# 2-20 - Chapter 4 Probabilities and Simulation Step 5 Find...

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Unformatted text preview: Chapter 4 -- Probabilities and Simulation Step 5. Find the Experimental Probability of t2e Event of Interest. l 75' ) What is the probability of getting at least one jack? Uén' (if) l0\000 5 “m“ “ “m .5 P(q+ 'eab‘l: 1&1le :30, 3% Simulation Software Steps The following steps illustrate how to use LifeStats to solve example 22. 1. Open “Part II Probability Modeling”, “Chapter 4 Probabilities & Simulation”, and then “Five-step Method”. 2. We are interested in counting the number of jacks that we get when we are dealt five cards. In a regular deck of cards there are 4 jacks and there are 48 cards that are not jacks. So for our box model, we have Count = “4" for Value = “1” and we have Count = “48” and Value = “0”. Then click “Next”. 3. Since we are drawing five cards without, replacement, “Draw n Without Replacement” and then let “n = 5”. Click “Next”. 4. We want to count the number of jacks, so our statistic of interest is “Sum”. Since we want at least one jack, our event of interest is “X>=a” where “a = 1”. Click \\ Next”. 5. Run “5000” simulations. What is the probability of getting at least one jack? Example 23: Cereal Boxes Suppose cereal boxes each contain an action ﬁgure and there are six total action ﬁgures. There is a 20% chance of getting Batman, a 10% chance of getting Spiderman, a 15% chance of getting Superman, a 25% chance of getting Wolverine, a 12% chance of getting the Hulk, and an 18% chance of getting Wonder Woman. How many cereal boxes do we expect we will need to buy in order to get all six action ﬁgures? Use LifeStats to answer this question. {Def/13 [o , 000 scmqlq-éi‘MS C517- 29 3 Z ' boxes- WW [2012; to [23 (56;) 2.5133 17.1353 (8 E11 3&(50100 until at“ X {Ethics appear . 501 : sample 51.336— 60 Example 24: Cereal Boxes, revisited Using the same setup as Example 23, what is the probability that it takes less than 25 boxes of cereal to find all six action figures? Use 10,000 simulations in LifeStats to estimate this probability. P( {355 ﬂ‘m" Z 5 loaves 6N: needlecw 3PCX<ZB> Cu“ 0. X“ 8%(07 61:25 Example 25: Normal Distribution Erin decides to draw 1,000 numbers at random from a standard normal distribution (u = 0, o = 1). She is curious as to what the minimum and maximum values will be. Let’s use LifeStats to help her out. By running 10,000 simulations in LifeStats, what will the minimum and maximum values be, on average, if we draw 1,000 numbers at random from a standard normal distribution? mm 21 '31’55 mus: We 7.3;wa WC\~—/“/‘ﬂ (generate velar; Prom a Normal 04575019“ 5'0” ”lb!” 0 iﬁ'ﬁm“ 5 Draco n with rep laccmeﬂf KL = 1000 SDI ‘- mm 5011 max What is the probability that the minimum value is less than —3? What is the probability that the maximum value is greater than 3? Run 10,000 simulations in LifeStats to estimate these probabilities. P(ml‘n'< "”6353 O. 73‘l3 Pémax > 33% 017%{1 What will the interquartile range of those 1,000 numbers be, on average? Run 10,000 simulations in LifeStats to estimate this average. IQ? 2: Laﬁgq W SDI: IQR Example 26: Roulette Nathan enjoys playing roulette. A roulette wheel has 38 pockets numbered 0, 00, 1, 2, 3, ..., 36. 0 and 00 are green. Half of the remaining numbers are red and the other half are black. Suppose Nathan bets \$3 on “Black.” The bet pays even money. This means that if the ball lands in a black pocket, Nathan gets his \$3 bet back, plus winnings of \$3. But if not, he . loses his \$3 bet. Suppose Nathan has the money and the patience to make this same bet 750 times. Using LifeStats, run 50,000 simulations to determine what, on average, will Nathan’s net gain be after those 750 plays of roulette (how much money will he be up or down after those 750 bets)? [’8 [E] 20 E531 Kill—75C) 03/ (epquEw’len‘é neﬁﬁqu/X ﬁsHQL-{é ~17,o Kalli?) y” llﬂ.‘7852flI8.oaoz\ *ll7.l“le-l\ "ll'7.lt—{0L( What is the probability that Nathan actually makes money after these 750 plays? Run 50,000 simulations to estimate this probability. 5 [6,000 Success aunts 6011: 26mm 7 H ‘ . é‘l7+70(o+ 7Zo+7lo+ 5 X > Q PCM‘ 34"“? ,2, Poem) ””5 W Example 27: Project Runway Julie is a big fan of Project Runway on Bravo. Suppose Bravo runs an all—day marathon of Project Runway with probability 1/11. If Julie is craving Project Runway and begins watching Bravo at some random time, how many days will it take, on average, before Bravo airs an all-day marathon of Project Runway (and Julie’s craving is satisfied)? Run 50,000 simulations in LifeStats to estimate this value. [I [EL-7 (O E ] Dram) unfl‘l 59515460} X Value occul‘s) X‘l 4‘ f ( Hm.) lens unfl‘l we Max/c a ﬂier" {Ac/l ) (hand/VA non-Mafel’l‘l’n SDI! Sample 5'IZC l; (app; amid/on meme" /0.Cl(otZ, 10.855, 11.0013, Il.1378)/0.95‘1¢( 771k? 4Vefa3¢t [0.9782‘! day |onaulef¢13€ ) un-{f/ a Pfojfc't Fun/nan)! MaMHM/L What is the probability that it will take between 7 and 10 days (inclusive) for Bravo airs an all-day marathon of Project Runway (and Julie’s craving is satisfied)? Run 50,000 simulations in LifeStats to estimate this probability. 501'. 6am|0l€ SlZC 5 [0,000 we¢e\$5 CouﬂéSf a<=X4=b lsae,i7sz,l797,rsqa,lséﬁ y a=7 , b=lO sum P(be‘5wcen 7 ml l0 days) as. 50,000 .— A ...
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