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...

(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) .)