A certain full year course has 20 online quizzes spread through the course. The quizzes are optional in the sense that if a student does any of the quizzes, then quizzes count for 10% of the student’s grade, with each quiz counting for 1/2 mark, but for a student who does no quizzes, that 10% is reallocated to other work. Each quiz has 5 multiple choice questions, and the quizzes are marked on a Pass/Fail basis: if a student answers at least 3 of the 5 questions correctly, the student gets the 1/2 mark for that quiz. Otherwise, the student gets 0 for that quiz. Students are allowed 2 attempts at each quiz, with the higher mark counting, so as long as a student gets at least 3 questions right on at least one attempt, the student gets the 1/2 mark for that quiz. Each question on each quiz tests a different concept from the course material. For each of these concepts, the quiz database contains a set of 4 different questions. When a quiz attempt is initiated, the computer randomly selects one of 4 questions for each of the 5 concepts being tested. Each question in the quiz database has 5 answer choices. Tony took this course during the year which ended in April. He decided to have the grading scheme without a quiz component apply for him. However, he wanted to see what the quiz questions were like, so each time a quiz was available, he initiated a quiz attempt. He didn’t answer any of the questions, and did not submit the quiz. Tony didn’t realize that when the quiz window closes, the computer automatically submits any unsubmitted quiz that has been initiated. All 20 of Tony’s quizzes were submitted. As far as the Professor is concerned, Tony did attempt the quizzes and therefore his final grade was calculated with 0 out of 10 as his quiz mark. Tony has been arguing with the Professor about this. Tony maintains that by never answering any quiz question, it was clear that his intention was not to do quizzes and therefore not have a quiz component in his grade. The Professor has decided to compromise with Tony, in the following way. He will reopen all the quiz windows for Tony, to give him his second attempt at each quiz. Tony will receive 1/2 mark for quizzes for each quiz that he gets at least 3 questions right on. Tony needs a quiz mark of at least 4 out of 10 to improve his grade to a pass. The Professor has made it clear that he will not bump Tony’s mark if he’s close – Tony must pass at least 8 of the 20 quizzes in order to receive a passing grade in the course. Tony is about to sit down at his laptop and do the 20 quizzes. The professor has forgotten that Tony can see his original quiz attempts, including being able to see which was the correct answer for each question. For each quiz question, Tony will look at his original attempt at that quiz on his phone, to see if it is a repeat question. If it is, he will select what he knows is the right answer choice. For any question that is not a repeat question, Tony will randomly choose one of the 5 answer choices, because he has forgotten what little he did know about the course material. (a) Consider any one question on any one quiz. What is the probability that Tony will answer the question correctly? (b) Consider any one of the 20 quizzes. What is the probability that Tony will answer at least 3 of the 5 questions correctly and earn the 1/2 mark for that quiz? (Express your answer exactly, which should be a number with 5 decimal places.) (c) What is the probability that Tony will pass at least 8 of the 20 quizzes? Hint: To answer this, you should be thinking about the number of quizzes that Tony passes, not the value of his “quiz mark” – and certainly not the number of quiz questions he answers correctly. Note: Your answer should be expressed to 4 decimal places. The answer should be as precise as possible, but you should definitely use the easiest method available to calculate that number.

(a) For each question, Tony knows the answer 25% of the time and is forced to guess 75% of the time. His probability of guessing correctly is 20% (1 in 5.)
So the probability that Tony gets a particular question correct on any given quiz is .25(1)+.75(.2)=0.4  (We multiply the probability of having seen the problem by the probability of getting it right; 1/4 of the time he has seen the problem and is guaranteed to get it right and 3/4 of the time he will not have seen the problem and must guess.)
(b) For a given quiz, consisting of 5 questions, we can determine the probability that Tony gets at least 3 questions correct. This is the sum of the probabilities that he gets 3 correct, 4 correct, and 5 correct.
Each of these is a binomial probability, where the probability of a success (getting an answer correct) is 0.4 and the total number of trials is 5:
P(3 correct)=_5C_3 (.4)^3(.6)^2=.2304 P(4 correct)=_5C_4 (.4)^4(.6)=.0768 P(5 correct)=_5C_5 (.4)^5=.01024
So the probability of getting at least 3 correct is .2304+.0768+.01024=.31744
(c) To find the probability that Tony passes at least 8 of the quizzes we recognize that this is also a binomial probability. The probability of success (passing the quiz) is .31744 and the number of trials is 20. The answer is the sum of P(8)+P(9)+...+P(20).
Computationally easier is to find the complement and subtract from 1: 1-[P(0)+P(1)+...+P(7)]
So we have:
1-(0.68256^20+20(.31744)(.68256)^19+_20C_2(.31744)^2(.68256)^18+...
...+_20C_7(.31744)^7(.68256)^13)~~.2836161405
So the probability that Tony passes at least 8 of the quizzes is about 0.2836
(Of course the easiest way to compute this is using technology if allowed. On a TI-83/84 graphing calculator we take 1- binomcdf(20,.31744,7) .)
http://mathworld.wolfram.com/BinomialDistribution.html