The teacher of a mathematics class has written up the final exam, but wants
the questions to be random for each student. 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.
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a. To determine the number of different final exams possible, we can use the concept of permutations. The number of ways to arrange the 100 questions for each student is given by the formula for permutations:
P(n, r) = n! / (n - r)!
where n is the total number of questions (100), and r is the number of questions on the exam.
In this case, since all 100 questions are on the exam for each student, we have:
P(100, 100) = 100! / (100 - 100)! = 100!
The number of different final exams possible would be 100!.
However, it is worth noting that there could be other methods to determine the number of outcomes. For example, if the order of questions within each exam doesn't matter, then we could use combinations instead of permutations. Since the order does matter in this scenario, we stick with permutations.
b. In this scenario, permutation would work better to determine the order of the questions.
Permutation calculates the number of different arrangements of a set of items where the order matters. In this case, the order of the questions in the exam matters because the teacher wants each student to have a different set of questions.
Combination, on the other hand, calculates the number of different subsets of a set where the order doesn't matter. It is typically used when we want to select a certain number of items from a larger set without considering their specific order.
c. If a student doesn't show up to take the test, it affects the number of outcomes. Since each student is assigned a different set of questions, if one student doesn't take the test, one set of questions would be left out. This would reduce the number of different final exams possible by one.
d. If the teacher doesn't end up randomizing the questions on the exam, it still doesn't affect the number of outcomes. The number of different final exams possible would still be the same, regardless of the order of the questions. However, it would impact the fairness of the exam and the variety of questions each student receives.
e. If each question on the test had the same answer (let's say C), the probability that students taking the test will get all the questions right would be 1.
Since each question has four possible answers and all questions are answered with C, there is only one correct answer per question. Therefore, if a student answers all questions with C, they would get all the questions right. With only one possible outcome of getting all the questions right, the probability is 1.