select five to get outlook. So 20 C five therefore probability of A will be 20 C. Five. 460 C five. So in solving this will get a resolution, next number of B. The number of these asking for at least one green. So at least one green can occur when we select agree to one Green chip or we can select then Out of 13 is elected one grand chief And remaining four will be selected from other that is 20 plus 10. 13 protect in C. Four. Then next year will be if you sell it two green from 13 In the morning will be selected will be three then three Green from 30. Then we'll have to select 50 C. Two. Then we'll select four green. In fact PC one and then select five green. So will not select anything from restaurant. So this is a number of B. So our probability of B will be number of P. Four. Number of days. So a number of B. So we'll take the model be here do what I do to buy it Number of years which is 60 c. five. Okay now let's move to the final one. That is probability of scenes for that. We need number of C. So you're saying at most one day, so you cannot have more than one day. So how many red chips we have? We have tendered chips out of 10 we can sell at zero and the rest Will be selected from just 50 chips, so five chips or we can have 12 from the red and the rest for will be From another one. So this is our number of years since we cannot have more than one number of 63, the probability of civil B 10 C zero 50 C five. 10 c 154. Uh huh, 60 C. Right. And that's a solution and


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