3. A student is preparing for an upcoming midterm exam. Professor Snape has give

Question

3. A student is preparing for an upcoming midterm exam. Professor Snape has given the class 40 practice questions and tells the class that he plans to select 12 of these questions at random to create the midterm exam. Suppose the student knows how to correctly answer 34 of the 40 questions (a) In how many ways can the professor select 12 questions for the midterm exam? That is, what is n(S)? (b) Let A represent the event that the student knows how to correctly answer exactly 10 of the 12 questions on the exam. What is n(A)? Explain your reasoning. (e) What is PA)? Justify your answer.

Explanation / Answer

(a)

The number of ways of selecting 12 questions out of a total of 40 is given by the combinatorics notation: 40C12 = 40!/(12! * 28!) = 5586853480

So,

n(S) = 5586853480

(b)

In order for the student to able to answer exactly 10 out of the 12 questions, then those 10 question must be from the 34 questions that the student knows, and the remaining 2 questions must be from those 6 question which the student does not know.

So, using this criteria, we have:

n(A) = 34C10*6C2 = (34!/(10! * 24!))*(6!/(2! * 4!)) = 1966922100

(c)

Using the definition of probability, we have:

P(A) = n(A)/n(S) = 1966922100/5586853480 = 0.352

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