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.

1) 2800

2) combination
3) 2700
4) becomes all the same, 1 outcome of final
5) 1/4

??????

a. To determine the number of different final exams possible, we need to calculate the number of ways to choose questions for each student. Since each question can have four possible answers, we have 4 choices for each question.

The number of ways to choose questions for one student is calculated using the concept of combinations. We have 100 questions, and we need to choose a certain number of them (let's say x) for one student. The number of ways to choose x questions out of 100 is given by the formula C(100, x) = 100! / (x! * (100 - x)!).

Since each student can have a different number of questions on their exam, we need to take into account all possible values of x. We sum up the combinations for each x from 0 to 100. Therefore, the total number of different final exams possible is the sum of C(100, x) for x = 0 to 100.

Besides the method of using combinations, another method to determine the number of outcomes would be to calculate the total number of permutations. However, since the order of questions does not matter in this case (as stated in the scenario), using combinations is the more appropriate method.

b. In this scenario, permutation is not suitable for determining the order of the questions because the order of the questions does not affect the outcome (i.e., the final exam). Each student will receive a different set of questions, but the order of the questions within their exam does not matter. Therefore, using permutation to determine the order of the questions would be unnecessary and more complex than needed. Combinations, on the other hand, take into account the selection of questions without considering the order.

c. If a student doesn't show up to take the test, it does not affect the number of outcomes. Each student's exam is independent of others, so if one student is absent, it only affects their own outcome. The total number of different final exams possible remains the same.

d. If the teacher doesn't randomize the questions on the exam, it does not affect the number of outcomes. The number of different final exams possible is determined by the choices made for each individual student, regardless of the order of the questions. However, not randomizing the questions may result in exams that are identical or have a similar pattern, which may not be desirable from an assessment perspective.

e. If each question on the test has the same answer (let's say C), the probability that students taking the test will get all the questions right can be calculated by considering each question independently. Since there are 100 questions and each question has 4 possible answers (one of which is the correct answer, C), the probability of getting a single question right is 1/4.

Since each question is independent and the probability of getting a question right is the same for all questions, the probability of getting all 100 questions right is calculated by multiplying the probabilities together. Therefore, the probability that students taking the test will get all the questions right is (1/4)^100 or approximately 7.8886 x 10^(-61).