Question

Suppose a professor has 3 separate lists of questions totaling 120 questions including:

List 1: includes 35 “Hard” questions. You have 40% likelihood of answering each hard question correctly

List 2: includes 50 “Intermediate” questions. You have 60% likelihood of answering each intermediate question correctly

List 3: includes 35 “Easy” questions. You have 80% likelihood of answering each easy question correctly

Professor lets you choose either one of the following 5 exams (You take only one of these 5):

Exam1: includes 30 randomly selected questions, all from list 1. If you choose this exam you can pass if you score 40% (12 out of 30) or higher in this exam.

Exam2:  includes 30 randomly selected questions, all from list 2. If you choose this exam you can pass if you score 60% (18 out of 30) or higher in this exam.

Exam3: includes 30 randomly selected questions, all from list 3. If you choose this exam you can pass if you score 80% (24 out of 30) or higher in this exam.

Exam4: including 30 questions that consist of 7 questions from the list 1, 13 questions from the list 2, and 10 questions from list 3 (all randomly selected). If you choose this exam you can pass if you score 60% (18 out of 30) or higher in this exam.

Exam5: includes 30 randomly selected questions from the list of all 120 questions of all 3 lists combined. If you choose this exam you can pass if you score 60% (18 out of 30) or higher in this exam.

Find probability of passing each exam i.e. fill out the empty column in the following table based on binomial distribution answers and highlight in the table the exam which you will choose to take. (Do not use binomial to Normal approximation for question 1, just solve it as binomial)

***Needs to be solved as Binomial***

P(passing the exam1)

P(passing the exam2)

P(passing the exam3)

P(passing the exam4)

P(passing the exam5)

Answer 1

Exam 1

Let X be the number questions answered correctly

X follows Binomial with p = 0.4 , n=30

= 0.5689

P(passing exam 1) = 0.5689

Exam 2

Let X be the number questions answered correctly

X follows Binomial with p = 0.6 , n=30

= 0.5785

P(passing exam 2) = 0.5785

Exam 3

Let X be the number questions answered correctly

X follows Binomial with p = 0.8 , n=30

= 0.607

P(passing exam 3) = 0.607

Exam 4

Will pass the exam if we can answer 18 correctly

From list 1 expected correct answers 7*0.4 =2.8,variance =1.68. From list 2 expected correct answers 13*0.6 =7.8 ,variance = 3.12 From list 3 expected correct answers 10*0.8 =8, variance = 1.6

Expected number of correct answer= 2.8+7.8+8 = 18.6

Variance = 6.4

Standard deviation = 2.53

Using normal approximation from binomial

X be the number of correct answers

X follows normal with mean =18.6 and standard deviation =2.53

then

= 0.5949

P(passing exam 4 ) =0.5949

Exam5

Will pass the exam if we can answer 18 correctly

expected number of questions from list 1 = (35/120)*30=8.75 , expected number of questions from list 2 = (50/120)*30=12.5 expected number of questions from list 3= (35/120)*30=8.75

From list 1 expected correct answers 8.75*0.4 =3.5, variance =2.1. From list 2 expected correct answers 12.5*0.6 =7.5 ,variance = 3    From list 3 expected correct answers 8.75*0.8 =7, variance = 1.4

Expected number of correct answer= 18

Variance = 6.5

Standard deviation = 2.55

Using normal approximation to binomial

X be the number of correct answers

X follows normal with mean =18 and standard deviation =2.55

then

= 0.50

P(passing exam 4 ) =0.50

We will choose to take exam 3 (as probability is highest)