Probability and Statistics: College Assignment on Random Experiments, Events, and Distributions

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Probability and Statistics: College Assignment on Random Experiments,Events, and DistributionsWeek 37.You flip a coin three times.The possible outcomes of this random experiment are,S ={(HHH), (HHT), (HTH), (THH), (HTT), (TTH), (THT),(TTT)}n (S) = 8(a) What is the probability of getting heads on onlyone of your flips?The cases favorable to the eventE1of gettingheads on only one of your flipsare,E1={(HTT), (TTH), (THT)}n (E1) = 3P (E1) = n (E1) / n (S) = 3/8 =0.375(b) What is the probability of getting heads on at least one flip?The cases favorable to the eventE2of gettingheads on only one of your flipsare,E2={(HHH), (HHT), (HTH), (THH), (HTT), (TTH), (THT)}n (E2) = 7P (E2) = n (E2) / n (S) =7/8 = 0.87525.You are to participate in an exam for which you had no chance to study, and for that reasoncannot do anything but guess for each question (all questions being of the multiple choice type,so the chance of guessing the correct answer for each question is 1/d, d being the number ofoptions (distractors) per question; so in case of a 4-choice question, your guess chance is 0.25).Your instructor offers you the opportunity to choose amongst the following exam formats: I. 6questions of the 4-choice type; you pass when 5 or more answers are correct; II. 5 questions ofthe 5-choice type; you pass when 4 or more answers are correct; III. 4 questions of the 10-choicetype; you pass when 3 or more answers are correct. Rank the three exam formats according to

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their attractiveness. It should be clear that the format with the highest probability to pass is themost attractive format. Which would you choose and why?RankExam FormatProbability1II 5660.250.750.25=0.00672I 4540.200.800.20=0.00463III 3440.100.900.10=0.003727. A refrigerator contains 6 apples, 5 oranges, 10 bananas, 3 pears, 7 peaches, 11plums, and 2mangos.a. Imagine you stick your hand in this refrigerator and pull out apiece of fruitat random. What isthe probability that you will pull out a pear?Total number of fruits in the refrigerator,n= 6 + 5 +10 + 3 + 7 + 11 + 2 = 44Number of pears,m= 3The required probability,p=m/n= 3/44 = 0.0682b.Imagine now that you put your hand in the refrigerator and pull out a pieceof fruit. Youdecide you do not want to eat that fruit so you put it back into therefrigerator and pull outanother piece of fruit. What is the probability that thefirst piece of fruit you pull out is a bananaand the second piece you pull out isan apple?Total number of fruits in the refrigerator,n= 6 + 5 +10 + 3 + 7 + 11 + 2 = 44Number ofbananas,m1= 10The required probabilityof selecting banana first,p1=m1/n=10/44 = 0.2273Number ofapples,m2= 6The required probabilityof selecting apple second,p2=m2/n=6/44 = 0.1364

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