the teacher of a mathematics exam has written up a final exam but wants the questions to be random for each student.

Question

There are 28 students in the class, and
the test has 100 questions. Each question has four possible answers.

a. After reading carefully reading the scenario, determine the number of different
final exams possible. Besides your method, are there other methods to determine
the number of outcomes?

b. For this scenario, which will work better to determine the order of the questions,
permutation or combination? Justify your answer.

c. If a student doesn’t show up to take the test, how does that affect the number of
outcomes?

d. Suppose the teacher doesn’t end up randomizing the questions on the exam. Does
this affect the number of outcomes?

e. If each question on the test had the same answer (let’s say C), what’s the probability
that students taking the test will get all the questions right? Justify your answer.

Answer 1

Note that the introductory statement is not very clear. Let assume that the exam has 100 questions, and the order of the 100 questions in the exam is different for each student (without repetition of any of the questions):

 

a. For the exam of a particular student, there are 100 possibilities for the first question, 99 possibilities for the second, and so on. Applying the basic counting rule (BCR), we get:

 

number of final exams possible= 100 x 99 x 98 x ...... x 1

 

b. Estimating the number of final exams possible as described above is possible, but will take forever. Therefore, a permutation rule can be applied.

 

number of final exams possible = _mP_m=m!/(m-m)!=m!  (If m is the total number of questions on the exam)

Therefore

number of final exams possible = 100! = 9.3x10¹⁵⁷

 

c. If we assume that an outcome is a group of 28 random exams selected from the total number of possible exams, than the total number of group of 28 exams is (applying the permutation rule again):

 

total number of group of exam: _{9.3x10¹⁵⁷} P ₂₈ (my calculator cannot handle this calculation).

 

But if a student does not show up, then only groups of 27 exams are needed. therefore: 

 

total number of group of exam:  _{9.3x10¹⁵⁷} P ₂₇

 

d. if the questions on the exam are not randomized, than there will be only 1 exam with the same order of questions for all the students

 

e. The probability of a student getting a question right is : 1/4

The probability of a student getting all the questions right is : (1/4)^100 = 6.2 x 10^-61

The probability of all 28 students getting all the questions right is: [(1/4)^100]^28 = (6.2 x 10^-61)^28≅0

In conclusion, the probability of all the student getting all the question right is almost zero.